AP Chemistry course outline
Current course outline · Fall 2024
Find a topic or learning goal, choose a lesson, then open a student prompt and its worked answer. The goal statements and codes below retain the official wording for this edition.
“Lesson provided” and “Practice provided” describe the available resources. Use students’ explanations and work to judge understanding. A resource marked as supporting part of a goal keeps that narrower scope.
Browse AP Chemistry by teaching topic · Teaching guidance
Unit 1: Atomic Structure and Properties
1.1 · Moles and Molar Mass
1.1.A · Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.
Lesson provided · Practice provided
Course and exam description p. 30
- Teach: Moles and Molar Mass
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Teaching passage: For a molecule, add the atomic masses multiplied by their subscripts. CO2 has one C and two O atoms, giving 44.01 g mol−1 using C = 12.01 and O =… -
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Worked example: A 5.55 g sample of anhydrous CaCl2 has M = 110.98 g mol−1. Find formula units and chloride ions.
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- Check understanding: A sample contains 0.125 mol Al2O3. Use M = 101.96 g mol−1, NA = 6.022 × 1023 mol−1 and 1 u ≈ 1.66054 × 10−24 g.Determine the sample mass and… · Worked answer for A sample contains 0.125 mol Al2O3. Use M = 101.96 g mol−1, NA = 6.022 × 1023 mol−1 and 1 u ≈ 1.66054 × 10−24 g.Determine the sample mass and…
- Check understanding: Equal-mass samples of Ne (20.18 g mol−1) and Ar (39.95 g mol−1) are compared. Without choosing a sample mass, find the ratio NNe/NAr and justify it. · Worked answer for Equal-mass samples of Ne (20.18 g mol−1) and Ar (39.95 g mol−1) are compared. Without choosing a sample mass, find the ratio NNe/NAr and justify it.
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Check understanding: What must you specify after saying “one mole”? · Worked answer for What must you specify after saying “one mole”? -
Supports part of this goal.
Check understanding: Which has more molecules: 10.0 g CO2 or 10.0 g H2O? · Worked answer for Which has more molecules: 10.0 g CO2 or 10.0 g H2O?
Required knowledge for 1.1.A
1.1.A.1 · One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes.
1.1.A.2 · Avogadro’s number (N_A = 6.022 × 10^23 mol−1) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance.
1.1.A.3 · Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains. EQN: n = m/M
1.2 · Mass Spectra of Elements
1.2.A · Explain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes.
Lesson provided · Practice provided
Course and exam description p. 31
- Teach: Mass Spectra of Elements
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- Teaching passage: Use average mass = Σ fimi, where each fraction fi is the fraction of atoms, not a mass fraction. A 75% isotope counts three times as much as a 25%…
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Supports part of this goal.
Worked example: An original simplified spectrum has peaks at 62.0 u and 64.0 u with relative intensities 100 and 40. What average atomic mass does it imply?
- Check understanding: An element has only two isotopes in the sample: 10.0 u and 11.0 u. Its measured average mass is 10.80 u. Determine both relative abundances and… · Worked answer for An element has only two isotopes in the sample: 10.0 u and 11.0 u. Its measured average mass is 10.80 u. Determine both relative abundances and…
- Check understanding: A singly charged elemental spectrum has peaks at 24, 25 and 26 u with counts 79, 10 and 11, respectively. A student claims that 25 u is the… · Worked answer for A singly charged elemental spectrum has peaks at 24, 25 and 26 u with counts 79, 10 and 11, respectively. A student claims that 25 u is the…
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Check understanding: A spectrum’s tallest peak is set to 100 and a second peak has intensity 50. Is the second isotope 50% abundant? · Worked answer for A spectrum’s tallest peak is set to 100 and a second peak has intensity 50. Is the second isotope 50% abundant? -
Supports part of this goal.
Check understanding: Why can an element have an average atomic mass of 35.45 u while having no isotope at 35.45 u? · Worked answer for Why can an element have an average atomic mass of 35.45 u while having no isotope at 35.45 u?
Required knowledge for 1.2.A
1.2.A.1 · The mass spectrum of a sample containing a single element can be used to determine the identity of the isotopes of that element and the relative abundance of each isotope in nature.
1.2.A.2 · The average atomic mass of an element can be estimated from the weighted average of the isotopic masses using the mass of each isotope and its relative abundance.
1.3 · Elemental Composition of Pure Substances
1.3.A · Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.
Lesson provided · Practice provided
Course and exam description p. 32
- Teach: Elemental Composition of Pure Substances
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Teaching passage: Imagine cutting a pure sample into many smaller samples. Each piece still has the same ratio of elements by mass, because it has the same… -
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Worked example: A pure binary iron oxide sample contains 2.80 g Fe and 1.20 g O. Use Fe = 55.85 and O = 16.00 g mol−1.
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- Check understanding: A pure compound contains 40.0% C, 6.71% H and 53.3% O by mass. Use 12.01, 1.008 and 16.00 g mol−1. Find its empirical formula and explain why… · Worked answer for A pure compound contains 40.0% C, 6.71% H and 53.3% O by mass. Use 12.01, 1.008 and 16.00 g mol−1. Find its empirical formula and explain why…
- Check understanding: Two samples contain 0.500 g Mg with 0.329 g O, and 1.00 g Mg with 0.658 g O. A student says this proves both are pure MgO. State what the data… · Worked answer for Two samples contain 0.500 g Mg with 0.329 g O, and 1.00 g Mg with 0.658 g O. A student says this proves both are pure MgO. State what the data…
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Check understanding: Why must you convert element masses to moles before finding an empirical formula? · Worked answer for Why must you convert element masses to moles before finding an empirical formula? -
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Check understanding: What does a 1:1.5 mole ratio become, and why not write a subscript 1.5? · Worked answer for What does a 1:1.5 mole ratio become, and why not write a subscript 1.5?
Required knowledge for 1.3.A
1.3.A.1 · Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit.
1.3.A.2 · According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same.
1.3.A.3 · The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula.
1.4 · Composition of Mixtures
1.4.A · Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.
Lesson provided · Practice provided
Course and exam description p. 33
- Teach: Composition of Mixtures
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Teaching passage: A pure substance has one chemical identity: one kind of atom, molecule or formula unit. A mixture contains two or more substances in proportions… -
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Worked example: A 2.00 g sample is known to contain only NaCl and chlorine-free sand. Elemental analysis finds 0.910 g Cl. NaCl has M = 58.44 g mol−1; chlorine…
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- Check understanding: A 10.00 g mixture contains only NaCl and KCl. Their chlorine mass fractions are 0.6066 and 0.4755. Analysis finds 5.542 g Cl. Determine the mass… · Worked answer for A 10.00 g mixture contains only NaCl and KCl. Their chlorine mass fractions are 0.6066 and 0.4755. Analysis finds 5.542 g Cl. Determine the mass…
- Check understanding: A 5.00 g sample labeled “pure compound X” contains less oxygen by mass than the formula of X predicts. A student reports an exact purity from this… · Worked answer for A 5.00 g sample labeled “pure compound X” contains less oxygen by mass than the formula of X predicts. A student reports an exact purity from this…
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Check understanding: Is pure CO2 a mixture because it contains carbon and oxygen? · Worked answer for Is pure CO2 a mixture because it contains carbon and oxygen? -
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Check understanding: A candidate substance is 40% X by mass and its only impurity contains no X. A 10.0 g sample has 3.00 g X. How much candidate is present? · Worked answer for A candidate substance is 40% X by mass and its only impurity contains no X. A 10.0 g sample has 3.00 g X. How much candidate is present?
Required knowledge for 1.4.A
1.4.A.1 · Pure substances contain atoms, molecules, or formula units of a single type. Mixtures contain atoms, molecules, or formula units of two or more types, whose relative proportions can vary.
1.4.A.2 · Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity.
1.5 · Atomic Structure and Electron Configuration
1.5.A · Represent the ground-state electron configuration of an atom of an element or its ions using the Aufbau principle.
Lesson provided · Practice provided
Course and exam description p. 34, 35
- Teach: Atomic Structure and Electron Configuration
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Teaching passage: Reviewed AP edition. This is newly written instruction and feedback aligned to the supplied Fall 2024 CED.Charge, nucleus and electron countAn… -
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Worked example: Write the configurations of S and S2−, describe the 3p orbital occupancy, and explain why core removal is harder than valence removal.
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- Check understanding: Use phosphorus-31, with atomic number Z = 15 and mass number 31.Give the full ground-state configurations for P and P3−. Create labeled… · Worked answer for Use phosphorus-31, with atomic number Z = 15 and mass number 31.Give the full ground-state configurations for P and P3−. Create labeled…
- Check understanding: A student writes Fe2+ as [Ar]4s23d4 from neutral Fe = [Ar]4s23d6. Evaluate the model and write the correct configuration. Explain why simply… · Worked answer for A student writes Fe2+ as [Ar]4s23d4 from neutral Fe = [Ar]4s23d6. Evaluate the model and write the correct configuration. Explain why simply…
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Check understanding: Write Fe3+ given neutral Fe = [Ar]4s23d6. · Worked answer for Write Fe3+ given neutral Fe = [Ar]4s23d6. -
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Check understanding: What is the difference between Hund’s rule and Pauli’s principle? · Worked answer for What is the difference between Hund’s rule and Pauli’s principle? -
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Check understanding: What happens to pairwise Coulombic force when the separation triples with charges unchanged? · Worked answer for What happens to pairwise Coulombic force when the separation triples with charges unchanged?
Required knowledge for 1.5.A
1.5.A.1 · The atom is composed of negatively charged electrons and a positively charged nucleus that is made of protons and neutrons.
1.5.A.2 · Coulomb’s law is used to calculate the force between two charged particles. EQN: F_coulombic ∝ q_1q_2/r^2
1.5.A.3 · In atoms and ions, the electrons can be thought of as being in “shells (energy levels)” and “subshells (sublevels),” as described by the ground-state electron configuration. Inner electrons are called core electrons, and outer electrons are called valence electrons. The electron configuration is explained by quantum mechanics, as delineated in the Aufbau principle and exemplified in the periodic table of the elements.
1.5.A.4 · The relative energy required to remove an electron from different subshells of an atom or ion or from the same subshell in different atoms or ions (ionization energy) can be estimated through a qualitative application of Coulomb’s law. This energy is related to the distance from the nucleus and the effective (shield) charge of the nucleus.
1.6 · Photoelectron Spectroscopy
1.6.A · Explain the relationship between the photoelectron spectrum of an atom or ion and: i. The ground-state electron configuration of the species. ii. The interactions between the electrons and the nucleus.
Lesson provided · Practice provided
Course and exam description p. 36
- Teach: Photoelectron Spectroscopy
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- Check understanding: An idealized neutral-element PES has three peaks at decreasing binding energies with populations 2, 2 and 4. A proposed model gives 1s22s42p2.… · Worked answer for An idealized neutral-element PES has three peaks at decreasing binding energies with populations 2, 2 and 4. A proposed model gives 1s22s42p2.…
- Check understanding: Two idealized PES plots have the same populations 2:2:6 and belong to Ne and Mg2+. Plot B places every corresponding peak at greater binding… · Worked answer for Two idealized PES plots have the same populations 2:2:6 and belong to Ne and Mg2+. Plot B places every corresponding peak at greater binding…
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Check understanding: A PES peak is three times taller than another in the stated ideal model. What does that mean? · Worked answer for A PES peak is three times taller than another in the stated ideal model. What does that mean? -
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Check understanding: Ne and Na+ each have 10 electrons. Which generally has higher corresponding binding energies? · Worked answer for Ne and Na+ each have 10 electrons. Which generally has higher corresponding binding energies?
Required knowledge for 1.6.A
1.6.A.1 · The energies of the electrons in a given shell can be measured experimentally with photoelectron spectroscopy (PES). The position of each peak in the PES spectrum is related to the energy required to remove an electron from the corresponding subshell, and the relative height of each peak is (ideally) proportional to the number of electrons in that subshell.
1.7 · Periodic Trends
1.7.A · Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity.
Lesson provided · Practice provided
Course and exam description p. 37
- Teach: Periodic Trends
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- Teaching passage: First ionization energy removes an electron from a gaseous atom and is positive. Successive values refer to increasingly positive gaseous ions. A…
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Worked example: Rank O2−, F−, Na+ and Mg2+ by increasing radius, and interpret a large ionization-energy jump after the second electron of an unknown main-group…
- Check understanding: Compare the first ionization energies of Mg and Ca using their valence configurations [Ne]3s2 and [Ar]4s2. Then compare radii of Na+ and Mg2+.… · Worked answer for Compare the first ionization energies of Mg and Ca using their valence configurations [Ne]3s2 and [Ar]4s2. Then compare radii of Na+ and Mg2+.…
- Check understanding: An unknown third-period main-group element has successive ionization energies 580, 1820, 2750 and 11,600 kJ mol−1. Identify its likely valence… · Worked answer for An unknown third-period main-group element has successive ionization energies 580, 1820, 2750 and 11,600 kJ mol−1. Identify its likely valence…
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- Check understanding: Define the electron-addition enthalpy as the change for X(g) + e− → X−(g). Negative values mean energy is released.Two anonymous atoms have values… · Worked answer for Define the electron-addition enthalpy as the change for X(g) + e− → X−(g). Negative values mean energy is released.Two anonymous atoms have values…
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Check understanding: Why does radius generally decrease across a period even though electron count increases? · Worked answer for Why does radius generally decrease across a period even though electron count increases? -
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Check understanding: IE values are 600, 1800, 14,800 and 21,000 kJ mol−1. What main-group valence count is suggested? · Worked answer for IE values are 600, 1800, 14,800 and 21,000 kJ mol−1. What main-group valence count is suggested? -
Supports part of this goal.
Check understanding: Why should “electron affinity increases” be qualified? · Worked answer for Why should “electron affinity increases” be qualified?
Required knowledge for 1.7.A
1.7.A.1 · The organization of the periodic table is based on patterns of recurring properties of the elements, which are explained by patterns of ground-state electron configurations and the presence of completely or partially filled shells (and subshells) of electrons in atoms.
1.7.A.2 · Trends in atomic properties within the periodic table (periodicity) can be predicted by the position of the element on the periodic table and qualitatively understood using Coulomb’s law, the shell model, and the concepts of shielding and effective nuclear charge. These properties include: i. Ionization energy ii. Atomic and ionic radii iii. Electron affinity iv. Electronegativity.
1.7.A.3 · The periodicity (in 1.7.A.2) is useful to predict/estimate values of properties in the absence of data.
1.8 · Valence Electrons and Ionic Compounds
1.8.A · Explain the relationship between trends in the reactivity of elements and periodicity.
Lesson provided · Practice provided
Course and exam description p. 38
- Teach: Valence Electrons and Ionic Compounds
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Worked example: Predict the formula formed by Ca and N, then the analogous formula formed by Mg and N.
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- Check understanding: A student argues that K should be less likely than Na to lose an electron because K has more protons. Evaluate the argument using electronic… · Worked answer for A student argues that K should be less likely than Na to lose an electron because K has more protons. Evaluate the argument using electronic…
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Check understanding: Use the familiar main-group monatomic-ion model for binary compounds. K and Na are in group 1, Ca and Mg in group 2, O in group 16 and F in group… · Worked answer for Use the familiar main-group monatomic-ion model for binary compounds. K and Na are in group 1, Ca and Mg in group 2, O in group 16 and F in group…
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Check understanding: Predict a formula for Al3+ and S2−. · Worked answer for Predict a formula for Al3+ and S2−. -
Supports part of this goal.
Check understanding: Why are NaCl and KCl analogous formulas? · Worked answer for Why are NaCl and KCl analogous formulas?
Required knowledge for 1.8.A
1.8.A.1 · The likelihood that two elements will form a chemical bond is determined by the interactions between the valence electrons and nuclei of elements.
1.8.A.2 · Elements in the same column of the periodic table tend to form analogous compounds.
1.8.A.3 · Typical charges of atoms in ionic compounds are governed by the number of valence electrons and predicted by their location on the periodic table.
Unit 2: Compound Structure and Properties
2.1 · Types of Chemical Bonds
2.1.A · Explain the relationship between the type of bonding and the properties of the elements participating in the bond.
Lesson provided · Practice provided
Course and exam description p. 44, 45
- Teach: Types of Chemical Bonds
More teaching passages in this lesson
- Teaching passage: A strong shift of electron density toward one partner gives substantial ionic character. The ionic–covalent distinction is a useful model on a…
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Worked example: Compare H–H, H–Cl and NaCl using supplied electronegativities H = 2.2, Cl = 3.2, Na = 0.9; then use conductivity data to check the NaCl model.
- Check understanding: Given electronegativities C = 2.6, H = 2.2, O = 3.4, describe C–H and O–H bonding in methanol. A second sample is a brittle metal–nonmetal solid… · Worked answer for Given electronegativities C = 2.6, H = 2.2, O = 3.4, describe C–H and O–H bonding in methanol. A second sample is a brittle metal–nonmetal solid…
- Check understanding: Samples A and B both conduct when molten. Only A conducts as a solid and can be hammered into a sheet; B is brittle. Propose bonding models and… · Worked answer for Samples A and B both conduct when molten. Only A conducts as a solid and can be hammered into a sheet; B is brittle. Propose bonding models and…
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Check understanding: Mark partial charges for an O–H bond. · Worked answer for Mark partial charges for an O–H bond. -
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Check understanding: What would solid conductivity suggest about a sample believed to be a metal? · Worked answer for What would solid conductivity suggest about a sample believed to be a metal?
Required knowledge for 2.1.A
2.1.A.1 · Electronegativity values for the representative elements increase going from left to right across a period and decrease going down a group. These trends can be understood qualitatively through the electronic structure of the atoms, the shell model, and Coulomb’s law.
2.1.A.2 · Valence electrons shared between atoms of similar electronegativity constitute a nonpolar covalent bond. For example, bonds between carbon and hydrogen are effectively nonpolar even though carbon is slightly more electronegative than hydrogen.
2.1.A.3 · Valence electrons shared between atoms of unequal electronegativity constitute a polar covalent bond. i. The atom with a higher electronegativity will develop a partial negative charge relative to the other atom in the bond. ii. In single bonds, greater differences in electronegativity lead to greater bond dipoles. iii. All polar bonds have some ionic character, and the difference between ionic and covalent bonding is not distinct but rather a continuum.
2.1.A.4 · The difference in electronegativity is not the only factor in determining if a bond should be designated as ionic or covalent. Generally, bonds between a metal and nonmetal are ionic, and bonds between two nonmetals are covalent. Examination of the properties of a compound is the best way to characterize the type of bonding.
2.1.A.5 · In a metallic solid, the valence electrons from the metal atoms are considered to be delocalized and not associated with any individual atom.
2.2 · Intramolecular Force and Potential Energy
2.2.A · Represent the relationship between potential energy and distance between atoms, based on factors that influence the interaction strength.
Lesson provided · Practice provided
Course and exam description p. 46
- Teach: Intramolecular Force and Potential Energy
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Teaching passage: For the same bonded elements, greater bond order generally makes a shorter, stronger bond: C≡C is shorter and requires more dissociation energy… -
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Worked example: Curve A has a minimum at 150 pm and −300 kJ mol−1; curve B at 120 pm and −500 kJ mol−1, relative to separated atoms. Compare bond lengths and…
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- Check understanding: Sketch two potential-energy curves with labeled axes and units and a separated-atom reference at zero. Mark minima X: (154 pm, −350 kJ mol−1) and… · Worked answer for Sketch two potential-energy curves with labeled axes and units and a separated-atom reference at zero. Mark minima X: (154 pm, −350 kJ mol−1) and…
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Check understanding: What physical quantities come from a potential curve’s minimum location and its depth? · Worked answer for What physical quantities come from a potential curve’s minimum location and its depth?
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Check understanding: Does bond breaking release energy? · Worked answer for Does bond breaking release energy?
Required knowledge for 2.2.A
2.2.A.1 · A graph of potential energy versus the distance between atoms (internuclear distance) is a useful representation for describing the interactions between atoms. Such graphs illustrate both the equilibrium bond length (the separation between atoms at which the potential energy is lowest) and the bond energy (the energy required to separate the atoms).
2.2.A.2 · In a covalent bond, the bond length is influenced by both the size of the atom’s core and the bond order (i.e., single, double, triple). Bonds with a higher order are shorter and have larger bond energies.
2.2.A.3 · Coulomb’s law can be used to understand the strength of interactions between cations and anions. i. Because the interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions. ii. Because the interaction strength increases as the distance between the centers of the ions (nuclei) decreases, smaller ions lead to stronger interactions.
2.3 · Structure of Ionic Solids
2.3.A · Represent an ionic solid with a particulate model that is consistent with Coulomb’s law and the properties of the constituent ions.
Lesson provided · Practice provided
Course and exam description p. 47
- Teach: Structure of Ionic Solids
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Worked example: Evaluate a proposed solid model for MgCl2 with four Mg2+ circles and four Cl− circles.
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- Check understanding: Sketch and label a corrected particle model for an ionic solid containing six Al3+ ions. How many O2− ions are required? Explain two features… · Worked answer for Sketch and label a corrected particle model for an ionic solid containing six Al3+ ions. How many O2− ions are required? Explain two features…
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Supports part of this goal.
Check understanding: A lattice drawing shows alternating unlike-charge layers. Explain how the same model accounts for a solid’s brittleness and its conductivity after… · Worked answer for A lattice drawing shows alternating unlike-charge layers. Explain how the same model accounts for a solid’s brittleness and its conductivity after…
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Check understanding: Why does a charged-ion solid usually fail to conduct? · Worked answer for Why does a charged-ion solid usually fail to conduct? -
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Check understanding: What charge ratio must a particle model of Al2O3 show? · Worked answer for What charge ratio must a particle model of Al2O3 show?
Required knowledge for 2.3.A
2.3.A.1 · The cations and anions in an ionic crystal are arranged in a systematic, periodic 3-D array that maximizes the attractive forces among cations and anions while minimizing the repulsive forces.
2.4 · Structure of Metals and Alloys
2.4.A · Represent a metallic solid and/or alloy using a model to show essential characteristics of the structure and interactions present in the substance.
Lesson provided · Practice provided
Course and exam description p. 48
- Teach: Structure of Metals and Alloys
More teaching passages in this lesson
- Teaching passage: An alloy is a mixture with metallic character. In a substitutional alloy, atoms of comparable radius replace some host atoms at their sites.…
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Worked example: Two schematic alloy models are proposed: A replaces some large host spheres with similarly sized guest spheres; B adds much smaller guest spheres…
- Check understanding: Guest element A has about the same atomic radius as a metal host, while guest B has a much smaller radius. Create two labeled particle sketches… · Worked answer for Guest element A has about the same atomic radius as a metal host, while guest B has a much smaller radius. Create two labeled particle sketches…
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Check understanding: A steel sample is harder to deform than a specified purer iron comparison but still conducts. Explain both observations using the interstitial… · Worked answer for A steel sample is harder to deform than a specified purer iron comparison but still conducts. Explain both observations using the interstitial…
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Check understanding: Why can an alloy conduct even though it contains more than one element? · Worked answer for Why can an alloy conduct even though it contains more than one element? -
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Check understanding: Would a similarly sized guest fit naturally into an interstitial gap? · Worked answer for Would a similarly sized guest fit naturally into an interstitial gap?
Required knowledge for 2.4.A
2.4.A.1 · Metallic bonding can be represented as an array of positive metal ions surrounded by delocalized valence electrons (i.e., a “sea of electrons”).
2.4.A.2 · Interstitial alloys form between atoms of significantly different radii, where the smaller atoms fill the interstitial spaces between the larger atoms (e.g., with steel in which carbon occupies the interstices in iron).
2.4.A.3 · Substitutional alloys form between atoms of comparable radius, where one atom substitutes for the other in the lattice. (e.g., in certain brass alloys, other elements, usually zinc, substitute for copper.)
2.5 · Lewis Diagrams
2.5.A · Represent a molecule with a Lewis diagram.
Lesson provided · Practice provided
Course and exam description p. 49
- Teach: Lewis Diagrams
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Worked example: CO2 has 4 + 2(6) = 16 valence electrons. Begin O–C–O: four electrons in bonds.
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- Check understanding: Construct the Lewis diagram for HCN. State the electron total, bond orders and lone-pair locations. Explain why H cannot be the central atom. · Worked answer for Construct the Lewis diagram for HCN. State the electron total, bond orders and lone-pair locations. Explain why H cannot be the central atom.
- Check understanding: A proposed NH4+ Lewis diagram has four N–H single bonds and one lone pair on N. Evaluate it by electron budget and local electron count; provide… · Worked answer for A proposed NH4+ Lewis diagram has four N–H single bonds and one lone pair on N. Evaluate it by electron budget and local electron count; provide…
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Check understanding: How many valence electrons are available in NO3−? · Worked answer for How many valence electrons are available in NO3−? -
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Check understanding: A double bond is drawn between two atoms. How many electrons does it use from the total budget? · Worked answer for A double bond is drawn between two atoms. How many electrons does it use from the total budget?
Required knowledge for 2.5.A
2.5.A.1 · Lewis diagrams can be constructed according to an established set of principles.
2.6 · Resonance and Formal Charge
2.6.A · Represent a molecule with a Lewis diagram that accounts for resonance between equivalent structures or that uses formal charge to select between nonequivalent structures.
Lesson provided · Practice provided
Course and exam description p. 50
- Teach: Resonance and Formal Charge
More teaching passages in this lesson
- Check understanding: For carbonate, CO32−, construct the equivalent ordinary-octet contributors, give formal charges in one contributor and predict relative C–O bond… · Worked answer for For carbonate, CO32−, construct the equivalent ordinary-octet contributors, give formal charges in one contributor and predict relative C–O bond…
- Check understanding: For CO2, compare an octet model O=C=O to an octet model with one C–O single bond and one C≡O triple bond. Give formal charges and identify the… · Worked answer for For CO2, compare an octet model O=C=O to an octet model with one C–O single bond and one C≡O triple bond. Give formal charges and identify the…
More checks for this goal
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Supports part of this goal.
Check understanding: Do resonance structures represent rapidly interconverting molecules? · Worked answer for Do resonance structures represent rapidly interconverting molecules? -
Supports part of this goal.
Check understanding: What must the formal charges of SO42− add to? · Worked answer for What must the formal charges of SO42− add to? -
Supports part of this goal.
Check understanding: Why can an ordinary all-paired octet model fail for NO? · Worked answer for Why can an ordinary all-paired octet model fail for NO?
Required knowledge for 2.6.A
2.6.A.1 · In cases where more than one equivalent Lewis structure can be constructed, resonance must be included as a refinement to the Lewis structure. In many such cases, this refinement is needed to provide qualitatively accurate predictions of molecular structure and properties.
2.6.A.2 · The octet rule and formal charge can be used as criteria for determining which of several possible valid Lewis diagrams provides the best model for predicting molecular structure and properties.
2.6.A.3 · As with any model, there are limitations to the use of the Lewis structure model, particularly in cases with an odd number of valence electrons.
2.7 · VSEPR and Hybridization
2.7.A · Based on the relationship between Lewis diagrams, VSEPR theory, bond orders, and bond polarities: i. Explain structural properties of molecules. ii. Explain electron properties of molecules.
Lesson provided · Practice provided
Course and exam description p. 51, 52
- Teach: VSEPR and Hybridization
More teaching passages in this lesson
- Teaching passage: Mark each bond dipole toward the more electronegative atom, then add the arrows as vectors in three dimensions. A vector has a magnitude and a…
- Teaching passage: For central atoms with two, three or four domains, AP uses sp, sp2 and sp3 nomenclature, with ideal domain angles 180°, 120° and 109.5°. This…
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Supports part of this goal.
Worked example: CO2: central C has two double bonds and no lone pairs → two domains → linear, 180°, sp. Its equal opposite C=O dipoles cancel.
- Check understanding: Using valid Lewis models, compare H2O, BF3 and NH3. Construct and label the NH3 Lewis diagram, including the lone pair. For each species give… · Worked answer for Using valid Lewis models, compare H2O, BF3 and NH3. Construct and label the NH3 Lewis diagram, including the lone pair. For each species give…
- Check understanding: The following central-atom models have (bond domains, lone-pair domains): PCl5 (5,0), SF4 (4,1), ClF3 (3,2), SF6 (6,0), BrF5 (5,1), XeF4 (4,2).… · Worked answer for The following central-atom models have (bond domains, lone-pair domains): PCl5 (5,0), SF4 (4,1), ClF3 (3,2), SF6 (6,0), BrF5 (5,1), XeF4 (4,2).…
More checks for this goal
- Check understanding: A student calls CH3Cl nonpolar because its central carbon is tetrahedral, and labels SO2 linear because each S=O is a double bond. Evaluate both… · Worked answer for A student calls CH3Cl nonpolar because its central carbon is tetrahedral, and labels SO2 linear because each S=O is a double bond. Evaluate both…
- Check understanding: Sketch and label ethyne, H–C≡C–H, showing its linear atom arrangement and bond orders. Give the domain count and hybridization at each carbon, the… · Worked answer for Sketch and label ethyne, H–C≡C–H, showing its linear atom arrangement and bond orders. Give the domain count and hybridization at each carbon, the…
- Check understanding: Why can 2-butene have geometric isomers while ethene cannot, and why does rotating about the C=C require disrupting bonding? Explain with… · Worked answer for Why can 2-butene have geometric isomers while ethene cannot, and why does rotating about the C=C require disrupting bonding? Explain with…
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Supports part of this goal.
Check understanding: Does a triple bond count as three VSEPR domains? · Worked answer for Does a triple bond count as three VSEPR domains? -
Supports part of this goal.
Check understanding: What shape follows six domains with two opposite lone pairs? · Worked answer for What shape follows six domains with two opposite lone pairs? -
Supports part of this goal.
Check understanding: Why is H2O polar but CO2 nonpolar despite polar bonds in both? · Worked answer for Why is H2O polar but CO2 nonpolar despite polar bonds in both? -
Supports part of this goal.
Check understanding: Can every alkene show two geometric isomers? · Worked answer for Can every alkene show two geometric isomers?
Required knowledge for 2.7.A
2.7.A.1 · VSEPR theory uses the Coulombic repulsion between electrons as a basis for predicting the arrangement of electron pairs around a central atom.
2.7.A.2 · Both Lewis diagrams and VSEPR theory must be used for predicting electronic and structural properties of many covalently bonded molecules and polyatomic ions, including the following: i. Molecular geometry (linear, trigonal planar, tetrahedral, trigonal pyramidal, bent, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, square planar) ii. Bond angles iii. Relative bond energies based on bond order iv. Relative bond lengths (multiple bonds, effects of atomic radius) v. Presence of a dipole moment vi. Hybridization of valence orbitals for atoms within a molecule or polyatomic ion
2.7.A.3 · The terms “hybridization” and “hybrid atomic orbital” are used to describe the arrangement of electrons around a central atom. When the central atom is sp hybridized, its ideal bond angles are 180°; for sp2 hybridized atoms the bond angles are 120°; and for sp3 hybridized atoms the bond angles are 109.5°.
2.7.A.4 · Bond formation is associated with overlap between atomic orbitals. In multiple bonds, such overlap leads to the formation of both sigma and pi bonds. The overlap is stronger in sigma than pi bonds, which is reflected in sigma bonds having greater bond energy than pi bonds. The presence of a pi bond also prevents the rotation of the bond and leads to geometric isomers.
Unit 3: Properties of Substances and Mixtures
3.1 · Intermolecular and Interparticle Forces
3.1.A · Explain the relationship between the chemical structures of molecules and the relative strength of their intermolecular forces when: i. The molecules are of the same chemical species. ii. The molecules are of two different chemical species.
Lesson provided · Practice provided
Course and exam description p. 58, 59
- Teach: Intermolecular and Interparticle Forces
More teaching passages in this lesson
- Teaching passage: Reviewed AP edition. This is newly written instruction and feedback aligned to the supplied Fall 2024 CED.Every molecular pair has dispersionEven…
- Teaching passage: In the AP model, a hydrogen covalently attached to N, O or F can be strongly attracted to a lone-pair-bearing N, O or F site in another molecule,…
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Supports part of this goal.
Worked example: Compare pure ethanol, pure dimethyl ether and an ether–water mixture. Both small organic molecules have formula C2H6O, but ethanol has O–H and the…
- Check understanding: A large nonpolar molecule boils above a much smaller polar molecule.Explain why this is consistent with interaction models. Identify the roles of… · Worked answer for A large nonpolar molecule boils above a much smaller polar molecule.Explain why this is consistent with interaction models. Identify the roles of…
- Check understanding: Compare intermolecular contacts in pure CH3OH, pure CH3OCH3, and CH3OCH3 mixed with water. Then describe water orientation around Cl−. Distinguish… · Worked answer for Compare intermolecular contacts in pure CH3OH, pure CH3OCH3, and CH3OCH3 mixed with water. Then describe water orientation around Cl−. Distinguish…
More checks for this goal
- Check understanding: A model protein has a polar site near a nonpolar aromatic region and oppositely charged sites on two distant chain segments. Explain a possible… · Worked answer for A model protein has a polar site near a nonpolar aromatic region and oppositely charged sites on two distant chain segments. Explain a possible…
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Supports part of this goal.
Check understanding: Do nonpolar molecules lack intermolecular forces? · Worked answer for Do nonpolar molecules lack intermolecular forces? -
Supports part of this goal.
Check understanding: Which end of water points toward Mg2+? · Worked answer for Which end of water points toward Mg2+? -
Supports part of this goal.
Check understanding: Can an ether accept a hydrogen bond from water even without an O–H bond of its own? · Worked answer for Can an ether accept a hydrogen bond from water even without an O–H bond of its own? -
Supports part of this goal.
Check understanding: Can noncovalent contacts within one protein influence its shape? · Worked answer for Can noncovalent contacts within one protein influence its shape?
Required knowledge for 3.1.A
3.1.A.1 · London dispersion forces are a result of the Coulombic interactions between temporary, fluctuating dipoles. London dispersion forces are often the strongest net intermolecular force between large molecules. Dispersion forces increase with increasing contact area between molecules and with increasing polarizability of the molecules. The polarizability of a molecule increases with an increasing number of electrons in the molecule and the size of the electron cloud. It is enhanced by the presence of pi bonding. iii. The term “London dispersion forces” should not be used synonymously with the term “van der Waals forces.”
3.1.A.2 · The dipole moment of a polar molecule leads to additional interactions with other chemical species. i. Dipole-induced dipole interactions are present between a polar and nonpolar molecule. These forces are always attractive. The strength of these forces increases with the magnitude of the dipole of the polar molecule and with the polarizability of the nonpolar molecule. ii. Dipole-dipole interactions are present between polar molecules. The interaction strength depends on the magnitudes of the dipoles and their relative orientation. Interactions between polar molecules are typically greater than those between nonpolar molecules of comparable size because these interactions act in addition to London dispersion forces. iii. Ion-dipole forces of attraction are present between ions and polar molecules. These tend to be stronger than dipole-dipole forces.
3.1.A.3 · The relative strength and orientation dependence of dipole-dipole and ion-dipole forces can be understood qualitatively by considering the sign of the partial charges responsible for the molecular dipole moment, and how these partial charges interact with an ion or with an adjacent dipole.
3.1.A.4 · Hydrogen bonding is a strong type of intermolecular interaction that exists when hydrogen atoms covalently bonded to the highly electronegative atoms (N, O, and F) are attracted to the negative end of a dipole formed by the electronegative atom (N, O, and F) in a different molecule, or a different part of the same molecule.
3.1.A.5 · In large biomolecules, noncovalent interactions may occur between different molecules or between different regions of the same large biomolecule.
3.2 · Properties of Solids
3.2.A · Explain the relationship among the macroscopic properties of a substance, the particulate-level structure of the substance, and the interactions between these particles.
Lesson provided · Practice provided
Course and exam description p. 60, 61
- Teach: Properties of Solids
More teaching passages in this lesson
- Teaching passage: Diamond’s three-dimensional covalent framework resists deformation. Graphite has covalent networks within layers, but weaker contacts between…
- Teaching passage: A large covalent chain still contains noncovalent contacts between regions or between neighboring chains. Those contacts can affect flexibility,…
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Supports part of this goal.
Worked example: A solid A is brittle, does not conduct as a solid but conducts when molten; B is soft, consists of carbon and conducts in layers; C is easily…
- Check understanding: Two molecular liquids have similar size and shape but liquid X forms extensive hydrogen bonds while Y lacks hydrogen-bond donors. At the same… · Worked answer for Two molecular liquids have similar size and shape but liquid X forms extensive hydrogen bonds while Y lacks hydrogen-bond donors. At the same…
- Check understanding: Use the many-particle models A–D below. Frames show portions of extended samples, not isolated formula units. Solid lines represent covalent… · Worked answer for Use the many-particle models A–D below. Frames show portions of extended samples, not isolated formula units. Solid lines represent covalent…
More checks for this goal
- Check understanding: A polymer remains chemically connected after heating, but becomes more flexible. Explain a possible particle-level reason and identify evidence… · Worked answer for A polymer remains chemically connected after heating, but becomes more flexible. Explain a possible particle-level reason and identify evidence…
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Supports part of this goal.
Check understanding: Why is a boiling-point comparison most meaningful at equal external pressure? · Worked answer for Why is a boiling-point comparison most meaningful at equal external pressure? -
Supports part of this goal.
Check understanding: Why is graphite softer than diamond? · Worked answer for Why is graphite softer than diamond? -
Supports part of this goal.
Check understanding: Can changing noncovalent contacts change protein function without breaking the backbone? · Worked answer for Can changing noncovalent contacts change protein function without breaking the backbone?
Required knowledge for 3.2.A
3.2.A.1 · Many properties of liquids and solids are determined by the strengths and types of intermolecular forces present. Because intermolecular interactions are overcome completely when a substance vaporizes, the vapor pressure and boiling point are directly related to the strength of those interactions. Melting points also tend to correlate with interaction strength, but because the interactions are only rearranged, in melting, the relations can be more subtle.
3.2.A.2 · Particulate-level representations, showing multiple interacting chemical species, are a useful means to communicate or understand how intermolecular interactions help to establish macroscopic properties.
3.2.A.3 · Due to strong interactions between ions, ionic solids tend to have low vapor pressures, high melting points, and high boiling points. They tend to be brittle due to the repulsion of like charges caused when one layer slides across another layer. They conduct electricity only when the ions are mobile, as when the ionic solid is melted (i.e., in a molten state) or dissolved in water or another solvent.
3.2.A.4 · In covalent network solids, the atoms are covalently bonded together into a three-dimensional network (e.g., diamond) or layers of two-dimensional networks (e.g., graphite). These are only formed from nonmetals and metalloids: elemental (e.g., diamond, graphite) or binary compounds (e.g., silicon dioxide and silicon carbide). Due to the strong covalent interactions, covalent solids have high melting points. Three-dimensional network solids are also rigid and hard, because the covalent bond angles are fixed. However, graphite is soft because adjacent layers can slide past each other relatively easily.
3.2.A.5 · Molecular solids are composed of distinct, individual units of covalently-bonded molecules attracted to each other through relatively weak intermolecular forces. Molecular solids generally have a low melting point because of the relatively weak intermolecular forces present between the molecules. They do not conduct electricity because their valence electrons are tightly held within the covalent bonds and the lone pairs of each constituent molecule. Molecular solids are sometimes composed of very large molecules or polymers.
3.2.A.6 · Metallic solids are good conductors of electricity and heat, due to the presence of free valence electrons. They also tend to be malleable and ductile, due to the ease with which the metal cores can rearrange their structure. In an interstitial alloy, interstitial atoms tend to make the lattice more rigid, decreasing malleability and ductility. Alloys typically retain a sea of mobile electrons and so remain conducting.
3.2.A.7 · In large biomolecules or polymers, noncovalent interactions may occur between different molecules or between different regions of the same large biomolecule. The functionality and properties of such molecules depend strongly on the shape of the molecule, which is largely dictated by noncovalent interactions.
3.3 · Solids, Liquids, and Gases
3.3.A · Represent the differences between solid, liquid, and gas phases using a particulate-level model.
Lesson provided · Practice provided
Course and exam description p. 62
- Teach: Solids, Liquids, and Gases
More teaching passages in this lesson
- Teaching passage: Gas particles travel through large separations with brief collisions. Weak interparticle effects in the ideal model let a gas adopt both the…
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Supports part of this goal.
Worked example: Sketch a fixed sample of a molecular substance as crystalline solid, liquid and gas; justify the drawings with volume and motion.
- Check understanding: A student draws a liquid as widely separated molecules, a gas as larger circles than the liquid, and an amorphous solid as particles moving freely… · Worked answer for A student draws a liquid as widely separated molecules, a gas as larger circles than the liquid, and an amorphous solid as particles moving freely…
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
More checks for this goal
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Supports part of this goal.
Check understanding: Does a solid’s ordered diagram mean its particles are motionless? · Worked answer for Does a solid’s ordered diagram mean its particles are motionless? -
Supports part of this goal.
Check understanding: Why are liquid and solid molar volumes generally much more similar than either is to a gas molar volume? · Worked answer for Why are liquid and solid molar volumes generally much more similar than either is to a gas molar volume?
Required knowledge for 3.3.A
3.3.A.1 · Solids can be crystalline, where the particles are arranged in a regular three-dimensional structure, or they can be amorphous, where the particles do not have a regular, orderly arrangement. In both cases, the motion of the individual particles is limited, and the particles do not undergo overall translation with respect to each other. The structure of the solid is influenced by interparticle interactions and the ability of the particles to pack together.
3.3.A.2 · The constituent particles in liquids are in close contact with each other, and they are continually moving and colliding. The arrangement and movement of particles are influenced by the nature and strength of the forces (e.g., polarity, hydrogen bonding, and temperature) between the particles.
3.3.A.3 · The solid and liquid phases for a particular substance typically have similar molar volume because, in both phases, the constituent particles are in close contact at all times.
3.3.A.4 · In the gas phase, the particles are in constant motion. Their frequencies of collision and the average spacing between them are dependent on temperature, pressure, and volume. Because of this constant motion, and minimal effects of forces between particles, a gas has neither a definite volume nor a definite shape.
3.4 · Ideal Gas Law
3.4.A · Explain the relationship between the macroscopic properties of a sample of gas or mixture of gases using the ideal gas law.
Lesson provided · Practice provided
Course and exam description p. 63
- Teach: Ideal Gas Law
More teaching passages in this lesson
- Teaching passage: For an ideal mixture, every component experiences the same T and V and contributes Pi = niRT/V. Sum the contributions to get total pressure. The…
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Supports part of this goal.
Worked example: A 2.00 L container at 300. K holds 0.0800 mol N2 and 0.0200 mol O2. Use R = 0.08206 L atm mol−1 K−1.
- Check understanding: A fixed ideal-gas sample occupies 1.50 L at 0.950 atm and 27.0 °C. It is moved to 3.00 L at 127.0 °C. Find final pressure and describe a linear… · Worked answer for A fixed ideal-gas sample occupies 1.50 L at 0.950 atm and 27.0 °C. It is moved to 3.00 L at 127.0 °C. Find final pressure and describe a linear…
- Check understanding: At fixed n,T a pressure-volume data set is (V/L,P/atm): (1.00,4.00), (2.00,2.00), (4.00,1.00). Choose a linear graph and determine its slope. A… · Worked answer for At fixed n,T a pressure-volume data set is (V/L,P/atm): (1.00,4.00), (2.00,2.00), (4.00,1.00). Choose a linear graph and determine its slope. A…
More checks for this goal
- Check understanding: A 5.00 L ideal mixture at 298 K contains 0.100 mol He and 0.200 mol Ne. Use R = 0.08206 L atm mol−1 K−1. Find total and He partial pressure. Then… · Worked answer for A 5.00 L ideal mixture at 298 K contains 0.100 mol He and 0.200 mol Ne. Use R = 0.08206 L atm mol−1 K−1. Find total and He partial pressure. Then…
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Supports part of this goal.
Check understanding: Why does doubling temperature mean doubling Kelvin temperature rather than Celsius temperature? · Worked answer for Why does doubling temperature mean doubling Kelvin temperature rather than Celsius temperature? -
Supports part of this goal.
Check understanding: What plot makes fixed-n,T pressure behavior linear? · Worked answer for What plot makes fixed-n,T pressure behavior linear? -
Supports part of this goal.
Check understanding: At fixed T,V, does adding He change the partial pressure of existing Ne in an ideal mixture? · Worked answer for At fixed T,V, does adding He change the partial pressure of existing Ne in an ideal mixture?
Required knowledge for 3.4.A
3.4.A.1 · The macroscopic properties of ideal gases are related through the ideal gas law: EQN: PV = nRT.
3.4.A.2 · In a sample containing a mixture of ideal gases, the pressure exerted by each component (the partial pressure) is independent of the other components. Therefore, the partial pressure of a gas within the mixture is proportional to its mole fraction (X), and the total pressure of the sample is the sum of the partial pressures. EQN: PA = Ptotal × XA, where XA = moles A/total moles; EQN: Ptotal = PA + PB + PC + …
3.4.A.3 · Graphical representations of the relationships between P, V, T, and n are useful to describe gas behavior.
3.5 · Kinetic Molecular Theory
3.5.A · Explain the relationship between the motion of particles and the macroscopic properties of gases with: i. The kinetic molecular theory (KMT). ii. A particulate model. iii. A graphical representation.
Lesson provided · Practice provided
Course and exam description p. 64
- Teach: Kinetic Molecular Theory
More teaching passages in this lesson
- Teaching passage: A Maxwell–Boltzmann distribution shows a spread of speeds or kinetic energies. Its horizontal axis must specify which of those quantities it…
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Supports part of this goal.
Worked example: At the same temperature, compare He (4.00 g mol−1) and Ne (20.18 g mol−1) and then predict the effect of heating He from 300 K to 600 K at fixed…
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Supports part of this goal.
Check understanding: The curves below are calculated ideal Maxwell–Boltzmann speed distributions for N₂ (28.02 g mol−1) at 300 K and 675 K. Each full probability… · Worked answer for The curves below are calculated ideal Maxwell–Boltzmann speed distributions for N₂ (28.02 g mol−1) at 300 K and 675 K. Each full probability… -
Supports part of this goal.
Check understanding: A student claims the speed distributions for H2 (2.00 g mol−1) and O2 (32.00 g mol−1) at 300 K must match because temperature is equal.Evaluate… · Worked answer for A student claims the speed distributions for H2 (2.00 g mol−1) and O2 (32.00 g mol−1) at 300 K must match because temperature is equal.Evaluate…
More checks for this goal
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Supports part of this goal.
Check understanding: At equal temperature, does a heavier ideal gas have greater average translational kinetic energy? · Worked answer for At equal temperature, does a heavier ideal gas have greater average translational kinetic energy? -
Supports part of this goal.
Check understanding: What does tripling a particle’s instantaneous speed do to its kinetic energy at fixed mass? · Worked answer for What does tripling a particle’s instantaneous speed do to its kinetic energy at fixed mass? -
Supports part of this goal.
Check understanding: Does a lower heated-distribution peak show fewer total particles? · Worked answer for Does a lower heated-distribution peak show fewer total particles?
Required knowledge for 3.5.A
3.5.A.1 · The kinetic molecular theory (KMT) relates the macroscopic properties of gases to motions of the particles in the gas. The Maxwell-Boltzmann distribution describes the distribution of the kinetic energies of particles at a given temperature.
3.5.A.2 · All the particles in a sample of matter are in continuous, random motion. The average kinetic energy of a particle is related to its average velocity by the equation: EQN: KE = ½ mv2.
3.5.A.3 · The Kelvin temperature of a sample of matter is proportional to the average kinetic energy of the particles in the sample.
3.5.A.4 · The Maxwell-Boltzmann distribution provides a graphical representation of the energies/velocities of particles at a given temperature.
3.6 · Deviation from Ideal Gas Law
3.6.A · Explain the relationship among non-ideal behaviors of gases, interparticle forces, and/or volumes.
Lesson provided · Practice provided
Course and exam description p. 65
- Teach: Deviation from Ideal Gas Law
More teaching passages in this lesson
- Teaching passage: At very high pressure, the intrinsic volumes of particles occupy a meaningful fraction of the container. The free volume available for center…
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Supports part of this goal.
Worked example: At fixed n,V,T the ideal calculation predicts 10.0 atm. Real sample A measures 9.20 atm near condensation; sample B under much stronger…
- Check understanding: An ideal calculation gives 4.00 atm for a sample at known n,V,T; the measured pressure is 3.60 atm and no liquid is present. Explain the deviation… · Worked answer for An ideal calculation gives 4.00 atm for a sample at known n,V,T; the measured pressure is 3.60 atm and no liquid is present. Explain the deviation…
- Check understanding: For matched n,V,T comparisons, gas X has PV/(nRT) = 0.82 at moderate pressure and 1.18 at very high pressure. A student says the second datum is… · Worked answer for For matched n,V,T comparisons, gas X has PV/(nRT) = 0.82 at moderate pressure and 1.18 at very high pressure. A student says the second datum is…
More checks for this goal
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Supports part of this goal.
Check understanding: Why are low temperature and high density often poor ideal-gas conditions? · Worked answer for Why are low temperature and high density often poor ideal-gas conditions? -
Supports part of this goal.
Check understanding: Must measured real-gas pressure always be below ideal pressure? · Worked answer for Must measured real-gas pressure always be below ideal pressure?
Required knowledge for 3.6.A
3.6.A.1 · The ideal gas law does not explain the actual behavior of real gases. Deviations from the ideal gas law may result from interparticle attractions among gas molecules, particularly at conditions that are close to those resulting in condensation. Deviations may also arise from particle volumes, particularly at extremely high pressures.
3.7 · Solutions and Mixtures
3.7.A · Calculate the number of solute particles, volume, or molarity of solutions.
Lesson provided · Practice provided
Course and exam description p. 66
- Teach: Solutions and Mixtures
More teaching passages in this lesson
- Teaching passage: In simple dilution with no reaction or solute loss, n before = n after, so M1V1 = M2V2. Use matching volume units in that ratio or liters when…
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Supports part of this goal.
Worked example: A solution contains 0.0150 mol Na2SO4 in a final volume of 250. mL. Assume complete dissociation and no other reactions.
- Check understanding: 0.0200 mol CaCl2 is dissolved to make 400. mL of final solution. Assume complete dissociation. Find analytical molarity, calcium and chloride… · Worked answer for 0.0200 mol CaCl2 is dissolved to make 400. mL of final solution. Assume complete dissociation. Find analytical molarity, calcium and chloride…
- Check understanding: A method says “dissolve 0.100 mol solute in 1.00 L water to make exactly 0.100 M solution.” Evaluate the exactness claim and give a better volume… · Worked answer for A method says “dissolve 0.100 mol solute in 1.00 L water to make exactly 0.100 M solution.” Evaluate the exactness claim and give a better volume…
More checks for this goal
- Check understanding: A 15.0 mL aliquot of a 0.200 M molecular-solute stock is diluted to a final 100.0 mL. Calculate final molarity. If the flask is filled above its… · Worked answer for A 15.0 mL aliquot of a 0.200 M molecular-solute stock is diluted to a final 100.0 mL. Calculate final molarity. If the flask is filled above its…
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Supports part of this goal.
Check understanding: What volume belongs in the molarity denominator? · Worked answer for What volume belongs in the molarity denominator? -
Supports part of this goal.
Check understanding: For completely dissociated 0.200 M CaCl2, what is [Cl−]? · Worked answer for For completely dissociated 0.200 M CaCl2, what is [Cl−]? -
Supports part of this goal.
Check understanding: Does doubling final volume by dilution double dissolved-solute particle count? · Worked answer for Does doubling final volume by dilution double dissolved-solute particle count?
Required knowledge for 3.7.A
3.7.A.1 · Solutions, also sometimes called homogeneous mixtures, can be solids, liquids, or gases. In a solution, the macroscopic properties do not vary throughout the sample. In a heterogeneous mixture, the macroscopic properties depend on location in the mixture.
3.7.A.2 · Solution composition can be expressed in a variety of ways; molarity is the most common method used in the laboratory. EQN: M = nsolute /Lsolution
3.8 · Representations of Solutions
3.8.A · Using particulate models for mixtures: i. Represent interactions between components. ii. Represent concentrations of components.
Lesson provided · Practice provided
Course and exam description p. 67
- Teach: Representations of Solutions
More teaching passages in this lesson
- Teaching passage: Solvent molecules are not passive dots. In water, Oδ− sides orient preferentially toward cations and Hδ+ sides toward anions. Thermal motion…
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Supports part of this goal.
Worked example: Equal-volume windows A and B represent completely dissociated CaCl2. A contains three Ca2+ icons and six Cl− icons; B has six Ca2+ and twelve Cl−,…
- Check understanding: Create and label two particle-window sketches of completely dissociated MgCl2 solution. Window 1 represents 1.0 volume unit with four Mg2+ icons… · Worked answer for Create and label two particle-window sketches of completely dissociated MgCl2 solution. Window 1 represents 1.0 volume unit with four Mg2+ icons…
- Check understanding: Create two labeled particle-window sketches for equal-volume solutions with the same analytical concentration of dissolved glucose and fully… · Worked answer for Create two labeled particle-window sketches for equal-volume solutions with the same analytical concentration of dissolved glucose and fully…
More checks for this goal
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Supports part of this goal.
Check understanding: A window has twice the solute icons but three times the volume of another. What is the concentration ratio? · Worked answer for A window has twice the solute icons but three times the volume of another. What is the concentration ratio? -
Supports part of this goal.
Check understanding: What is wrong with a dissolved Na2SO4 diagram showing free S atoms and O2− ions? · Worked answer for What is wrong with a dissolved Na2SO4 diagram showing free S atoms and O2− ions? -
Supports part of this goal.
Check understanding: What does a hydration drawing’s fixed water orientation mean? · Worked answer for What does a hydration drawing’s fixed water orientation mean?
Required knowledge for 3.8.A
3.8.A.1 · Particulate representations of solutions communicate the structure and properties of solutions, by illustration of the relative concentrations of the components in the solution and/or drawings that show interactions among the components.
3.9 · Separation of Solutions and Mixtures
3.9.A · Explain the results of a separation experiment based on intermolecular interactions.
Lesson provided · Practice provided
Course and exam description p. 68
- Teach: Separation of Solutions and Mixtures
More teaching passages in this lesson
- Teaching passage: Measure travel from the origin/baseline, not from the bottom of the sheet. A useful ratio is Rf = solute travel / solvent-front travel under the…
- Teaching passage: Distillation separates volatile components through differences in their vapor pressures and intermolecular interactions. The more volatile…
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Supports part of this goal.
Worked example: On the specified polar stationary phase and relatively nonpolar mobile phase, A travels 1.2 cm and B 4.8 cm while the solvent front travels 6.0…
- Check understanding: A mixture of two dissolved dyes is run on polar silica with a relatively nonpolar mobile phase. Dye A travels 2.0 cm, dye B 5.0 cm, solvent front… · Worked answer for A mixture of two dissolved dyes is run on polar silica with a relatively nonpolar mobile phase. Dye A travels 2.0 cm, dye B 5.0 cm, solvent front…
- Check understanding: A sample contains an insoluble solid suspended in a homogeneous solution of two volatile liquids. Propose a sequence to isolate the solid and… · Worked answer for A sample contains an insoluble solid suspended in a homogeneous solution of two volatile liquids. Propose a sequence to isolate the solid and…
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- Check understanding: A student compares a dye’s Rf from yesterday’s solvent to a reference obtained in another solvent, declaring different values prove different… · Worked answer for A student compares a dye’s Rf from yesterday’s solvent to a reference obtained in another solvent, declaring different values prove different…
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Supports part of this goal.
Check understanding: Can ordinary filtration separate dissolved ethanol from water? · Worked answer for Can ordinary filtration separate dissolved ethanol from water? -
Supports part of this goal.
Check understanding: On a polar stationary phase with relatively nonpolar mobile phase, what does a lower Rf suggest? · Worked answer for On a polar stationary phase with relatively nonpolar mobile phase, what does a lower Rf suggest? -
Supports part of this goal.
Check understanding: Does the first condensed vapor from a two-liquid solution have to be a pure component? · Worked answer for Does the first condensed vapor from a two-liquid solution have to be a pure component?
Required knowledge for 3.9.A
3.9.A.1 · The components of a liquid solution cannot be separated by filtration. They can, however, be separated using processes that take advantage of differences in the intermolecular interactions of the components. i. Chromatography (paper, thin-layer, and column) separates chemical species by taking advantage of the differential strength of intermolecular interactions between and among the components of the solution (the mobile phase) and with the surface components of the stationary phase. The resulting chromatogram can be used to infer the relative polarities of components in a mixture. ii. Distillation separates chemical species by taking advantage of the differential strength of intermolecular interactions between and among the components and the effects these interactions have on the vapor pressures of the components in the mixture.
3.10 · Solubility
3.10.A · Explain the relationship between the solubility of ionic and molecular compounds in aqueous and nonaqueous solvents, and the intermolecular interactions between particles.
Lesson provided · Practice provided
Course and exam description p. 69
- Teach: Solubility
More teaching passages in this lesson
- Teaching passage: A molecular compound may contain both polar sites and a nonpolar hydrocarbon region. As a nonpolar region grows, similar numbers of…
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Supports part of this goal.
Worked example: Predict which solvent, water or a nonpolar hydrocarbon, is a better match for a small hydroxyl-rich molecular solute and for a hydrocarbon solute.…
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Supports part of this goal.
Check understanding: Compare a short-chain alcohol and a much longer-chain alcohol, each with one O–H group, for relative water solubility. Then predict which would… · Worked answer for Compare a short-chain alcohol and a much longer-chain alcohol, each with one O–H group, for relative water solubility. Then predict which would… -
Supports part of this goal.
Check understanding: At 25 °C, compare equal small portions of NaCl(s) added separately to ample liquid water and to liquid n-hexane, C6H14. Water is polar and… · Worked answer for At 25 °C, compare equal small portions of NaCl(s) added separately to ample liquid water and to liquid n-hexane, C6H14. Water is polar and…
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Supports part of this goal.
Check understanding: Why does water not dissolve every salt well? · Worked answer for Why does water not dissolve every salt well? -
Supports part of this goal.
Check understanding: Why might adding a long nonpolar tail reduce water solubility despite retaining one O–H group? · Worked answer for Why might adding a long nonpolar tail reduce water solubility despite retaining one O–H group? -
Supports part of this goal.
Check understanding: Does an observed cooling during dissolution prove the solute cannot dissolve? · Worked answer for Does an observed cooling during dissolution prove the solute cannot dissolve?
Required knowledge for 3.10.A
3.10.A.1 · Substances with similar intermolecular interactions tend to be miscible or soluble in one another.
3.11 · Spectroscopy and the Electromagnetic Spectrum
3.11.A · Explain the relationship between a region of the electromagnetic spectrum and the types of molecular or electronic transitions associated with that region.
Lesson provided · Practice provided
Course and exam description p. 70
- Teach: Spectroscopy and the Electromagnetic Spectrum
More teaching passages in this lesson
- Check understanding: A data set assigns feature A to microwave radiation, B to infrared and C to visible radiation. Match each to a typical transition class and… · Worked answer for A data set assigns feature A to microwave radiation, B to infrared and C to visible radiation. Match each to a typical transition class and…
- Check understanding: A student says an IR band is evidence of a new electronic configuration and a UV/visible band means an isotope became heavier. Evaluate both… · Worked answer for A student says an IR band is evidence of a new electronic configuration and a UV/visible band means an isotope became heavier. Evaluate both…
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Supports part of this goal.
Check understanding: Which region is typically used to probe molecular vibrations? · Worked answer for Which region is typically used to probe molecular vibrations? -
Supports part of this goal.
Check understanding: Does a visible absorption band automatically prove electrons were ejected? · Worked answer for Does a visible absorption band automatically prove electrons were ejected?
Required knowledge for 3.11.A
3.11.A.1 · Differences in absorption or emission of photons in different spectral regions are related to the different types of molecular motion or electronic transition: i. Microwave radiation is associated with transitions in molecular rotational levels. ii. Infrared radiation is associated with transitions in molecular vibrational levels. iii. Ultraviolet/visible radiation is associated with transitions in electronic energy levels.
3.12 · Properties of Photons
3.12.A · Explain the properties of an absorbed or emitted photon in relationship to an electronic transition in an atom or molecule.
Lesson provided · Practice provided
Course and exam description p. 71
- Teach: Properties of Photons
More teaching passages in this lesson
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Supports part of this goal.
Teaching passage: An absorbed photon increases the species’ energy by the photon’s energy. An emitted photon decreases it by the same amount. For an allowed… -
Supports part of this goal.
Worked example: An allowed transition absorbs 500. nm light. Use h = 6.626 × 10−34 J s, c = 3.00 × 108 m s−1, and NA = 6.022 × 1023 mol−1.
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- Check understanding: An atom emits a 4.50 × 10−19 J photon. Using h = 6.626 × 10−34 J s and c = 3.00 × 108 m s−1, calculate frequency and wavelength in nm. State the… · Worked answer for An atom emits a 4.50 × 10−19 J photon. Using h = 6.626 × 10−34 J s and c = 3.00 × 108 m s−1, calculate frequency and wavelength in nm. State the…
- Check understanding: An idealized species has allowed level energies 0, 3.0 × 10−19 and 5.0 × 10−19 J. Starting in the highest level, compare photon wavelengths for… · Worked answer for An idealized species has allowed level energies 0, 3.0 × 10−19 and 5.0 × 10−19 J. Starting in the highest level, compare photon wavelengths for…
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Supports part of this goal.
Check understanding: If wavelength doubles in vacuum, what happens to frequency and photon energy? · Worked answer for If wavelength doubles in vacuum, what happens to frequency and photon energy? -
Supports part of this goal.
Check understanding: Does brighter monochromatic light necessarily have more energetic photons? · Worked answer for Does brighter monochromatic light necessarily have more energetic photons? -
Supports part of this goal.
Check understanding: What sign is the species’ energy change in emission? · Worked answer for What sign is the species’ energy change in emission?
Required knowledge for 3.12.A
3.12.A.1 · When a photon is absorbed (or emitted) by an atom or molecule, the energy of the species is increased (or decreased) by an amount equal to the energy of the photon.
3.12.A.2 · The wavelength of the electromagnetic wave is related to its frequency and the speed of light by the equation: EQN: c = λν. The energy of a photon is related to the frequency of the electromagnetic wave through Planck’s equation: EQN: E = ℎν.
3.13 · Beer-Lambert Law
3.13.A · Explain the amount of light absorbed by a solution of molecules or ions in relationship to the concentration, path length, and molar absorptivity.
Lesson provided · Practice provided
Course and exam description p. 72
- Teach: Beer-Lambert Law
More teaching passages in this lesson
- Teaching passage: Reviewed AP edition. This is newly written instruction and feedback aligned to the supplied Fall 2024 CED.More absorbing particles along the light…
- Teaching passage: If a dirty surface, bubble or suspended particle removes additional light from the beam, the instrument can report too large an apparent…
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Supports part of this goal.
Worked example: An original calibration at fixed wavelength and 1.00 cm path has the following data. An unknown has A = 0.395 ± 0.006; treat slope and intercept…
- Check understanding: A species has ε = 250. L mol−1 cm−1 at the chosen wavelength. For b = 1.00 cm and c = 0.00200 mol L−1, find A and transmittance I/I0. Predict A if… · Worked answer for A species has ε = 250. L mol−1 cm−1 at the chosen wavelength. For b = 1.00 cm and c = 0.00200 mol L−1, find A and transmittance I/I0. Predict A if…
- Check understanding: The graph and table give an original blank-corrected absorption spectrum for one dilute analyte in a 1.00 cm cuvette at fixed concentration.… · Worked answer for The graph and table give an original blank-corrected absorption spectrum for one dilute analyte in a 1.00 cm cuvette at fixed concentration.…
More checks for this goal
- Check understanding: A calibration at fixed wavelength and path is A = (400. L mol−1)c + 0.015. A diluted unknown gives A = 0.255 ± 0.008. It was prepared by diluting… · Worked answer for A calibration at fixed wavelength and path is A = (400. L mol−1)c + 0.015. A diluted unknown gives A = 0.255 ± 0.008. It was prepared by diluting…
- Check understanding: A blank-corrected calibration is linear through 0.0030 M, but a 0.0060 M standard gives less absorbance than an extrapolated line predicts. An… · Worked answer for A blank-corrected calibration is linear through 0.0030 M, but a 0.0060 M standard gives less absorbance than an extrapolated line predicts. An…
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Supports part of this goal.
Check understanding: At fixed wavelength, what happens to A if b doubles and c halves? · Worked answer for At fixed wavelength, what happens to A if b doubles and c halves? -
Supports part of this goal.
Check understanding: Why hold the wavelength fixed after calibration? · Worked answer for Why hold the wavelength fixed after calibration? -
Supports part of this goal.
Check understanding: Should a nonzero measured intercept always be hidden by forcing the line through zero? · Worked answer for Should a nonzero measured intercept always be hidden by forcing the line through zero? -
Supports part of this goal.
Check understanding: If an unknown’s cuvette is dirty and attenuates extra light, what likely happens to inferred concentration? · Worked answer for If an unknown’s cuvette is dirty and attenuates extra light, what likely happens to inferred concentration?
Required knowledge for 3.13.A
3.13.A.1 · The Beer-Lambert law relates the absorption of light by a solution to three variables according to the equation: EQN: A = εbc. The molar absorptivity, ε, describes how intensely a chemical species absorbs light of a specific wavelength. The path length, b, and concentration, c, are proportional to the number of light-absorbing particles in the light path.
3.13.A.2 · In most experiments the path length and wavelength of light are held constant. In such cases, the absorbance is proportional only to the concentration of absorbing molecules or ions. The spectrophotometer is typically set to the wavelength of maximum absorbance (optimum wavelength) for the species being analyzed to ensure the maximum sensitivity of measurement.
Unit 4: Chemical Reactions
4.1 · Introduction for Reactions
4.1.A · Identify evidence of chemical and physical changes in matter.
Lesson provided · Practice provided
Course and exam description p. 78
- Teach: Introduction for Reactions
More teaching passages in this lesson
- Teaching passage: A chemical change produces different chemical species by rearranging atoms and electrons. A new solid, gas, color, heat, or light can be a useful…
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Supports part of this goal.
Worked example: A clear liquid forms bubbles on warming. Has a reaction occurred? Let’s slow down before naming it.
- Check understanding: Two clear aqueous samples are mixed. The temperature falls 3 °C, but no solid or gas appears. A student says a chemical reaction is proved.… · Worked answer for Two clear aqueous samples are mixed. The temperature falls 3 °C, but no solid or gas appears. A student says a chemical reaction is proved.…
- Check understanding: A sealed container holds liquid water and water vapor. Heating increases the vapor amount while the total mass stays fixed. State a particle… · Worked answer for A sealed container holds liquid water and water vapor. Heating increases the vapor amount while the total mass stays fixed. State a particle…
More checks for this goal
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Supports part of this goal.
Check understanding: A salt solution becomes more concentrated when water evaporates. What species changed identity? · Worked answer for A salt solution becomes more concentrated when water evaporates. What species changed identity? -
Supports part of this goal.
Check understanding: Why is a temperature rise only possible evidence of reaction? · Worked answer for Why is a temperature rise only possible evidence of reaction?
Required knowledge for 4.1.A
4.1.A.1 · A physical change occurs when a substance undergoes a change in properties but not a change in composition. Changes in the phase of a substance (solid, liquid, gas) or formation/separation of mixtures of substances are common physical changes.
4.1.A.2 · A chemical change occurs when substances are transformed into new substances, typically with different compositions. Production of heat or light, formation of a gas, formation of a precipitate, and/or color change provide possible evidence that a chemical change has occurred.
4.2 · Net Ionic Equations
4.2.A · Represent changes in matter with a balanced chemical or net ionic equation: i. For physical changes. ii. For given information about the identity of the reactants and/or product. iii. For ions in a given chemical reaction.
Lesson provided · Practice provided
Course and exam description p. 79
- Teach: Net Ionic Equations
More teaching passages in this lesson
- Teaching passage: For a chemical reaction, change coefficients until every element has the same atom count on both sides. Do not edit subscripts: changing H2O to…
- Teaching passage: A molecular equation records reagent formulas. A complete ionic equation separates dissolved strong electrolytes into ions. Keep solids, liquids,…
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Supports part of this goal.
Worked example: Given that AgCl is sparingly soluble, represent mixing AgNO3(aq) and NaCl(aq).
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- Multipart prompt: only the expressly relevant actions receive credit; all parts share the existing question anchor.
- Check understanding: Given that CaCO3 is sparingly soluble, write molecular, complete ionic, and net ionic equations for CaCl2(aq) mixed with Na2CO3(aq). Identify… · Worked answer for Given that CaCO3 is sparingly soluble, write molecular, complete ionic, and net ionic equations for CaCl2(aq) mixed with Na2CO3(aq). Identify…
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Supports part of this goal.
Check understanding: Write vaporization of ethanol, C2H5OH. · Worked answer for Write vaporization of ethanol, C2H5OH. -
Supports part of this goal.
Check understanding: Why is AgCl(s) not split into Ag+(aq) and Cl−(aq)? · Worked answer for Why is AgCl(s) not split into Ag+(aq) and Cl−(aq)?
Required knowledge for 4.2.A
4.2.A.1 · All physical and chemical processes can be represented symbolically by balanced equations.
4.2.A.2 · Chemical equations represent chemical changes. These changes are the result of a rearrangement of atoms into new combinations; thus, any representation of a chemical change must contain equal numbers of atoms of every element before and after the change occurred. Equations thus demonstrate that mass and charge are conserved in chemical reactions.
4.2.A.3 · Balanced molecular, complete ionic, and net ionic equations are differing symbolic forms used to represent a chemical reaction. The form used to represent the reaction depends on the context in which it is to be used.
4.3 · Representations of Reactions
4.3.A · Represent a given chemical reaction or physical process with a consistent particulate model.
Lesson provided · Practice provided
Course and exam description p. 80
- Teach: Representations of Reactions
More teaching passages in this lesson
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Supports part of this goal.
Check understanding: For the model reaction 2A2 + B2 → 2A2B, a closed whole-sample box starts with seven A2 and three B2 molecules. The reaction proceeds completely… · Worked answer for For the model reaction 2A2 + B2 → 2A2B, a closed whole-sample box starts with seven A2 and three B2 molecules. The reaction proceeds completely… -
Supports part of this goal.
Check understanding: A whole-sample solution model initially contains two Mg2+, four NO3−, six Na+ and six OH− ions. Mg(OH)2 is sparingly soluble. For this… · Worked answer for A whole-sample solution model initially contains two Mg2+, four NO3−, six Na+ and six OH− ions. Mg(OH)2 is sparingly soluble. For this…
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Supports part of this goal.
Check understanding: Five H2 and two O2 react completely as far as possible. What remains? · Worked answer for Five H2 and two O2 react completely as far as possible. What remains? -
Supports part of this goal.
Check understanding: How should NaCl(aq) be drawn? · Worked answer for How should NaCl(aq) be drawn?
Required knowledge for 4.3.A
4.3.A.1 · Balanced chemical equations in their various forms can be translated into symbolic particulate representations.
4.4 · Physical and Chemical Changes
4.4.A · Explain the relationship between macroscopic characteristics and bond interactions for: i. Chemical processes. ii. Physical processes.
Lesson provided · Practice provided
Course and exam description p. 81
- Teach: Physical and Chemical Changes
More teaching passages in this lesson
- Check understanding: A student says “evaporation and NaCl dissolution are physical, so no bonds or attractions change.” Evaluate each process at the particle level and… · Worked answer for A student says “evaporation and NaCl dissolution are physical, so no bonds or attractions change.” Evaluate each process at the particle level and…
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- Independently identified cross-topic support; its locator retains the task's actual teaching owner.
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Supports part of this goal.
Check understanding: What bonds stay intact when liquid ethanol evaporates? · Worked answer for What bonds stay intact when liquid ethanol evaporates? -
Supports part of this goal.
Check understanding: Why is “nothing breaks during dissolution” too strong? · Worked answer for Why is “nothing breaks during dissolution” too strong?
Required knowledge for 4.4.A
4.4.A.1 · Processes that involve the breaking and/or formation of chemical bonds are typically classified as chemical processes. Processes that involve only changes in intermolecular interactions, such as phase changes, are typically classified as physical processes.
4.4.A.2 · Sometimes physical processes involve the breaking of chemical bonds. For example, plausible arguments could be made for the dissolution of a salt in water, as either a physical or chemical process, involves breaking of ionic bonds, and the formation of ion-dipole interactions between ions and solvent.
4.5 · Stoichiometry
4.5.A · Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.
Lesson provided · Practice provided
Course and exam description p. 82
- Teach: Stoichiometry
More teaching passages in this lesson
- Teaching passage: For 2Al + 3Cl2 → 2AlCl3, three moles of chlorine react with two moles of aluminum. The ratio does not mean 3 g react with 2 g because different…
- Teaching passage: For solutions, n = MV with V in liters gives the number of available moles. For gases, n = PV/RT supplies the same bridge, with temperature in…
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Supports part of this goal.
Worked example: Mix 25.0 mL of 0.200 M CaCl2 and 40.0 mL of 0.150 M Na2CO3. Assume complete CaCO3 precipitation, additive volumes, and molar mass 100.09 g mol−1. -
Supports part of this goal.
Worked example: Mix 25.0 mL of 0.200 M CaCl2 and 40.0 mL of 0.150 M Na2CO3. Assume complete CaCO3 precipitation, additive volumes, and molar mass 100.09 g mol−1.
- Check understanding: For N2 + 3H2 → 2NH3, 0.400 mol N2 and 0.900 mol H2 react as far as possible. Find NH3 produced and excess reactant remaining. Then predict the… · Worked answer for For N2 + 3H2 → 2NH3, 0.400 mol N2 and 0.900 mol H2 react as far as possible. Find NH3 produced and excess reactant remaining. Then predict the…
- Check understanding: 30.0 mL of 0.120 M BaCl2 is mixed with 20.0 mL of 0.150 M Na2SO4. BaSO4 is sparingly soluble; assume complete precipitation and additive volumes.… · Worked answer for 30.0 mL of 0.120 M BaCl2 is mixed with 20.0 mL of 0.150 M Na2SO4. BaSO4 is sparingly soluble; assume complete precipitation and additive volumes.…
More checks for this goal
- Check understanding: A 0.0150 mol sample of CaCO3 reacts completely with excess acid: CaCO3(s) + 2H+(aq) → Ca2+(aq) + CO2(g) + H2O(l). CO2 is collected over water at… · Worked answer for A 0.0150 mol sample of CaCO3 reacts completely with excess acid: CaCO3(s) + 2H+(aq) → Ca2+(aq) + CO2(g) + H2O(l). CO2 is collected over water at…
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Supports part of this goal.
Check understanding: Why can’t you compare 2.0 g A and 3.0 g B to decide the limiting reactant directly? · Worked answer for Why can’t you compare 2.0 g A and 3.0 g B to decide the limiting reactant directly? -
Supports part of this goal.
Check understanding: What happens if you double the limiting reagent while keeping all others sufficiently in excess? · Worked answer for What happens if you double the limiting reagent while keeping all others sufficiently in excess? -
Supports part of this goal.
Check understanding: Why does M×V require volume in liters when M is mol L⁻¹? How do you use those moles in a reaction? · Worked answer for Why does M×V require volume in liters when M is mol L⁻¹? How do you use those moles in a reaction? -
Supports part of this goal.
Check understanding: A gas is collected over water. Which pressure belongs in n =PV/RT for the dry product gas? · Worked answer for A gas is collected over water. Which pressure belongs in n =PV/RT for the dry product gas?
Required knowledge for 4.5.A
4.5.A.1 · Because atoms must be conserved during a chemical process, it is possible to calculate product amounts by using known reactant amounts, or to calculate reactant amounts given known product amounts.
4.5.A.2 · Coefficients of balanced chemical equations contain information regarding the proportionality of the amounts of substances involved in the reaction. These values can be used in chemical calculations involving the mole concept.
4.5.A.3 · Stoichiometric calculations can be combined with the ideal gas law and calculations involving molarity to quantitatively study gases and solutions.
4.6 · Introduction to Titration
4.6.A · Identify the equivalence point in a titration based on the amounts of the titrant and analyte, assuming the titration reaction goes to completion.
Lesson provided · Practice provided
Course and exam description p. 83
- Teach: Introduction to Titration
More teaching passages in this lesson
- Check understanding: A 25.00 mL sample of monoprotic acid HA is titrated by 0.1250 M NaOH. Buret readings are 1.35 mL initially and 19.75 mL at the indicator endpoint,… · Worked answer for A 25.00 mL sample of monoprotic acid HA is titrated by 0.1250 M NaOH. Buret readings are 1.35 mL initially and 19.75 mL at the indicator endpoint,…
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Supports part of this goal.
Check understanding: A titration of an unknown acid uses known base concentration. Three endpoints agree closely, but all are reached after adding a little excess… · Worked answer for A titration of an unknown acid uses known base concentration. Three endpoints agree closely, but all are reached after adding a little excess…
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Supports part of this goal.
Check understanding: Are endpoint and equivalence point identical by definition? · Worked answer for Are endpoint and equivalence point identical by definition? -
Supports part of this goal.
Check understanding: At equivalence for H2A + 2OH−, what amount ratio is needed? · Worked answer for At equivalence for H2A + 2OH−, what amount ratio is needed?
Required knowledge for 4.6.A
4.6.A.1 · Titrations may be used to determine the amount of an analyte in solution. The titrant has a known concentration of a species that reacts specifically and quantitatively with the analyte. The equivalence point of the titration occurs when the analyte is totally consumed by the reacting species in the titrant. The equivalence point is often indicated by a change in a property (such as color) that occurs when the equivalence point is reached. This observable event is called the endpoint of the titration.
4.7 · Types of Chemical Reactions
4.7.A · Identify a reaction as acid-base, oxidation-reduction, or precipitation.
Lesson provided · Practice provided
Course and exam description p. 84, 85
- Teach: Types of Chemical Reactions
More teaching passages in this lesson
- Teaching passage: A redox reaction changes oxidation numbers. Complete hydrocarbon combustion produces CO2 and H2O while O2 is reduced. This is electron bookkeeping…
- Teaching passage: Oxidation is loss of electrons and increasing oxidation number; reduction is gain and decreasing oxidation number. They are coupled: the electrons…
- Teaching passage: To assign oxidation numbers, begin with elemental species at zero and monatomic ions at their actual charge. Use familiar assignments such as…
- Teaching passage: Precipitation takes dissolved ions into a sparingly soluble solid lattice. Sodium, potassium, ammonium, and nitrate salts are soluble in the AP…
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Supports part of this goal.
Worked example: Classify NH3 + H2O ⇌ NH4+ + OH−, Zn + Cu2+ → Zn2+ + Cu, and Ag+ + Cl− → AgCl(s).
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- Multipart prompt: only the expressly relevant actions receive credit; all parts share the existing question anchor.
- Check understanding: For C2H6 + O2 → CO2 + H2O, balance complete combustion and find the oxidation-number change per carbon and electrons transferred per two ethane… · Worked answer for For C2H6 + O2 → CO2 + H2O, balance complete combustion and find the oxidation-number change per carbon and electrons transferred per two ethane…
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Supports part of this goal.
Check understanding: In NH₄⁺ +OH⁻ →NH₃ +H₂O, what moves, and which species donates and accepts it? · Worked answer for In NH₄⁺ +OH⁻ →NH₃ +H₂O, what moves, and which species donates and accepts it? -
Supports part of this goal.
Check understanding: Is combustion redox if the products are neutral molecules? · Worked answer for Is combustion redox if the products are neutral molecules? -
Supports part of this goal.
Check understanding: For Zn +Cu²⁺ →Zn²⁺ +Cu, how many electrons are transferred and which species loses them? · Worked answer for For Zn +Cu²⁺ →Zn²⁺ +Cu, how many electrons are transferred and which species loses them? -
Supports part of this goal.
Check understanding: Which AP solubility patterns must be available without a long table? · Worked answer for Which AP solubility patterns must be available without a long table?
Required knowledge for 4.7.A
4.7.A.1 · Acid-base reactions involve transfer of one or more protons (H+ ions) between chemical species.
4.7.A.2 · Oxidation-reduction (redox) reactions involve transfer of one or more electrons between chemical species, as indicated by changes in oxidation numbers of the involved species. Combustion is an important subclass of oxidation-reduction reactions, in which a species reacts with oxygen gas. In the case of hydrocarbons, carbon dioxide and water are products of complete combustion.
4.7.A.3 · In a redox reaction, electrons are transferred from the species that is oxidized to the species that is reduced.
4.7.A.4 · Oxidation numbers may be assigned to each of the atoms in the reactants and products; this is often an effective way to identify the oxidized and reduced species in a redox reaction.
4.7.A.5 · Precipitation reactions frequently involve mixing ions in aqueous solution to produce an insoluble or sparingly soluble ionic compound. All sodium, potassium, ammonium, and nitrate salts are soluble in water.
4.8 · Introduction to Acid-Base Reactions
4.8.A · Identify species as Brønsted-Lowry acids, bases, and/or conjugate acid-base pairs, based on proton-transfer involving those species.
Lesson provided · Practice provided
Course and exam description p. 86
- Teach: Introduction to Acid-Base Reactions
More teaching passages in this lesson
- Teaching passage: In aqueous acid–base chemistry, water is a participant, not always a background spectator. It accepts a proton from HA to form H3O+, and it…
- Teaching passage: HA + H2O ⇌ A− + H3O+ pairs HA with A− and H2O with H3O+. A stronger acid gives a weaker conjugate base: if HA readily gives up a proton, A− has…
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Supports part of this goal.
Worked example: For HCO3− + H2O ⇌ CO32− + H3O+, identify the roles and pairs.
- Check understanding: In CH3COOH + H2O ⇌ CH3COO− + H3O+, label acids, bases and conjugate pairs. Given that H3O+ is a stronger acid than CH3COOH, predict which side is… · Worked answer for In CH3COOH + H2O ⇌ CH3COO− + H3O+, label acids, bases and conjugate pairs. Given that H3O+ is a stronger acid than CH3COOH, predict which side is…
- Check understanding: Two diagrams show water reacting: model I, H2O becomes H3O+ while HX becomes X−; model II, H2O becomes OH− while B becomes BH+. Explain water’s… · Worked answer for Two diagrams show water reacting: model I, H2O becomes H3O+ while HX becomes X−; model II, H2O becomes OH− while B becomes BH+. Explain water’s…
More checks for this goal
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Supports part of this goal.
Check understanding: What is the conjugate acid of CO32−? · Worked answer for What is the conjugate acid of CO32−? -
Supports part of this goal.
Check understanding: How does water act differently in HA +H₂O ⇌A⁻ +H₃O⁺ and B +H₂O ⇌BH⁺ +OH⁻? · Worked answer for How does water act differently in HA +H₂O ⇌A⁻ +H₃O⁺ and B +H₂O ⇌BH⁺ +OH⁻? -
Supports part of this goal.
Check understanding: If acid HA is stronger than HB, how do their conjugate bases compare? · Worked answer for If acid HA is stronger than HB, how do their conjugate bases compare?
Required knowledge for 4.8.A
4.8.A.1 · By definition, a Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.
4.8.A.2 · Only in aqueous solutions, water plays an important role in many acid-base reactions, as its molecular structure allows it to accept protons from and donate protons to dissolved species.
4.8.A.3 · When an acid or base ionizes in water, the conjugate acid-base pairs can be identified and their relative strengths compared.
4.9 · Oxidation-Reduction (Redox) Reactions
4.9.A · Represent a balanced redox reaction equation using half-reactions.
Lesson provided · Practice provided
Course and exam description p. 87
- Teach: Oxidation-Reduction (Redox) Reactions
More teaching passages in this lesson
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Supports part of this goal.
Worked example: Balance permanganate oxidation of Fe2+ to Fe3+ in acidic solution, with MnO4− reduced to Mn2+.
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- Check understanding: Use half-reactions to balance Cr2O72− + I− → Cr3+ + I2 in acidic solution. Verify total charge. · Worked answer for Use half-reactions to balance Cr2O72− + I− → Cr3+ + I2 in acidic solution. Verify total charge.
- Check understanding: Balance Cl2 → Cl− + ClO− in basic aqueous solution using half-reactions. Explain why one element can be both oxidized and reduced. · Worked answer for Balance Cl2 → Cl− + ClO− in basic aqueous solution using half-reactions. Explain why one element can be both oxidized and reduced.
More checks for this goal
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Supports part of this goal.
Check understanding: Why multiply one half-reaction before adding? · Worked answer for Why multiply one half-reaction before adding? -
Supports part of this goal.
Check understanding: Can electrons remain in the final chemical equation? · Worked answer for Can electrons remain in the final chemical equation?
Required knowledge for 4.9.A
4.9.A.1 · Balanced chemical equations for redox reactions can be constructed from half-reactions.
Unit 5: Kinetics
5.1 · Reaction Rates
5.1.A · Explain the relationship between the rate of a chemical reaction and experimental parameters.
Lesson provided · Practice provided
Course and exam description p. 94
- Teach: Reaction Rates
More teaching passages in this lesson
- Teaching passage: Species do not all change at the same numerical rate. For 2A → B, two A units disappear for each B formed. The reaction rate is r = −½ Δ[A]/Δt =…
- Teaching passage: Higher reactant concentration often increases collision frequency; higher temperature usually increases the fraction of sufficiently energetic…
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Supports part of this goal.
Worked example: For 2NO2(g) → 2NO(g) + O2(g) in a constant-volume container, [NO2] falls from 0.0800 M to 0.0680 M in 30.0 s. -
Supports part of this goal.
Worked example: A supplied comparison uses equal solid mass and identical solution at 25 °C, with only solid particle size changed. Powder produces the same final…
- Check understanding: For 2A → 3B at fixed volume, [A] changes from 0.0500 M to 0.0380 M over 40.0 s. Calculate A’s average disappearance rate, the common reaction… · Worked answer for For 2A → 3B at fixed volume, [A] changes from 0.0500 M to 0.0380 M over 40.0 s. Calculate A’s average disappearance rate, the common reaction…
- Check understanding: Supplied observations show the same mass of a powdered solid reacts faster with a solution than a single piece.Suggest a controlled comparison… · Worked answer for Supplied observations show the same mass of a powdered solid reacts faster with a solution than a single piece.Suggest a controlled comparison…
More checks for this goal
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Supports part of this goal.
Check understanding: For 3A → 2B, A disappears at 0.060 M s−1. What is B’s formation rate? · Worked answer for For 3A → 2B, A disappears at 0.060 M s−1. What is B’s formation rate? -
Supports part of this goal.
Check understanding: Why might powdered solid react faster than the same mass of one chunk? · Worked answer for Why might powdered solid react faster than the same mass of one chunk?
Required knowledge for 5.1.A
5.1.A.1 · The kinetics of a chemical reaction is defined as the rate at which an amount of reactants is converted to products per unit of time.
5.1.A.2 · The rates of change of reactant and product concentrations are determined by the stoichiometry in the balanced chemical equation.
5.1.A.3 · The rate of a reaction is influenced by reactant concentrations, temperature, surface area, catalysts, and other environmental factors.
5.2 · Introduction to Rate Law
5.2.A · Represent experimental data with a consistent rate law expression.
Lesson provided · Practice provided
Course and exam description p. 95
- Teach: Introduction to Rate Law
More teaching passages in this lesson
- Teaching passage: An empirical rate law, r = k[A]m[B]n, summarizes how the measured rate responds to concentrations under the tested conditions. The exponents come…
- Teaching passage: The order in A is m; the overall order is m+n. If m =2, doubling A with B fixed multiplies rate by four. If n =0, changing B in that regime leaves…
- Teaching passage: The rate constant k belongs to a specified reaction model and temperature. If rate units are M s−1, total order z gives units M1−z s−1 for k.…
- Teaching passage: Compare two initial-rate trials that differ in only one concentration. Divide rates so k and fixed factors cancel, then solve the remaining power.…
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Supports part of this goal.
Worked example: A calibrated colored-reactant signal has absorbance 0.500 at 0.0200 M and 0.450 after 10.0 s. Assume only this reactant absorbs, fixed path… -
Supports part of this goal.
Worked example: At one temperature, trials give ([A], [B], r): (0.100 M,0.100 M,2.00×10−4 M s−1), (0.200,0.100,8.00×10−4), (0.100,0.300,6.00×10−4).
- Check understanding: At fixed temperature, initial-rate data are:Initial rates for reaction of A and BTrial[A] / M[B] / Mr / M… · Worked answer for At fixed temperature, initial-rate data are:Initial rates for reaction of A and BTrial[A] / M[B] / Mr / M…
- Check understanding: For an overall reaction A + 2B → C, doubling [A] leaves the measured initial rate unchanged; tripling [B] triples it. State the empirical law and… · Worked answer for For an overall reaction A + 2B → C, doubling [A] leaves the measured initial rate unchanged; tripling [B] triples it. State the empirical law and…
More checks for this goal
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Supports part of this goal.
Check understanding: A colored reactant is monitored by absorbance at a fixed wavelength. A student doubles both reactant concentrations and sees four times the… · Worked answer for A colored reactant is monitored by absorbance at a fixed wavelength. A student doubles both reactant concentrations and sees four times the… -
Supports part of this goal.
Check understanding: For r =k[A]2[B], what if A doubles and B halves? · Worked answer for For r =k[A]2[B], what if A doubles and B halves? -
Supports part of this goal.
Check understanding: Can 2A + B → C alone establish r =k[A]2[B]? · Worked answer for Can 2A + B → C alone establish r =k[A]2[B]?
Required knowledge for 5.2.A
5.2.A.1 · Experimental methods can be used to monitor the amounts of reactants and/or products of a reaction over time and to determine the rate of the reaction.
5.2.A.2 · The rate law expresses the rate of a reaction as proportional to the concentration of each reactant raised to a power.
5.2.A.3 · The power of each reactant in the rate law is the order of the reaction with respect to that reactant. The sum of the powers of the reactant concentrations in the rate law is the overall order of the reaction.
5.2.A.4 · The proportionality constant in the rate law is called the rate constant. The value of this constant is temperature dependent and the units reflect the overall reaction order.
5.2.A.5 · Comparing initial rates of a reaction is a method to determine the order with respect to each reactant.
5.3 · Concentration Changes Over Time
5.3.A · Identify the rate law expression of a chemical reaction using data that show how the concentrations of reaction species change over time.
Lesson provided · Practice provided
Course and exam description p. 96, 97
- Teach: Concentration Changes Over Time
More teaching passages in this lesson
- Teaching passage: For first order, ln([A]t/[A]0) = −kt. A plot of ln([A]/1 M) versus t is linear with slope −k; dividing by a reference unit makes the logarithm…
- Teaching passage: For second order in A, 1/[A]t − 1/[A]0 = kt. The reciprocal plot is linear with positive slope k, measured in M−1 s−1. Raw concentration…
- Teaching passage: Zero order: [A]t =[A]0 −kt and raw slope −k, with k in M s−1. First order: log slope −k, with k in s−1. Second order: reciprocal slope +k, with k…
- Teaching passage: First-order half-life is t½ = ln2/k ≈0.693/k. Every halving takes the same time because the fractional disappearance rate is constant. After two…
- Teaching passage: Radioactive decay illustrates first-order kinetics: each nucleus has a constant probability of decaying per time, and the sample’s average number…
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Supports part of this goal.
Worked example: For a zero-order regime, [A] falls from 0.0500 to 0.0400 M in 20.0 s and to 0.0300 M in another 20.0 s. -
Supports part of this goal.
Worked example: For a first-order disappearance of A, [A]0 =0.0800 M and [A]60s =0.0400 M; multiple measurements establish a linear log plot. -
Supports part of this goal.
Worked example: In a second-order regime, [A] is 0.100 M initially and 0.0500 M at 100 s; additional measurements establish a straight reciprocal plot. -
Supports part of this goal.
Worked example: A supplied isotope has half-life 5.00 h and starts with 48.0 μg undecayed isotope.
- Check understanding: For A disappearance at fixed temperature, measurements are t =0, 40.0, 80.0, 120.0 s with [A] =0.160, 0.0800, 0.0400, 0.0200 M. Identify the… · Worked answer for For A disappearance at fixed temperature, measurements are t =0, 40.0, 80.0, 120.0 s with [A] =0.160, 0.0800, 0.0400, 0.0200 M. Identify the…
- Check understanding: Data for A: at 0, 50.0, 100.0 s, concentrations are 0.1000, 0.06667, 0.05000 M. The reciprocal plot is linear within measurement precision. Find… · Worked answer for Data for A: at 0, 50.0, 100.0 s, concentrations are 0.1000, 0.06667, 0.05000 M. The reciprocal plot is linear within measurement precision. Find…
More checks for this goal
- Check understanding: A monitored reactant decreases linearly from 0.0600 M at t =0 to 0.0480 M at 30.0 s and 0.0360 M at 60.0 s. State the supported model and k. A… · Worked answer for A monitored reactant decreases linearly from 0.0600 M at t =0 to 0.0480 M at 30.0 s and 0.0360 M at 60.0 s. State the supported model and k. A…
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
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Supports part of this goal.
Check understanding: Which plot should be straight for second-order disappearance of A? · Worked answer for Which plot should be straight for second-order disappearance of A? -
Supports part of this goal.
Check understanding: Is half of the original amount removed in every first-order half-life? · Worked answer for Is half of the original amount removed in every first-order half-life?
Required knowledge for 5.3.A
5.3.A.1 · The order of a reaction can be inferred from a graph of concentration of reactant versus time.
5.3.A.2 · If a reaction is first order with respect to a reactant being monitored, a plot of the natural log (ln) of the reactant concentration as a function of time will be linear.
5.3.A.3 · If a reaction is second order with respect to a reactant being monitored, a plot of the reciprocal of the concentration of that reactant versus time will be linear.
5.3.A.4 · The slopes of the concentration versus time data for zeroth, first, and second order reactions can be used to determine the rate constant for the reaction. Zeroth order: EQN: [A]t − [A]0 = −kt First order: EQN: ln[A]t − ln[A]0 = −kt Second order: EQN: 1/[A]t − 1/[A]0 = kt
5.3.A.5 · Half-life is a critical parameter for first order reactions because the half-life is constant and related to the rate constant for the reaction by the equation: EQN: t1/2 = 0.693/k.
5.3.A.6 · Radioactive decay processes provide an important illustration of first order kinetics.
5.4 · Elementary Reactions
5.4.A · Represent an elementary reaction as a rate law expression using stoichiometry.
Lesson provided · Practice provided
Course and exam description p. 98
- Teach: Elementary Reactions
More teaching passages in this lesson
- Check understanding: A mechanism proposes the elementary step 2NO2 → NO3 + NO. Give the elementary rate law and molecularity. Another student models the overall 2NO +… · Worked answer for A mechanism proposes the elementary step 2NO2 → NO3 + NO. Give the elementary rate law and molecularity. Another student models the overall 2NO +…
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Supports part of this goal.
Check understanding: For an elementary 2A → B step, what law follows? · Worked answer for For an elementary 2A → B step, what law follows?
More checks for this goal
Required knowledge for 5.4.A
5.4.A.1 · The rate law of an elementary reaction can be inferred from the stoichiometry of the particles participating in a collision.
5.4.A.2 · Elementary reactions involving the simultaneous collision of three or more particles are rare.
5.5 · Collision Model
5.5.A · Explain the relationship between the rate of an elementary reaction and the frequency, energy, and orientation of particle collisions.
Lesson provided · Practice provided
Course and exam description p. 99
- Teach: Collision Model
More teaching passages in this lesson
- Teaching passage: Two requirements matter together: sufficient collision energy to cross the activation barrier and suitable orientation of reacting sites. A…
- Teaching passage: The Maxwell–Boltzmann distribution shows that particles at one temperature have many energies, not one shared energy. Draw energy on the…
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Supports part of this goal.
Worked example: Imagine the end of one molecule that must bond to a specific atom of another. Two equal-energy collisions can hit different ends. -
Supports part of this goal.
Worked example: The height at one energy is a density. Count a fraction by the area over an energy range. Heating changes the distribution; a catalyst offers a…
- Check understanding: At the same concentration, warming a reacting gas greatly increases the observed rate. Draw two particle-energy distributions on the same axes,… · Worked answer for At the same concentration, warming a reacting gas greatly increases the observed rate. Draw two particle-energy distributions on the same axes,…
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
More checks for this goal
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Supports part of this goal.
Check understanding: Does heating make every particle exceed Ea? · Worked answer for Does heating make every particle exceed Ea? -
Supports part of this goal.
Check understanding: What changes if a catalyst lowers the relevant barrier at fixed temperature? · Worked answer for What changes if a catalyst lowers the relevant barrier at fixed temperature?
Required knowledge for 5.5.A
5.5.A.1 · For an elementary reaction to successfully produce products, reactants must successfully collide to initiate bond-breaking and bond-making events.
5.5.A.2 · In most reactions, only a small fraction of the collisions leads to a reaction. Successful collisions have both sufficient energy to overcome the activation energy requirements and orientations that allow the bonds to rearrange in the required manner.
5.5.A.3 · The Maxwell-Boltzmann distribution curve describes the distribution of particle energies; this distribution can be used to gain a qualitative estimate of the fraction of collisions with sufficient energy to lead to a reaction, and also how that fraction depends on temperature.
5.6 · Reaction Energy Profile
5.6.A · Represent the activation energy and overall energy change in an elementary reaction using a reaction energy profile.
Lesson provided · Practice provided
Course and exam description p. 100
- Teach: Reaction Energy Profile
More teaching passages in this lesson
- Teaching passage: The horizontal reaction coordinate represents progress through the set of particle motions involved in rearrangement. It is not elapsed time and…
- Teaching passage: The forward activation energy is Etransition −Ereactants. The reverse barrier is Etransition −Eproducts. The overall energy change is Eproducts…
- Teaching passage: For a fixed elementary pathway, increasing temperature increases the fraction of collisions that can reach the transition-state region, so k…
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Supports part of this goal.
Worked example: In the schematic elementary event A–B +C–D →A–C +B–D, compare the starting bonded pairs with the final pairs. -
Supports part of this goal.
Worked example: An elementary-step energy profile places reactants at 25 kJ mol−1, the transition-state peak at 95 kJ mol−1, and products at 5 kJ mol−1, relative… -
Supports part of this goal.
Worked example: For a fixed elementary pathway, the reactant and peak energies stay the same in the qualitative model while temperature rises.
- Check understanding: A hypothetical elementary rearrangement A–B +C–D →A–C +B–D has reactants at 10, transition state at 75, and products at 40 kJ mol⁻¹ on one energy… · Worked answer for A hypothetical elementary rearrangement A–B +C–D →A–C +B–D has reactants at 10, transition state at 75, and products at 40 kJ mol⁻¹ on one energy…
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Supports part of this goal.
Check understanding: Is reaction coordinate a clock? · Worked answer for Is reaction coordinate a clock?
More checks for this goal
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Supports part of this goal.
Check understanding: Why can an exothermic reaction be slow? · Worked answer for Why can an exothermic reaction be slow?
Required knowledge for 5.6.A
5.6.A.1 · Elementary reactions typically involve the breaking of some bonds and the forming of new ones.
5.6.A.2 · The reaction coordinate is the axis along which the complex set of motions involved in rearranging reactants to form products can be plotted.
5.6.A.3 · The energy profile gives the energy along the reaction coordinate, which typically proceeds from reactants, through a transition state, to products. The energy difference between the reactants and the transition state is the activation energy for the forward reaction.
5.6.A.4 · The rate of an elementary reaction is temperature dependent because the proportion of particle collisions that are energetic enough to reach the transition state varies with temperature. The Arrhenius equation relates the temperature dependence of the rate of an elementary reaction to the activation energy needed by molecular collisions to reach the transition state.
5.7 · Introduction to Reaction Mechanisms
5.7.A · Identify the components of a reaction mechanism.
Lesson provided · Practice provided
Course and exam description p. 101
- Teach: Introduction to Reaction Mechanisms
More teaching passages in this lesson
- Teaching passage: Add all elementary steps algebraically. Cancel a species that appears in equal amounts on both sides; the remaining equation must match the…
- Teaching passage: An intermediate is produced in one step and consumed in another, so it may be present transiently even though absent from the net equation. It…
- Teaching passage: If independent supplied evidence detects a proposed intermediate, that supports a mechanism containing it. If another proposed route does not…
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Supports part of this goal.
Worked example: Proposed steps: NO2 + NO2 → NO3 + NO; then NO3 + CO → NO2 + CO2. -
Supports part of this goal.
Worked example: Mechanism M predicts intermediate I; mechanism N predicts no I. A supplied composition record detects transient I before product accumulation.
- Check understanding: Proposed steps: NO + O3 → NO2 + O2; NO2 + O → NO + O2. Identify the catalyst and intermediate, give the net equation, and explain how supplied… · Worked answer for Proposed steps: NO + O3 → NO2 + O2; NO2 + O → NO + O2. Identify the catalyst and intermediate, give the net equation, and explain how supplied…
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Supports part of this goal.
Check understanding: How does a catalyst differ from an intermediate? · Worked answer for How does a catalyst differ from an intermediate?
More checks for this goal
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Supports part of this goal.
Check understanding: Does a correct sum prove a mechanism? · Worked answer for Does a correct sum prove a mechanism?
Required knowledge for 5.7.A
5.7.A.1 · A reaction mechanism consists of a series of elementary reactions, or steps, that occur in sequence. The components may include reactants, intermediates, products, and catalysts.
5.7.A.2 · The elementary steps when combined should align with the overall balanced equation of a chemical reaction.
5.7.A.3 · A reaction intermediate is produced by some elementary steps and consumed by others, such that it is present only while a reaction is occurring.
5.7.A.4 · Experimental detection of a reaction intermediate is a common way to build evidence in support of one reaction mechanism over an alternative mechanism.
5.8 · Reaction Mechanism and Rate Law
5.8.A · Identify the rate law for a reaction from a mechanism in which the first step is rate limiting.
Lesson provided · Practice provided
Course and exam description p. 102
- Teach: Reaction Mechanism and Rate Law
More teaching passages in this lesson
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Supports part of this goal.
Worked example: Step 1: 2NO2 → NO3 + NO, slow. Step 2: NO3 + CO → NO2 + CO2, fast.
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- Check understanding: A proposed mechanism has slow elementary A + A → I, followed by fast I + B → P; labels denote conserved chemical units. The net equation is 2A + B… · Worked answer for A proposed mechanism has slow elementary A + A → I, followed by fast I + B → P; labels denote conserved chemical units. The net equation is 2A + B…
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Supports part of this goal.
Check understanding: Why may a net reactant be missing from the rate law? · Worked answer for Why may a net reactant be missing from the rate law?
More checks for this goal
Required knowledge for 5.8.A
5.8.A.1 · For reaction mechanisms in which each elementary step is irreversible, or in which the first step is rate limiting, the rate law of the reaction is set by the molecularity of the slowest elementary step (i.e., the rate-limiting step).
5.9 · Pre-Equilibrium Approximation
5.9.A · Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting.
Lesson provided · Practice provided
Course and exam description p. 103
- Teach: Pre-Equilibrium Approximation
More teaching passages in this lesson
- Check understanding: Mechanism: NO + Br2 ⇌ NOBr2 (rapid pre-equilibrium); NOBr2 + NO → 2NOBr (slow). Derive the reactant-only rate law and predict the rate factor if… · Worked answer for Mechanism: NO + Br2 ⇌ NOBr2 (rapid pre-equilibrium); NOBr2 + NO → 2NOBr (slow). Derive the reactant-only rate law and predict the rate factor if…
- Check understanding: Two mechanisms have the same net reaction A + B → P. M1: slow elementary A + B → P. M2: rapid A ⇌ I followed by slow I + B → P, with… · Worked answer for Two mechanisms have the same net reaction A + B → P. M1: slow elementary A + B → P. M2: rapid A ⇌ I followed by slow I + B → P, with…
More checks for this goal
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Supports part of this goal.
Check understanding: Why not leave r =k2[I][C] as the final observed reactant law? · Worked answer for Why not leave r =k2[I][C] as the final observed reactant law? -
Supports part of this goal.
Check understanding: What assumption permits equality of the fast forward and reverse rates? · Worked answer for What assumption permits equality of the fast forward and reverse rates?
Required knowledge for 5.9.A
5.9.A.1 · If the first elementary reaction is not rate limiting, approximations (such as pre-equilibrium) must be made to determine a rate law expression.
5.10 · Multistep Reaction Energy Profile
5.10.A · Represent the activation energy and overall energy change in a multistep reaction with a reaction energy profile.
Lesson provided · Practice provided
Course and exam description p. 104
- Teach: Multistep Reaction Energy Profile
More teaching passages in this lesson
- Check understanding: Construct a labeled three-step energy profile using reactants0, TS1 35, I1 10, TS2 60, I2 −25, TS3 20 and products −40 kJ mol⁻¹. Label the energy… · Worked answer for Construct a labeled three-step energy profile using reactants0, TS1 35, I1 10, TS2 60, I2 −25, TS3 20 and products −40 kJ mol⁻¹. Label the energy…
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Supports part of this goal.
Check understanding: Can the highest plotted peak alone identify the slowest step? · Worked answer for Can the highest plotted peak alone identify the slowest step?
More checks for this goal
Required knowledge for 5.10.A
5.10.A.1 · Knowledge of the energetics of each elementary reaction in a mechanism allows for the construction of an energy profile for a multistep reaction.
5.11 · Catalysis
5.11.A · Explain the relationship between the effect of a catalyst on a reaction and changes in the reaction mechanism.
Lesson provided · Practice provided
Course and exam description p. 105
- Teach: Catalysis
More teaching passages in this lesson
- Teaching passage: A catalyst participates in elementary steps and is regenerated. For E + A ⇌ EA followed by EA → E + P, E is used to bind A and returned when…
- Teaching passage: Binding can position reacting groups together and make a suitable reaction geometry more likely. Enzymes commonly form bound intermediates that…
- Teaching passage: In acid–base catalysis, a proton can be donated to create a more reactive intermediate and later removed to regenerate the catalytic acid, or…
- Teaching passage: In surface catalysis, reactants bind at accessible sites, can be brought near one another or have bonds weakened, react while bound, and products…
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Supports part of this goal.
Worked example: Model: E + A ⇌ EA; EA → EP; EP → E + P. Compare it with uncatalyzed A → P. -
Supports part of this goal.
Worked example: Two supplied cycles illustrate distinct catalytic mechanisms.
- Check understanding: An enzyme route is E + S ⇌ ES, ES → EP, EP → E + P.Identify catalyst and intermediates; explain how rate can increase, whether ΔH changes, and… · Worked answer for An enzyme route is E + S ⇌ ES, ES → EP, EP → E + P.Identify catalyst and intermediates; explain how rate can increase, whether ΔH changes, and…
- Check understanding: Compare two supplied catalytic models: acid H+ is added to a reactant to form a protonated intermediate then returned during product formation; a… · Worked answer for Compare two supplied catalytic models: acid H+ is added to a reactant to form a protonated intermediate then returned during product formation; a…
More checks for this goal
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Supports part of this goal.
Check understanding: What must the mechanism show to support calling a species a catalyst? · Worked answer for What must the mechanism show to support calling a species a catalyst? -
Supports part of this goal.
Check understanding: Why can adding surface catalyst stop giving a proportional rate increase? · Worked answer for Why can adding surface catalyst stop giving a proportional rate increase?
Required knowledge for 5.11.A
5.11.A.1 · In order for a catalyst to increase the rate of a reaction, the addition of the catalyst must increase the number of effective collisions and/or provide a reaction path with a lower activation energy relative to the original reaction coordinate.
5.11.A.2 · In a reaction mechanism containing a catalyst, the net concentration of the catalyst is constant. However, the catalyst will frequently be consumed in the rate-determining step of the reaction, only to be regenerated in a subsequent step in the mechanism.
5.11.A.3 · Some catalysts accelerate a reaction by binding to the reactant(s). The reactants are either oriented more favorably or react with lower activation energy. There is often a new reaction intermediate in which the catalyst is bound to the reactant(s). Many enzymes function in this manner.
5.11.A.4 · Some catalysts involve covalent bonding between the catalyst and the reactant(s). An example is acid-base catalysis, in which a reactant or intermediate either gains or loses a proton. This introduces a new reaction intermediate and new elementary reactions involving that intermediate.
5.11.A.5 · In surface catalysis, a reactant or intermediate binds to, or forms a covalent bond with, the surface. This introduces elementary reactions involving these new bound reaction intermediate(s).
Unit 6: Thermochemistry
6.1 · Endothermic and Exothermic Processes
6.1.A · Explain the relationship between experimental observations and energy changes associated with a chemical or physical transformation.
Lesson provided · Practice provided
Course and exam description p. 112
- Teach: Endothermic and Exothermic Processes
More teaching passages in this lesson
- Teaching passage: Endothermic means energy is transferred into the chosen process/system as heat; exothermic means it transfers energy out. Heating a nonreacting…
- Teaching passage: When an exothermic reaction reduces the chemical potential energy of the reacting species, that energy can increase surrounding particles’ thermal…
- Teaching passage: Dissolution has competing energetic contributions. Separating solute particles costs energy; making space/rearranging solvent interactions costs…
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Supports part of this goal.
Worked example: A supplied heating record gives a flat temperature while a pure solid melts at fixed pressure. -
Supports part of this goal.
Worked example: A supplied insulated-cup record shows that dissolving a solid lowers solution temperature from 24.0 to 19.5 °C, with cup effects temporarily…
- Check understanding: A supplied dissolution experiment in an insulated cup lowers the solution temperature. Identify energy-flow direction for the dissolution and for… · Worked answer for A supplied dissolution experiment in an insulated cup lowers the solution temperature. Identify energy-flow direction for the dissolution and for…
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Supports part of this goal.
Check understanding: Can energy enter a substance while its temperature stays constant? · Worked answer for Can energy enter a substance while its temperature stays constant?
More checks for this goal
Required knowledge for 6.1.A
6.1.A.1 · Temperature changes in a system indicate energy changes.
6.1.A.2 · Energy changes in a system can be described as endothermic and exothermic processes such as the heating or cooling of a substance, phase changes, or chemical transformations.
6.1.A.3 · When a chemical reaction occurs, the energy of the system either decreases (exothermic reaction), increases (endothermic reaction), or remains the same. For exothermic reactions, the energy lost by the reacting species (system) is gained by the surroundings, as heat transfer from or work done by the system. Likewise, for endothermic reactions, the system gains energy from the surroundings by heat transfer to or work done on the system.
6.1.A.4 · The formation of a solution may be an exothermic or endothermic process, depending on the relative strengths of intermolecular/interparticle interactions before and after the dissolution process.
6.2 · Energy Diagrams
6.2.A · Represent a chemical or physical transformation with an energy diagram.
Lesson provided · Practice provided
Course and exam description p. 113
- Teach: Energy Diagrams
More teaching passages in this lesson
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Supports part of this goal.
Worked example: For a process with initial level 12 and final level 47 kJ mol−1, construct a two-level diagram.
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- Check understanding: A molar physical transformation has initial enthalpy 8.0 and final enthalpy 30.0 kJ mol−1 on a common scale. Construct an energy diagram: label… · Worked answer for A molar physical transformation has initial enthalpy 8.0 and final enthalpy 30.0 kJ mol−1 on a common scale. Construct an energy diagram: label…
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Supports part of this goal.
Check understanding: If final particles have a lower enthalpy than initial particles, what sign is ΔH? · Worked answer for If final particles have a lower enthalpy than initial particles, what sign is ΔH?
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Required knowledge for 6.2.A
6.2.A.1 · A physical or chemical process can be described with an energy diagram that shows the endothermic or exothermic nature of that process.
6.3 · Heat Transfer and Thermal Equilibrium
6.3.A · Explain the relationship between the transfer of thermal energy and molecular collisions.
Lesson provided · Practice provided
Course and exam description p. 114
- Teach: Heat Transfer and Thermal Equilibrium
More teaching passages in this lesson
- Teaching passage: When bodies make thermal contact, particle interactions and collisions exchange energy. On average there is net transfer from higher temperature…
- Teaching passage: At thermal equilibrium their temperatures are equal, and there is no net heat flow between them. Microscopic exchanges can continue in both…
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Supports part of this goal.
Worked example: Supplied data: 40.0 g copper at 80.0 °C and 40.0 g water at 20.0 °C exchange heat in an ideal insulated container. cCu =0.385 and cwater =4.184 J…
- Check understanding: Two isolated bodies with constant heat capacities Chot =30.0 J °C−1 at 70.0 °C and Ccool =90.0 J °C−1 at 10.0 °C reach thermal equilibrium without… · Worked answer for Two isolated bodies with constant heat capacities Chot =30.0 J °C−1 at 70.0 °C and Ccool =90.0 J °C−1 at 10.0 °C reach thermal equilibrium without…
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Supports part of this goal.
Check understanding: At thermal equilibrium do collisions stop? · Worked answer for At thermal equilibrium do collisions stop?
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Required knowledge for 6.3.A
6.3.A.1 · The particles in a warmer body have a greater average kinetic energy than those in a cooler body.
6.3.A.2 · Collisions between particles in thermal contact can result in the transfer of energy. This process is called “heat transfer,” “heat exchange,” or “transfer of energy as heat.”
6.3.A.3 · Eventually, thermal equilibrium is reached as the particles continue to collide. At thermal equilibrium, the average kinetic energy of both bodies is the same, and hence, their temperatures are the same.
6.4 · Heat Capacity and Calorimetry
6.4.A · Calculate the heat q absorbed or released by a system undergoing heating/cooling based on the amount of the substance, the heat capacity, and the change in temperature.
Lesson provided · Practice provided
Course and exam description p. 115, 116
- Teach: Heat Capacity and Calorimetry
More teaching passages in this lesson
- Teaching passage: Energy conservation sets the heat balance. In an insulated constant-pressure reaction calorimeter, qreaction +qsolution +qcup =0. If external heat…
- Teaching passage: Specific heat capacity c states energy per gram per degree. For the same positive q and equal masses, the material with larger c has the smaller…
- Teaching passage: For a sample undergoing only heating/cooling, energy entering raises its thermal energy and energy leaving lowers it. The sign follows the named…
- Teaching passage: Molar heat capacity Cm uses moles rather than grams: q =nCmΔT. If molar mass is Mmolar in g mol−1, then Cm =cMmolar. The whole-sample heat…
- Teaching passage: Separate three contributions when several processes occur: sensible heating/cooling, phase transition, and chemical transformation. The first uses…
- Teaching passage: For dissolution data, warming of the mixture indicates an exothermic dissolution under the chosen heat-balance model; cooling indicates…
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Supports part of this goal.
Worked example: A supplied constant-pressure trial reacts 0.0200 mol of a limiting species. The 100.0 g solution warms by 2.50 °C; c =4.184 J g−1 °C−1 and cup… -
Supports part of this goal.
Worked example: Give 100 J to 5.00 g of sample A with c =0.500 J g⁻¹ °C⁻¹; its molar mass is 40.0 g mol⁻¹. -
Supports part of this goal.
Worked example: A supplied thermal record includes warming a solid, melting it, and then reacting the liquid at constant pressure.
- Check understanding: Read the initial and corrected-final temperatures from these two thermometer scales. Record each value with its unit and a sensible estimated… · Worked answer for Read the initial and corrected-final temperatures from these two thermometer scales. Record each value with its unit and a sensible estimated…
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
More checks for this goal
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
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Supports part of this goal.
Check understanding: A sample cools from 30.0 to 20.0 °C. What is ΔT and q’s sign for that sample? · Worked answer for A sample cools from 30.0 to 20.0 °C. What is ΔT and q’s sign for that sample? -
Supports part of this goal.
Check understanding: Why does ignoring cup heating bias an exothermic ΔH toward zero? · Worked answer for Why does ignoring cup heating bias an exothermic ΔH toward zero? -
Supports part of this goal.
Check understanding: What units distinguish c, Cm, and whole-sample C? · Worked answer for What units distinguish c, Cm, and whole-sample C?
Required knowledge for 6.4.A
6.4.A.1 · The heating of a cool body by a warmer body is an important form of energy transfer between two systems. The amount of heat transferred between two bodies may be quantified by the heat transfer equation: EQN: q = mcΔT. Calorimetry experiments are used to measure the transfer of heat.
6.4.A.2 · The first law of thermodynamics states that energy is conserved in chemical and physical processes.
6.4.A.3 · The transfer of a given amount of thermal energy will not produce the same temperature change in equal masses of matter with differing specific heat capacities.
6.4.A.4 · Heating a system increases the energy of the system, while cooling a system decreases the energy of the system.
6.4.A.5 · The specific heat capacity of a substance and the molar heat capacity are both used in energy calculations.
6.4.A.6 · Chemical systems change their energy through three main processes: heating/cooling, phase transitions, and chemical reactions.
6.4.A.7 · In calorimetry experiments involving dissolution, temperature changes of the mixture within the calorimeter can be used to determine the direction of energy flow. If the temperature of the mixture increases, thermal energy is released by the dissolution process (exothermic). If the temperature of the mixture decreases, thermal energy is absorbed by the dissolution process (endothermic).
6.5 · Energy of Phase Changes
6.5.A · Explain changes in the heat q absorbed or released by a system undergoing a phase transition based on the amount of the substance in moles and the molar enthalpy of the phase transition.
Lesson provided · Practice provided
Course and exam description p. 117
- Teach: Energy of Phase Changes
More teaching passages in this lesson
- Teaching passage: For a phase segment, q =nΔHtransition. If ΔHvap is positive, condensation at the same conditions has ΔHcond =−ΔHvap; freezing similarly reverses…
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Supports part of this goal.
Worked example: At its melting temperature, 18.0 g ice melts completely. Given molar mass 18.0 g mol−1 and ΔHfus =6.01 kJ mol−1, find heat. Then the liquid warms…
- Check understanding: Consider this ideal heating sequence: begin with 36.0 g ice at −10.0 °C; melt it at 0 °C, then warm the liquid to 20.0 °C. Use cice =2.10, cwater… · Worked answer for Consider this ideal heating sequence: begin with 36.0 g ice at −10.0 °C; melt it at 0 °C, then warm the liquid to 20.0 °C. Use cice =2.10, cwater…
- Check understanding: At the specified boiling temperature/pressure, a substance has ΔHvap =30.0 kJ mol−1. A 0.250 mol vapor sample condenses completely without further… · Worked answer for At the specified boiling temperature/pressure, a substance has ΔHvap =30.0 kJ mol−1. A 0.250 mol vapor sample condenses completely without further…
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Supports part of this goal.
Check understanding: How can a melting plateau absorb energy at constant temperature? · Worked answer for How can a melting plateau absorb energy at constant temperature? -
Supports part of this goal.
Check understanding: If vaporizing 0.50 mol absorbs 20 kJ, what does condensing 1.00 mol release under the same conditions? · Worked answer for If vaporizing 0.50 mol absorbs 20 kJ, what does condensing 1.00 mol release under the same conditions?
Required knowledge for 6.5.A
6.5.A.1 · Energy must be transferred to a system to cause a substance to melt (or boil). The energy of the system therefore increases as the system undergoes a solid-to-liquid (or liquid-to-gas) phase transition. Likewise, a system releases energy when it freezes (or condenses). The energy of the system decreases as the system undergoes a liquid-to-solid (or gas-to-liquid) phase transition. The temperature of a pure substance remains constant during a phase change.
6.5.A.2 · The energy absorbed during a phase change is equal to the energy released during a complementary phase change in the opposite direction. For example, the molar enthalpy of condensation of a substance is equal to the negative of its molar enthalpy of vaporization. Similarly, the molar enthalpy of fusion can be used to calculate the energy absorbed when melting a substance and the energy released when freezing a substance.
6.6 · Introduction to Enthalpy of Reaction
6.6.A · Calculate the heat q absorbed or released by a system undergoing a chemical reaction in relationship to the amount of the reacting substance in moles and the molar enthalpy of reaction.
Lesson provided · Practice provided
Course and exam description p. 118
- Teach: Introduction to Enthalpy of Reaction
More teaching passages in this lesson
- Teaching passage: An exothermic reaction creates products with lower chemical potential energy than reactants. Some of that difference can appear as increased…
- Teaching passage: Bond breaking requires energy input and bond formation releases energy. The net change depends on both, not on breaking alone. At the same final…
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Supports part of this goal.
Worked example: For 2H2(g) + O2(g) → 2H2O(l), ΔH =−571.6 kJ. A supplied reaction consumes 0.150 mol H2 with sufficient oxygen. -
Supports part of this goal.
Worked example: A reaction’s new bond formation releases more energy than the initial bond breaking consumes.
- Check understanding: For 2CO(g) + O2(g) → 2CO2(g), ΔH = −566.0 kJ. A supplied sample contains 0.300 mol CO and 0.100 mol O2.Find the heat as the reaction proceeds to… · Worked answer for For 2CO(g) + O2(g) → 2CO2(g), ΔH = −566.0 kJ. A supplied sample contains 0.300 mol CO and 0.100 mol O2.Find the heat as the reaction proceeds to…
- Check understanding: A balanced reaction is 2A → B with ΔH =+80.0 kJ for the equation as written. If 0.250 mol B is formed, find q and state ΔH for the reversed… · Worked answer for A balanced reaction is 2A → B with ΔH =+80.0 kJ for the equation as written. If 0.250 mol B is formed, find q and state ΔH for the reversed…
More checks for this goal
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Supports part of this goal.
Check understanding: If an equation’s coefficients are doubled, what happens to its ΔH? · Worked answer for If an equation’s coefficients are doubled, what happens to its ΔH? -
Supports part of this goal.
Check understanding: Why can bond breaking not be the reason an exothermic reaction releases heat? · Worked answer for Why can bond breaking not be the reason an exothermic reaction releases heat?
Required knowledge for 6.6.A
6.6.A.1 · The enthalpy change of a reaction gives the amount of heat energy released (for negative values) or absorbed (for positive values) by a chemical reaction at constant pressure.
6.6.A.2 · When the products of a reaction are at a different temperature than their surroundings, they exchange energy with the surroundings to reach thermal equilibrium. Thermal energy is transferred to the surroundings as the reactants convert to products in an exothermic reaction. Thermal energy is transferred from the surroundings as the reactants convert to products in an endothermic reaction.
6.6.A.3 · The chemical potential energy of the products of a reaction is different from that of the reactants because of the breaking and forming of bonds. The energy difference results in a change in the kinetic energy of the particles, which manifests as a temperature change.
6.7 · Bond Enthalpies
6.7.A · Calculate the enthalpy change of a reaction based on the average bond energies of bonds broken and formed in the reaction.
Lesson provided · Practice provided
Course and exam description p. 119
- Teach: Bond Enthalpies
More teaching passages in this lesson
- Teaching passage: Estimate reaction enthalpy with ΔH ≈ΣDbroken −ΣDformed. Multiply bond counts by equation coefficients. Unchanged bonds can cancel if treated…
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Supports part of this goal.
Worked example: Estimate H2(g) + Cl2(g) → 2HCl(g) with D(H–H) =436, D(Cl–Cl) =243, D(H–Cl) =431 kJ mol−1 of bonds.
- Check understanding: Estimate C2H4(g) + H2(g) → C2H6(g). Ethene has one C=C and four C–H bonds; ethane has one C–C and six C–H bonds. Average D values: C=C 614, H–H… · Worked answer for Estimate C2H4(g) + H2(g) → C2H6(g). Ethene has one C=C and four C–H bonds; ethane has one C–C and six C–H bonds. Average D values: C=C 614, H–H…
- Check understanding: For N2(g) + 3H2(g) → 2NH3(g), use average D(N≡N) =945, D(H–H) =436, D(N–H) =391 kJ mol−1. Estimate ΔH and separate the breaking input from… · Worked answer for For N2(g) + 3H2(g) → 2NH3(g), use average D(N≡N) =945, D(H–H) =436, D(N–H) =391 kJ mol−1. Estimate ΔH and separate the breaking input from…
More checks for this goal
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Supports part of this goal.
Check understanding: What sign does the bond-breaking sum have in an energy ledger? · Worked answer for What sign does the bond-breaking sum have in an energy ledger? -
Supports part of this goal.
Check understanding: Why is a bond-enthalpy result usually an estimate? · Worked answer for Why is a bond-enthalpy result usually an estimate?
Required knowledge for 6.7.A
6.7.A.1 · During a chemical reaction, bonds are broken and/or formed, and these events change the potential energy of the system.
6.7.A.2 · The average energy required to break all of the bonds in the reactant molecules can be estimated by adding up the average bond energies of all the bonds in the reactant molecules. Likewise, the average energy released in forming the bonds in the product molecules can be estimated. If the energy released is greater than the energy required, the reaction is exothermic. If the energy required is greater than the energy released, the reaction is endothermic.
6.8 · Enthalpy of Formation
6.8.A · Calculate the enthalpy change for a chemical or physical process based on the standard enthalpies of formation.
Lesson provided · Practice provided
Course and exam description p. 120
- Teach: Enthalpy of Formation
More teaching passages in this lesson
- Check understanding: For CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), supplied 298 K ΔHf° values in kJ mol−1 are CH4 −74.8, CO2 −393.5, H2O(l) −285.8, O2 0. Find ΔH°.… · Worked answer for For CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), supplied 298 K ΔHf° values in kJ mol−1 are CH4 −74.8, CO2 −393.5, H2O(l) −285.8, O2 0. Find ΔH°.…
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Supports part of this goal.
Check understanding: Write the standard formation reaction for exactly one mole of C2H5OH(l) at 298 K using carbon graphite, H2(g), and O2(g). Explain why formation… · Worked answer for Write the standard formation reaction for exactly one mole of C2H5OH(l) at 298 K using carbon graphite, H2(g), and O2(g). Explain why formation…
More checks for this goal
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Supports part of this goal.
Check understanding: Is ΔHf° zero for every elemental form? · Worked answer for Is ΔHf° zero for every elemental form? -
Supports part of this goal.
Check understanding: Why must liquid and vapor product labels be matched to the data table? · Worked answer for Why must liquid and vapor product labels be matched to the data table?
Required knowledge for 6.8.A
6.8.A.1 · Tables of standard enthalpies of formation can be used to calculate the standard enthalpies of reactions. EQN: ΔH° reaction = ΣΔHf ° products − ΣΔHf ° reactants
6.9 · Hess’s Law
6.9.A · Represent a chemical or physical process as a sequence of steps.
Lesson provided · Practice provided
Course and exam description p. 121
- Teach: Hess’s Law
More teaching passages in this lesson
- Check understanding: Find ΔH for 2C(graphite,s) +2H2(g) → C2H4(g) using: C +O2 → CO2, −393.5 kJ; H2 +½O2 → H2O(l), −285.8 kJ; C2H4 +3O2 → 2CO2 +2H2O(l), −1411.0 kJ.… · Worked answer for Find ΔH for 2C(graphite,s) +2H2(g) → C2H4(g) using: C +O2 → CO2, −393.5 kJ; H2 +½O2 → H2O(l), −285.8 kJ; C2H4 +3O2 → 2CO2 +2H2O(l), −1411.0 kJ.…
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Supports part of this goal.
Check understanding: An accounting path for A(g) → B(l) has A(g) → B(g), ΔH =+25.0 kJ mol−1, and B(g) → B(l), ΔH =−40.0 kJ mol−1, at compatible stated conditions. Find… · Worked answer for An accounting path for A(g) → B(l) has A(g) → B(g), ΔH =+25.0 kJ mol−1, and B(g) → B(l), ΔH =−40.0 kJ mol−1, at compatible stated conditions. Find…
Required knowledge for 6.9.A
6.9.A.1 · Many processes can be broken down into a series of steps. Each step in the series has its own energy change.
6.9.B · Explain the relationship between the enthalpy of a chemical or physical process and the sum of the enthalpies of the individual steps.
Lesson provided · Practice provided
Course and exam description p. 121
- Teach: Hess’s Law
More teaching passages in this lesson
- Teaching passage: Three operations keep the energy ledger aligned with the species ledger. Reverse a reaction and reverse ΔH’s sign. Multiply every coefficient by a…
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Supports part of this goal.
Worked example: Find C(graphite,s) +½O2(g) → CO(g), using C +O2 → CO2, ΔH =−393.5 kJ, and CO +½O2 → CO2, ΔH =−283.0 kJ.
- Check understanding: Find ΔH for 2C(graphite,s) +2H2(g) → C2H4(g) using: C +O2 → CO2, −393.5 kJ; H2 +½O2 → H2O(l), −285.8 kJ; C2H4 +3O2 → 2CO2 +2H2O(l), −1411.0 kJ.… · Worked answer for Find ΔH for 2C(graphite,s) +2H2(g) → C2H4(g) using: C +O2 → CO2, −393.5 kJ; H2 +½O2 → H2O(l), −285.8 kJ; C2H4 +3O2 → 2CO2 +2H2O(l), −1411.0 kJ.…
- Check understanding: An accounting path for A(g) → B(l) has A(g) → B(g), ΔH =+25.0 kJ mol−1, and B(g) → B(l), ΔH =−40.0 kJ mol−1, at compatible stated conditions. Find… · Worked answer for An accounting path for A(g) → B(l) has A(g) → B(g), ΔH =+25.0 kJ mol−1, and B(g) → B(l), ΔH =−40.0 kJ mol−1, at compatible stated conditions. Find…
More checks for this goal
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Supports part of this goal.
Check understanding: Why do step enthalpies add? · Worked answer for Why do step enthalpies add? -
Supports part of this goal.
Check understanding: What two things change when you reverse a thermochemical equation? · Worked answer for What two things change when you reverse a thermochemical equation? -
Supports part of this goal.
Check understanding: Can liquid water and vapor water cancel in a Hess sum? · Worked answer for Can liquid water and vapor water cancel in a Hess sum?
Required knowledge for 6.9.B
6.9.B.1 · Because total energy is conserved (first law of thermodynamics), and each individual reaction in a sequence transfers thermal energy to or from the surroundings, the net thermal energy transferred in the sequence will be equal to the sum of the thermal energy transfers in each of the steps. These thermal energy transfers are the result of potential energy changes among the species in the reaction sequence; thus, at constant pressure, the enthalpy change of the overall process is equal to the sum of the enthalpy changes of the individual steps.
6.9.B.2 · The following are essential principles of Hess’s law: i. When a reaction is reversed, the enthalpy change stays constant in magnitude but becomes reversed in mathematical sign. ii. When a reaction is multiplied by a factor c, the enthalpy change is multiplied by the same factor c. iii. When two (or more) reactions are added to obtain an overall reaction, the individual enthalpy changes of each reaction are added to obtain the net enthalpy change of the overall reaction.
Unit 7: Equilibrium
7.1 · Introduction to Equilibrium
7.1.A · Explain the relationship between the occurrence of a reversible chemical or physical process, and the establishment of equilibrium, to experimental observations.
Lesson provided · Practice provided
Course and exam description p. 128
- Teach: Introduction to Equilibrium
More teaching passages in this lesson
- Teaching passage: At equilibrium, reactants and products coexist and their concentrations or partial pressures stay constant under fixed conditions. That sounds…
- Teaching passage: On a concentration–time graph, equilibrium appears as plateaus, usually at different heights. On a forward-rate/reverse-rate graph, the curves…
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Supports part of this goal.
Worked example: For A ⇌ B, begin with 0.90 M A and 0.10 M B in a sealed, constant-volume flask. Measurements eventually settle at 0.60 M A and 0.40 M B.
- Check understanding: A sealed flask at constant temperature contains A(g) ⇌ B(g). A sensor records [A] = 0.80, 0.62, 0.53, 0.50, 0.50 M at 0, 10, 20, 40, 60 s; [B] =… · Worked answer for A sealed flask at constant temperature contains A(g) ⇌ B(g). A sensor records [A] = 0.80, 0.62, 0.53, 0.50, 0.50 M at 0, 10, 20, 40, 60 s; [B] =…
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Supports part of this goal.
Check understanding: A sealed bottle keeps the same liquid level. Has evaporation stopped? · Worked answer for A sealed bottle keeps the same liquid level. Has evaporation stopped?
More checks for this goal
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Supports part of this goal.
Check understanding: Does an equilibrium mixture need equal amounts of reactants and products? · Worked answer for Does an equilibrium mixture need equal amounts of reactants and products? -
Supports part of this goal.
Check understanding: How do you distinguish equilibrium on a concentration–time graph and on a forward/reverse-rate graph? · Worked answer for How do you distinguish equilibrium on a concentration–time graph and on a forward/reverse-rate graph?
Required knowledge for 7.1.A
7.1.A.1 · Many observable processes are reversible. Examples include evaporation and condensation of water, absorption and desorption of a gas, or dissolution and precipitation of a salt. Some important reversible chemical processes include the transfer of protons in acid-base reactions and the transfer of electrons in redox reactions.
7.1.A.2 · When equilibrium is reached, no observable changes occur in the system. Reactants and products are simultaneously present, and the concentrations or partial pressures of all species remain constant.
7.1.A.3 · The equilibrium state is dynamic. The forward and reverse processes continue to occur at equal rates, resulting in no net observable change.
7.1.A.4 · Graphs of concentration, partial pressure, or rate of reaction versus time for simple chemical reactions can be used to understand the establishment of chemical equilibrium.
7.2 · Direction of Reversible Reactions
7.2.A · Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.
Lesson provided · Practice provided
Course and exam description p. 129
- Teach: Direction of Reversible Reactions
More teaching passages in this lesson
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Supports part of this goal.
Worked example: In a model interval, 12 A → B events and 5 B → A events occur for A ⇌ B.
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- Check understanding: For A(g) ⇌ B(g), the forward and reverse rates at one instant are 0.030 and 0.012 mol L−1 s−1. Determine the signs and magnitudes of the… · Worked answer for For A(g) ⇌ B(g), the forward and reverse rates at one instant are 0.030 and 0.012 mol L−1 s−1. Determine the signs and magnitudes of the…
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Supports part of this goal.
Check understanding: Which way does the mixture move if the reverse rate is larger? · Worked answer for Which way does the mixture move if the reverse rate is larger?
More checks for this goal
Required knowledge for 7.2.A
7.2.A.1 · If the rate of the forward reaction is greater than the reverse reaction, then there is a net conversion of reactants to products. If the rate of the reverse reaction is greater than that of the forward reaction, then there is a net conversion of products to reactants. An equilibrium state is reached when these rates are equal.
7.3 · Reaction Quotient and Equilibrium Constant
7.3.A · Represent the reaction quotient Q_c or Q_p, for a reversible reaction, and the corresponding equilibrium expressions K_c = Q_c or K_p = Q_p.
Lesson provided · Practice provided
Course and exam description p. 130, 131
- Teach: Reaction Quotient and Equilibrium Constant
More teaching passages in this lesson
- Check understanding: Write Qc for CaCO3(s) ⇌ CaO(s) + CO2(g), and Qp for N2(g) + 3H2(g) ⇌ 2NH3(g). Explain each omitted term and exponent. When can either quotient be… · Worked answer for Write Qc for CaCO3(s) ⇌ CaO(s) + CO2(g), and Qp for N2(g) + 3H2(g) ⇌ 2NH3(g). Explain each omitted term and exponent. When can either quotient be…
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Supports part of this goal.
Check understanding: Does H2O(g) disappear from a quotient? · Worked answer for Does H2O(g) disappear from a quotient?
More checks for this goal
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Supports part of this goal.
Check understanding: Why omit a pure solid? · Worked answer for Why omit a pure solid?
Required knowledge for 7.3.A
7.3.A.1 · The reaction quotient Q_c describes the relative concentrations of reaction species at any time. For gas phase reactions, the reaction quotient may instead be written in terms of partial pressures as Q_p. The reaction quotient tends toward the equilibrium constant such that at equilibrium K_c = Q_c and K_p = Q_p. As examples, for the reaction a A + b B ⇌ c C + d D the law of mass action indicates that the equilibrium expression for (K_c, Q_c) is EQN: K_c = [C]^c[D]^d/([A]^a[B]^b) and that for (K_p, Q_p) is EQN: K_p = (P_C)^c(P_D)^d/((P_A)^a(P_B)^b)
7.3.A.2 · The reaction quotient does not include substances whose concentrations (or partial pressures) are independent of the amount, such as for solids and pure liquids.
7.4 · Calculating the Equilibrium Constant
7.4.A · Calculate K_c or K_p based on experimental observations of concentrations or pressures at equilibrium.
Lesson provided · Practice provided
Course and exam description p. 132
- Teach: Calculating the Equilibrium Constant
More teaching passages in this lesson
- Check understanding: For 2NO2(g) ⇌ N2O4(g), equilibrium samples at one temperature give [NO2] = 0.0400 M and [N2O4] = 0.120 M. Calculate Kc. A second run reports… · Worked answer for For 2NO2(g) ⇌ N2O4(g), equilibrium samples at one temperature give [NO2] = 0.0400 M and [N2O4] = 0.120 M. Calculate Kc. A second run reports…
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Supports part of this goal.
Check understanding: May you insert equilibrium moles directly into Kc? · Worked answer for May you insert equilibrium moles directly into Kc?
More checks for this goal
Required knowledge for 7.4.A
7.4.A.1 · Equilibrium constants can be determined from experimental measurements of the concentrations or partial pressures of the reactants and products at equilibrium.
7.5 · Magnitude of the Equilibrium Constant
7.5.A · Explain the relationship between very large or very small values of K and the relative concentrations of chemical species at equilibrium.
Lesson provided · Practice provided
Course and exam description p. 133
- Teach: Magnitude of the Equilibrium Constant
More teaching passages in this lesson
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Supports part of this goal.
Worked example: Write [B] = 99.0[A]. The mass balance becomes 100[A] = 0.200 M.
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- Check understanding: Two 1:1 isomerizations A ⇌ B begin with only A. At the same temperature Kc is 2.0 × 105 for one and 4.0 × 10−6 for the other. Compare their… · Worked answer for Two 1:1 isomerizations A ⇌ B begin with only A. At the same temperature Kc is 2.0 × 105 for one and 4.0 × 10−6 for the other. Compare their…
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Supports part of this goal.
Check understanding: Does K = 108 mean zero reactant remains? · Worked answer for Does K = 108 mean zero reactant remains?
Required knowledge for 7.5.A
7.5.A.1 · Some equilibrium reactions have very large K values and proceed essentially to completion. Others have very small K values and barely proceed at all.
7.6 · Properties of the Equilibrium Constant
7.6.A · Represent a multistep process with an overall equilibrium expression, using the constituent K expressions for each individual reaction.
Lesson provided · Practice provided
Course and exam description p. 134
- Teach: Properties of the Equilibrium Constant
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- Multipart prompt: only the expressly relevant actions receive credit; all parts share the existing question anchor.
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Supports part of this goal.
Check understanding: What happens to K when a reaction is reversed? · Worked answer for What happens to K when a reaction is reversed?
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Supports part of this goal.
Check understanding: A ⇌ B has current Q = 3 and equilibrium K = 5. What are Q and K for 2A ⇌ 2B? · Worked answer for A ⇌ B has current Q = 3 and equilibrium K = 5. What are Q and K for 2A ⇌ 2B? -
Supports part of this goal.
Check understanding: Why multiply K values when equations add? · Worked answer for Why multiply K values when equations add?
Required knowledge for 7.6.A
7.6.A.1 · When a reaction is reversed, K is inverted.
7.6.A.2 · When the stoichiometric coefficients of a reaction are multiplied by a factor c, K is raised to the power c.
7.6.A.3 · When reactions are added together, the K of the resulting overall reaction is the product of the K’s for the reactions that were summed.
7.6.A.4 · Since the expressions for K and Q have identical mathematical forms, all valid algebraic manipulations of K also apply to Q.
7.7 · Calculating Equilibrium Concentrations
7.7.A · Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.
Lesson provided · Practice provided
Course and exam description p. 135
- Teach: Calculating Equilibrium Concentrations
More teaching passages in this lesson
- Teaching passage: An initial–change–equilibrium table is bookkeeping with a chemical reason. Choose x as the amount of reaction progress per liter, or a consistent…
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Supports part of this goal.
Worked example: A ⇌ B has Kc = 3.00, initially [A] = 0.200 M and [B] = 0.100 M. Q = 0.500 < K, so choose a forward x.
- Check understanding: A(g) ⇌ B(g) has Kc = 4.00 at a fixed temperature. A sealed constant-volume flask initially has [A] = 0.300 M and [B] = 0.200 M. Determine the net… · Worked answer for A(g) ⇌ B(g) has Kc = 4.00 at a fixed temperature. A sealed constant-volume flask initially has [A] = 0.300 M and [B] = 0.200 M. Determine the net…
- Check understanding: For N2O4(g) ⇌ 2NO2(g), Kc = 0.100 at the stated temperature. Initially [N2O4] = 0.100 M and [NO2] = 0. Calculate both equilibrium concentrations.… · Worked answer for For N2O4(g) ⇌ 2NO2(g), Kc = 0.100 at the stated temperature. Initially [N2O4] = 0.100 M and [NO2] = 0. Calculate both equilibrium concentrations.…
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Supports part of this goal.
Check understanding: For A ⇌ 2B, is the product change +x? · Worked answer for For A ⇌ 2B, is the product change +x? -
Supports part of this goal.
Check understanding: What checks belong after solving? · Worked answer for What checks belong after solving?
Required knowledge for 7.7.A
7.7.A.1 · The concentrations or partial pressures of species at equilibrium can be predicted given the balanced reaction, initial concentrations, and the appropriate K.
7.7.A.2 · When Q < K, the reaction will proceed with a net consumption of reactants and generation of products. When Q > K, the reaction will proceed with a net consumption of products and generation of reactants. When Q = K, the system is at dynamic equilibrium; both forward and reverse reactions proceed at the same rate, and the proportion of reactants and products remains constant.
7.8 · Representations of Equilibrium
7.8.A · Represent a system undergoing a reversible reaction with a particulate model.
Lesson provided · Practice provided
Course and exam description p. 136
- Teach: Representations of Equilibrium
More teaching passages in this lesson
- Check understanding: In equal-volume boxes for the isomerization A ⇌ B, box I contains 8 A and 2 B particles, box II contains 4 A and 6 B, and box III contains 2 A and… · Worked answer for In equal-volume boxes for the isomerization A ⇌ B, box I contains 8 A and 2 B particles, box II contains 4 A and 6 B, and box III contains 2 A and…
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Supports part of this goal.
Check understanding: For 2A ⇌ A2, what must a diagram conserve? · Worked answer for For 2A ⇌ A2, what must a diagram conserve?
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Required knowledge for 7.8.A
7.8.A.1 · Particulate representations can be used to describe the relative numbers of reactant and product particles present prior to and at equilibrium, and the value of the equilibrium constant.
7.9 · Introduction to Le Châtelier’s Principle
7.9.A · Identify the response of a system at equilibrium to an external stress, using Le Châtelier's principle.
Lesson provided · Practice provided
Course and exam description p. 137
- Teach: Introduction to Le Châtelier’s Principle
More teaching passages in this lesson
- Teaching passage: Connect the predicted response to what is measured. If a colored complex forms, a net increase of that complex can deepen its color at a fixed…
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Supports part of this goal.
Worked example: For Fe3+(aq) + SCN−(aq) ⇌ FeSCN2+(aq), the complex is red. Consider a small addition of SCN− that changes volume negligibly.
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- Multipart prompt: only the expressly relevant actions receive credit; all parts share the existing question anchor.
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Supports part of this goal.
Check understanding: Does adding an inert gas at constant volume always shift equilibrium? · Worked answer for Does adding an inert gas at constant volume always shift equilibrium?
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Supports part of this goal.
Check understanding: What does a catalyst change? · Worked answer for What does a catalyst change? -
Supports part of this goal.
Check understanding: A red complex forms by A(aq) + B(aq) ⇌ AB(aq). A small addition of B negligibly changes volume. Predict the response and color at a fixed light path. · Worked answer for A red complex forms by A(aq) + B(aq) ⇌ AB(aq). A small addition of B negligibly changes volume. Predict the response and color at a fixed light path.
Required knowledge for 7.9.A
7.9.A.1 · Le Châtelier’s principle can be used to predict the response of a system to stresses such as addition or removal of a chemical species, change in temperature, change in volume/pressure of a gas-phase system, or dilution of a reaction system.
7.9.A.2 · Le Châtelier’s principle can be used to predict the effect that a stress will have on experimentally measurable properties such as pH, temperature, and color of a solution.
7.10 · Reaction Quotient and Le Châtelier’s Principle
7.10.A · Explain the relationships between Q, K, and the direction in which a reversible reaction will proceed to reach equilibrium.
Lesson provided · Practice provided
Course and exam description p. 138
- Teach: Reaction Quotient and Le Châtelier’s Principle
More teaching passages in this lesson
- Teaching passage: At fixed temperature, adding or removing species, compressing gases, or diluting solutions changes Q but leaves K unchanged. For A ⇌ 2B, doubling…
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Supports part of this goal.
Worked example: A + B ⇌ C is at equilibrium. Dilute the solution to twice its volume at constant T; every concentration is halved immediately.
- Check understanding: A(g) ⇌ 2B(g) is at equilibrium with Kp = 0.800. The volume is doubled instantly at constant temperature. Determine the new Qp in terms of its old… · Worked answer for A(g) ⇌ 2B(g) is at equilibrium with Kp = 0.800. The volume is doubled instantly at constant temperature. Determine the new Qp in terms of its old…
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Supports part of this goal.
Check understanding: Why does compression not shift A ⇌ B in the ideal gas model? · Worked answer for Why does compression not shift A ⇌ B in the ideal gas model?
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Required knowledge for 7.10.A
7.10.A.1 · A disturbance to a system at equilibrium causes Q to differ from K, thereby taking the system out of equilibrium. The system responds by bringing Q back into agreement with K, thereby establishing a new equilibrium state.
7.10.A.2 · Some stresses, such as changes in concentration, cause a change in Q only. A change in temperature causes a change in K. In either case, the concentrations or partial pressures of species redistribute to bring Q and K back into equality.
7.11 · Introduction to Solubility Equilibria
7.11.A · Calculate the solubility of a salt based on the value of K_sp for the salt.
Lesson provided · Practice provided
Course and exam description p. 139
- Teach: Introduction to Solubility Equilibria
More teaching passages in this lesson
- Check understanding: At a stated temperature, MX2(s) ⇌ M2+(aq) + 2X−(aq) has Ksp = 3.20 × 10−11. Find its molar solubility in pure water, ignoring side reactions. Then… · Worked answer for At a stated temperature, MX2(s) ⇌ M2+(aq) + 2X−(aq) has Ksp = 3.20 × 10−11. Find its molar solubility in pure water, ignoring side reactions. Then…
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Supports part of this goal.
- Credit is limited to the action/knowledge actually elicited; neighboring clauses and solution feedback do not enlarge the student task.
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Supports part of this goal.
Check understanding: Qsp is below Ksp with no solid in the container. Must precipitation occur? · Worked answer for Qsp is below Ksp with no solid in the container. Must precipitation occur? -
Supports part of this goal.
Check understanding: Is s equal to [X−] for MX2? · Worked answer for Is s equal to [X−] for MX2?
Required knowledge for 7.11.A
7.11.A.1 · The dissolution of a salt is a reversible process whose extent can be described by K_sp, the solubility-product constant.
7.11.A.2 · The solubility of a substance can be calculated from the K_sp for the dissolution process. This relationship can also be used to predict the relative solubility of different substances.
7.11.A.3 · The solubility rules (see 4.7.A.5) can be quantitatively related to K_sp, in which K_sp values >1 correspond to soluble salts.
7.11.A.4 · The molar solubility of one or more species in a saturated solution can be used to calculate the K_sp of a substance.
7.12 · Common-Ion Effect
7.12.A · Identify the solubility of a salt, and/or the value of K_sp for the salt, based on the concentration of a common ion already present in solution.
Lesson provided · Practice provided
Course and exam description p. 140
- Teach: Common-Ion Effect
More teaching passages in this lesson
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Supports part of this goal.
Worked example: MX2 has Ksp = 1.00 × 10−12 and dissolves into a solution initially 0.0100 M in X−.
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- Check understanding: MX(s) ⇌ M+(aq) + X−(aq) has Ksp = 4.00 × 10−10 at 25 °C. Compare its molar solubility in pure water with that in 0.0200 M NaX. Show and check an… · Worked answer for MX(s) ⇌ M+(aq) + X−(aq) has Ksp = 4.00 × 10−10 at 25 °C. Compare its molar solubility in pure water with that in 0.0200 M NaX. Show and check an…
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Supports part of this goal.
Check understanding: What changes on adding a common ion at fixed temperature? · Worked answer for What changes on adding a common ion at fixed temperature?
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Required knowledge for 7.12.A
7.12.A.1 · The solubility of a salt is reduced when it is dissolved into a solution that already contains one of the ions present in the salt. The impact of this “common-ion effect” on solubility can be understood qualitatively using Le Châtelier’s principle or calculated from the K_sp for the dissolution process.
Unit 8: Acids and Bases
8.1 · Introduction to Acids and Bases
8.1.A · Calculate the values of pH and pOH, based on K_w and the concentration of all species present in a neutral solution of water.
Lesson provided · Practice provided
Course and exam description p. 146
- Teach: Introduction to Acids and Bases
More teaching passages in this lesson
- Teaching passage: Water itself transfers protons: 2H2O(l) ⇌ H3O+(aq) + OH−(aq). Its equilibrium constant is Kw = [H3O+][OH−], equal to 1.0 × 10−14 at 25 °C. Both…
- Teaching passage: Neutral means [H3O+] = [OH−]. In pure water both equal √Kw. At 25 °C they are 1.0 × 10−7 M, giving pH = pOH = 7.00 and pH + pOH = pKw = 14.00. At…
- Check understanding: At a particular temperature, Kw = 4.00 × 10−14. Calculate [H3O+], [OH−], pH, and pOH in pure water at that temperature. A student calls this water… · Worked answer for At a particular temperature, Kw = 4.00 × 10−14. Calculate [H3O+], [OH−], pH, and pOH in pure water at that temperature. A student calls this water…
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Supports part of this goal.
Check understanding: Does H+(aq) mean an unhydrated proton? · Worked answer for Does H+(aq) mean an unhydrated proton?
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Supports part of this goal.
Check understanding: What is neutral at any temperature? · Worked answer for What is neutral at any temperature?
Required knowledge for 8.1.A
8.1.A.1 · The concentrations of hydronium ion and hydroxide ion are often reported as pH and pOH, respectively. EQN: pH = −log[H_3O^+] EQN: pOH = −log[OH^−] The terms “hydrogen ion” and “hydronium ion” and the symbols H^+(aq) and H_3O^+(aq) are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and H_3O^+(aq) are preferred, but H^+(aq) is also accepted on the AP Exam.
8.1.A.2 · Water autoionizes with an equilibrium constant K_w. EQN: K_w = [H3O+][OH−] = 1.0 × 10−14 at 25°C
8.1.A.3 · In pure water, pH = pOH is called a neutral solution. At 25°C, pK_w = 14.0 and thus pH = pOH = 7.0. EQN: pK_w = 14 = pH + pOH at 25°C
8.1.A.4 · The value of K_w is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.
8.2 · pH and pOH of Strong Acids and Bases
8.2.A · Calculate pH and pOH based on concentrations of all species in a solution of a strong acid or a strong base.
Lesson provided · Practice provided
Course and exam description p. 147
- Teach: pH and pOH of Strong Acids and Bases
More teaching passages in this lesson
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- Multipart prompt: only the expressly relevant actions receive credit; all parts share the existing question anchor.
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Supports part of this goal.
Check understanding: Is a 0.001 M HCl solution weak because it is dilute? · Worked answer for Is a 0.001 M HCl solution weak because it is dilute?
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Supports part of this goal.
Check understanding: Why double C for dissolved Ca(OH)2? · Worked answer for Why double C for dissolved Ca(OH)2?
Required knowledge for 8.2.A
8.2.A.1 · Molecules of a strong acid (e.g., HCl, HBr, HI, HClO4, H2SO4, and HNO3) will completely ionize in aqueous solution to produce hydronium ions and the conjugate base of the acid. As such, the concentration of H3O+ in a strong acid solution is equal to the initial concentration of the strong acid, and thus the pH of the strong acid solution is easily calculated.
8.2.A.2 · When dissolved in solution, strong bases (e.g., group I and II hydroxides) completely dissociate to produce hydroxide ions. As such, the concentration of OH− in a strong base solution is equal to the initial concentration of a group I hydroxide and double the initial concentration of a group II hydroxide, and thus the pOH (and pH) of the strong base solution is easily calculated.
8.3 · Weak Acid and Base Equilibria
8.3.A · Explain the relationship among pH, pOH, and concentrations of all species in a solution of a monoprotic weak acid or weak base.
Lesson provided · Practice provided
Course and exam description p. 148, 149
- Teach: Weak Acid and Base Equilibria
More teaching passages in this lesson
- Teaching passage: A weak base accepts a proton from water: B + H2O ⇌ HB+ + OH−. Its constant is Kb = [HB+][OH−]/[B], and pKb = −log Kb. Set the formed hydroxide…
- Teaching passage: Percent ionization is the amount converted divided by the initial amount, multiplied by 100%. Under the simple single-solute assumptions it is…
- Teaching passage: A conjugate pair differs by one proton. At a given temperature its constants satisfy KaKb = Kw and pKa + pKb = pKw. A stronger acid therefore has…
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Supports part of this goal.
Worked example: A weak acid has C = 0.0500 M and Ka = 2.00 × 10−4. Write Ka = x2/(0.0500 − x). -
Supports part of this goal.
Worked example: Worked weak-base example at 25 °C: C(B) = 0.0100 M and Kb = 1.00 × 10−5.
- Check understanding: A 0.100 M solution of a monoprotic weak acid HA has Ka = 1.00 × 10−5 at 25 °C. Determine pH, percent ionization, the major acid-related species,… · Worked answer for A 0.100 M solution of a monoprotic weak acid HA has Ka = 1.00 × 10−5 at 25 °C. Determine pH, percent ionization, the major acid-related species,…
- Check understanding: A 0.0200 M weak base B has Kb = 2.00 × 10−5 at 25 °C. Estimate pOH, pH, percent protonation, and Ka of its conjugate acid HB+. Write the… · Worked answer for A 0.0200 M weak base B has Kb = 2.00 × 10−5 at 25 °C. Estimate pOH, pH, percent protonation, and Ka of its conjugate acid HB+. Write the…
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Supports part of this goal.
Check understanding: Why is pH not −log C for a weak acid? · Worked answer for Why is pH not −log C for a weak acid? -
Supports part of this goal.
Check understanding: Which base belongs in KaKb = Kw for HA? · Worked answer for Which base belongs in KaKb = Kw for HA?
Required knowledge for 8.3.A
8.3.A.1 · Weak acids react with water to produce hydronium ions. However, only a small percentage of molecules of a weak acid will ionize in this way. Thus, the concentration of H_3O^+ is much less than the initial concentration of the molecular acid, and the vast majority of the acid molecules remain un-ionized.
8.3.A.2 · A solution of a weak acid involves equilibrium between an un-ionized acid and its conjugate base. The equilibrium constant for this reaction is K_a, often reported as pK_a. The pH of a weak acid solution can be determined from the initial acid concentration and the pK_a. EQN: K_a = [H_3O^+][A^−]/[HA] EQN: pK_a = −log K_a
8.3.A.3 · Weak bases react with water to produce hydroxide ions in solution. However, ordinarily just a small percentage of the molecules of a weak base in solution will ionize in this way. Thus, the concentration of OH− in the solution does not equal the initial concentration of the base, and the vast majority of the base molecules remain un-ionized.
8.3.A.4 · A solution of a weak base involves equilibrium between an un-ionized base and its conjugate acid. The equilibrium constant for this reaction is K_b, often reported as pK_b. The pH of a weak base solution can be determined from the initial base concentration and the pK_b. EQN: K_b = [OH^−][HB^+]/[B] EQN: pK_b = −log K_b
8.3.A.5 · The percent ionization of a weak acid (or base) can be calculated from its pK_a (pK_b) and the initial concentration of the acid (base). The percent ionization can also be calculated from the initial concentration of the acid (base) and the equilibrium concentration of any of the species in the equilibrium expression.
8.3.A.6 · For any conjugate acid-base pair, the acid ionization constant and base ionization constant are related by K_w: EQN: K_w = K_a × K_b EQN: pK_w = pK_a + pK_b
8.4 · Acid-Base Reactions and Buffers
8.4.A · Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.
Lesson provided · Practice provided
Course and exam description p. 150, 151
- Teach: Acid-Base Reactions and Buffers
More teaching passages in this lesson
- Teaching passage: Weak acid plus strong base also reacts essentially quantitatively in the usual AP case: HA + OH− → A− + H2O. If HA remains alongside the A− made,…
- Teaching passage: Weak base plus strong acid follows the mirror logic: B + H3O+ → HB+ + H2O. Excess B leaves a B/HB+ buffer. At equivalence the conjugate acid HB+…
- Teaching passage: Weak acid plus weak base instead needs its own equilibrium: HA + B ⇌ A− + HB+. The reaction favors the side containing the weaker acid and weaker…
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Supports part of this goal.
Worked example: 20.0 mL 0.100 M HCl supplies 2.00 mmol hydronium. 30.0 mL 0.0500 M NaOH supplies 1.50 mmol hydroxide. -
Supports part of this goal.
Worked example: Worked weak-acid/strong-base regimes at 25 °C: separate 20.0 mL samples of 0.100 M HA, Ka = 1.00 × 10−5, receive 0.100 M NaOH. -
Supports part of this goal.
Worked example: At 25 °C, mix 20.0 mL 0.100 M B with 20.0 mL 0.100 M HCl. B + H3O+ → HB+ + H2O consumes 2.00 mmol of each. -
Supports part of this goal.
Worked example: Worked weak-acid/weak-base equilibrium model: HA + B ⇌ A− + HB+, Ka(HA) = 1.00 × 10−6 and Ka(HB+) = 1.00 × 10−4.
- Check understanding: At 25 °C, 25.0 mL of 0.120 M HCl is mixed with 35.0 mL of 0.100 M NaOH. Determine the excess reagent, [OH−], pH, and the major spectator-ion… · Worked answer for At 25 °C, 25.0 mL of 0.120 M HCl is mixed with 35.0 mL of 0.100 M NaOH. Determine the excess reagent, [OH−], pH, and the major spectator-ion…
- Check understanding: Three separate 25.0 mL samples of 0.100 M HA, Ka = 1.00 × 10−5, receive 0.100 M NaOH at 25 °C: sample I gets 10.0 mL, II gets 25.0 mL, III gets… · Worked answer for Three separate 25.0 mL samples of 0.100 M HA, Ka = 1.00 × 10−5, receive 0.100 M NaOH at 25 °C: sample I gets 10.0 mL, II gets 25.0 mL, III gets…
More checks for this goal
- Check understanding: A weak base B has Kb = 1.00 × 10−5. Explain the dominant species and pH method when a 20.0 mL 0.100 M B sample receives 0.100 M HCl volumes of… · Worked answer for A weak base B has Kb = 1.00 × 10−5. Explain the dominant species and pH method when a 20.0 mL 0.100 M B sample receives 0.100 M HCl volumes of…
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Supports part of this goal.
Check understanding: After HA reacts with strong hydroxide, what remaining species distinguish a buffer from equivalence and excess-base conditions? · Worked answer for After HA reacts with strong hydroxide, what remaining species distinguish a buffer from equivalence and excess-base conditions? -
Supports part of this goal.
Check understanding: What determines pH beyond weak-acid/strong-base equivalence? · Worked answer for What determines pH beyond weak-acid/strong-base equivalence?
Required knowledge for 8.4.A
8.4.A.1 · When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: H+(aq) + OH−(aq) → H2O(l). The pH of the resulting solution may be determined from the concentration of excess reagent.
8.4.A.2 · When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: HA(aq) + OH^−(aq) ⇌ A^−(aq) H_2O(l). If the weak acid is in excess, then a buffer solution is formed, and the pH can be determined from the Henderson-Hasselbalch (H−H) equation (see 8.9.A.1). If the strong base is in excess, then the pH can be determined from the moles of excess hydroxide ion and the total volume of solution. If they are equimolar, then the (slightly basic) pH can be determined from the equilibrium represented by the equation: A^−(aq) + H_2O(l) ⇌ HA(aq) + OH^−(aq).
8.4.A.3 · When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation: B(aq) + H3O+(aq) HB+(aq) + H2O(l). If the weak base is in excess, then a buffer solution is formed, and the pH can be determined from the H−H equation. If the strong acid is in excess, then the pH can be determined from the moles of excess hydronium ion and the total volume of solution. If they are equimolar, then the (slightly acidic) pH can be determined from the equilibrium represented by the equation: HB+(aq) + H2O(l) B(aq) + H3O+(aq).
8.4.A.4 · When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation: HA(aq) + B(aq) A−(aq) + HB+(aq).
8.5 · Acid-Base Titrations
8.5.A · Explain results from the titration of a mono- or polyprotic acid or base solution, in relation to the properties of the solution and its components.
Lesson provided · Practice provided
Course and exam description p. 152, 153
- Teach: Acid-Base Titrations
More teaching passages in this lesson
- Teaching passage: For a weak monoprotic acid, the curve starts with weak-acid equilibrium, passes through a HA/A− buffer region, reaches a conjugate-base solution…
- Teaching passage: A polyprotic curve can show successive steps. Each additional equivalent of strong base removes another proton; identify the dominant forms as…
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Supports part of this goal.
Worked example: A 20.00 mL weak-base sample requires 30.00 mL 0.1000 M HCl to reach its single equivalence point.
- Check understanding: At 25 °C, a 25.00 mL monoprotic weak-acid sample is titrated with 0.1000 M NaOH. Supplied data are: added volume 0.00, 10.00, 19.50, 20.00, 20.50,… · Worked answer for At 25 °C, a 25.00 mL monoprotic weak-acid sample is titrated with 0.1000 M NaOH. Supplied data are: added volume 0.00, 10.00, 19.50, 20.00, 20.50,…
- Check understanding: A hypothetical diprotic acid H2A is titrated with strong base. Its resolved curve has equivalence points at 15.0 and 30.0 mL; pH values at 7.50… · Worked answer for A hypothetical diprotic acid H2A is titrated with strong base. Its resolved curve has equivalence points at 15.0 and 30.0 mL; pH values at 7.50…
More checks for this goal
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Supports part of this goal.
Check understanding: Is equivalence the same as endpoint? · Worked answer for Is equivalence the same as endpoint? -
Supports part of this goal.
Check understanding: At weak-acid half-equivalence, what lets you read pKa? · Worked answer for At weak-acid half-equivalence, what lets you read pKa?
Required knowledge for 8.5.A
8.5.A.1 · An acid-base reaction can be carried out under controlled conditions in a titration. A titration curve, plotting pH against the volume of titrant added, is useful for summarizing results from a titration.
8.5.A.2 · At the equivalence point for titrations of monoprotic acids or bases, the number of moles of titrant added is equal to the number of moles of analyte originally present. This relationship can be used to obtain the concentration of the analyte. This is the case for titrations of strong acids/bases and weak acids/bases.
8.5.A.3 · For titrations of weak acids/bases, it is useful to consider the point halfway to the equivalence point, that is, the half-equivalence point. At this point, there are equal concentrations of each species in the conjugate acid-base pair, for example, for a weak acid [HA] = [A−]. Because pH = pK_a when the conjugate acid and base have equal concentrations, the pK_a can be determined from the pH at the half-equivalence point in a titration.
8.5.A.4 · At the equivalence point, pH is determined by the major species in solution. Strong acid and strong base titrations result in neutral pH at the equivalence point. However, in titrations of weak acids (weak bases), the conjugate base of the weak acid (conjugate acid of the weak base) is present at the equivalence point and can undergo proton-transfer reactions with the surrounding water, producing basic (acidic) solutions.
8.5.A.5 · For polyprotic acids, titration curves can be used to determine the number of acidic protons. In doing so, the major species present at any point along the curve can be identified, along with the pK_a associated with each proton in a weak polyprotic acid.
8.6 · Molecular Structure of Acids and Bases
8.6.A · Explain the relationship between the strength of an acid or base and the structure of the molecule or ion.
Lesson provided · Practice provided
Course and exam description p. 154
- Teach: Molecular Structure of Acids and Bases
More teaching passages in this lesson
- Check understanding: Explain why CH3COOH is more acidic than CH3CH2OH and why ClCH2COOH is more acidic than CH3COOH. Identify the transferred proton and compare the… · Worked answer for Explain why CH3COOH is more acidic than CH3CH2OH and why ClCH2COOH is more acidic than CH3COOH. Identify the transferred proton and compare the…
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Supports part of this goal.
Check understanding: Why can an oxygen-rich conjugate base be weak? · Worked answer for Why can an oxygen-rich conjugate base be weak?
More checks for this goal
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Supports part of this goal.
Check understanding: Which site accepts a proton in NH3? · Worked answer for Which site accepts a proton in NH3?
Required knowledge for 8.6.A
8.6.A.1 · The protons on a molecule that will participate in acid-base reactions, and the relative strength of these protons, can be inferred from the molecular structure. i. Strong acids (such as HCl, HBr, HI, HClO4, H2SO4, and HNO3) have very weak conjugate bases that are stabilized by electronegativity, inductive effects, resonance, or some combination thereof. ii. Carboxylic acids are one common class of weak acid. iii. Strong bases (such as group I and II hydroxides) have very weak conjugate acids. iv. Common weak bases include nitrogenous bases such as ammonia as well as carboxylate ions. v. Electronegative elements tend to stabilize the conjugate base relative to the conjugate acid, and so increase acid strength.
8.7 · pH and pKa
8.7.A · Explain the relationship between the predominant form of a weak acid or base in solution at a given pH and the pK_a of the conjugate acid or the pK_b of the conjugate base.
Lesson provided · Practice provided
Course and exam description p. 155
- Teach: pH and pKa
More teaching passages in this lesson
- Teaching passage: An acid–base indicator is itself a conjugate pair whose forms have different observable properties, often different colors. Its color follows the…
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Supports part of this goal.
Worked example: A drug-like model base B has conjugate-acid pKa = 9.00. At pH 7.00, [B]/[HB+] = 10^(7.00 − 9.00) = 0.0100.
- Check understanding: A weak acid indicator has pKa = 6.50. Its protonated HIn form is yellow and deprotonated In− form is blue. Predict the predominant form at pH 5.50… · Worked answer for A weak acid indicator has pKa = 6.50. Its protonated HIn form is yellow and deprotonated In− form is blue. Predict the predominant form at pH 5.50…
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Supports part of this goal.
Check understanding: If pH is two units below pKa, which form dominates? · Worked answer for If pH is two units below pKa, which form dominates?
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Supports part of this goal.
Check understanding: Why select pKa near equivalence pH? · Worked answer for Why select pKa near equivalence pH?
Required knowledge for 8.7.A
8.7.A.1 · The protonation state of an acid or base (i.e., the relative concentrations of HA and A−) can be predicted by comparing the pH of a solution to the pK_a of the acid in that solution. When solution pH < acid pK_a, the acid form has a higher concentration than the base form. When solution pH > acid pK_a, the base form has a higher concentration than the acid form.
8.7.A.2 · Acid-base indicators are substances that exhibit different properties (such as color) in their protonated versus deprotonated state, making that property respond to the pH of a solution.
8.7.A.3 · To ensure accurate results in a titration experiment, acid-base indicators should be selected that have a pK_a close to the pH at the equivalence point.
8.8 · Properties of Buffers
8.8.A · Explain the relationship between the ability of a buffer to stabilize pH and the reactions that occur when an acid or a base is added to a buffered solution.
Lesson provided · Practice provided
Course and exam description p. 156
- Teach: Properties of Buffers
More teaching passages in this lesson
- Check understanding: Three equal-volume solutions contain (I) 0.100 M HA only, (II) 0.100 M HA and 0.100 M A−, or (III) 0.100 M HCl and 0.100 M NaCl. HA is a weak… · Worked answer for Three equal-volume solutions contain (I) 0.100 M HA only, (II) 0.100 M HA and 0.100 M A−, or (III) 0.100 M HCl and 0.100 M NaCl. HA is a weak…
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Supports part of this goal.
Check understanding: Which member removes added hydroxide? · Worked answer for Which member removes added hydroxide?
More checks for this goal
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Supports part of this goal.
Check understanding: Why does a buffer eventually fail? · Worked answer for Why does a buffer eventually fail?
Required knowledge for 8.8.A
8.8.A.1 · A buffer solution contains a large concentration of both members in a conjugate acid-base pair. The conjugate acid reacts with added base and the conjugate base reacts with added acid. These reactions are responsible for the ability of a buffer to stabilize pH.
8.9 · Henderson-Hasselbalch Equation
8.9.A · Identify the pH of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer.
Lesson provided · Practice provided
Course and exam description p. 157
- Teach: Henderson-Hasselbalch Equation
More teaching passages in this lesson
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Supports part of this goal.
Worked example: A buffer pair has pKa = 9.20. In one final solution, nHB+ = 0.0200 mol and nB = 0.0300 mol.
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- Check understanding: A buffer has pKa(HA) = 4.80, [HA] = 0.200 M, and [A−] = 0.0500 M. Calculate its current pH and identify the more abundant form. To prepare a… · Worked answer for A buffer has pKa(HA) = 4.80, [HA] = 0.200 M, and [A−] = 0.0500 M. Calculate its current pH and identify the more abundant form. To prepare a…
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Supports part of this goal.
Check understanding: Which form is on top in the logarithm? · Worked answer for Which form is on top in the logarithm?
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Required knowledge for 8.9.A
8.9.A.1 · The pH of the buffer is related to the pK_a of the acid and the concentration ratio of the conjugate acid-base pair. This relation is a consequence of the equilibrium expression associated with the dissociation of a weak acid, and is described by the Henderson-Hasselbalch equation. Adding small amounts of acid or base to a buffered solution does not significantly change the ratio of [A−]/[HA] and thus does not significantly change the solution pH. The change in pH on addition of acid or base to a buffered solution is therefore much less than it would have been in the absence of the buffer. EQN: pH = pK_a + log([A^−]/[HA])
8.10 · Buffer Capacity
8.10.A · Explain the relationship between the buffer capacity of a solution and the relative concentrations of the conjugate acid and conjugate base components of the solution.
Lesson provided · Practice provided
Course and exam description p. 158
- Teach: Buffer Capacity
More teaching passages in this lesson
- Teaching passage: Capacity can be directional. More HA than A− gives a larger reserve to consume added base, because HA donates the needed protons. More A− gives a…
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Supports part of this goal.
Worked example: Two 50.0 mL samples have pKa = 4.80 and [A−]/[HA] = 2.00. Sample I has [HA] = 0.100 M; II has [HA] = 0.0200 M.
- Check understanding: Equal 100 mL buffers A and B have the same HA/A− pair. A contains 0.100 M of each component; B contains 0.0100 M of each. Compare their initial pH… · Worked answer for Equal 100 mL buffers A and B have the same HA/A− pair. A contains 0.100 M of each component; B contains 0.0100 M of each. Compare their initial pH…
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Supports part of this goal.
Check understanding: Why can diluted buffer retain its pH but lose capacity? · Worked answer for Why can diluted buffer retain its pH but lose capacity?
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Required knowledge for 8.10.A
8.10.A.1 · Increasing the concentration of the buffer components (while keeping the ratio of these concentrations constant) keeps the pH of the buffer the same but increases the capacity of the buffer to neutralize added acid or base.
8.10.A.2 · When a buffer has more conjugate acid than base, it has a greater buffer capacity for addition of added base than acid. When a buffer has more conjugate base than acid, it has a greater buffer capacity for addition of added acid than base.
8.11 · pH and Solubility
8.11.A · Identify the qualitative effect of changes in pH on the solubility of a salt.
Lesson provided · Practice provided
Course and exam description p. 159
- Teach: pH and Solubility
More teaching passages in this lesson
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Supports part of this goal.
Worked example: For MC2O4(s) ⇌ M2+ + C2O42−, suppose the anion is a weak base and no other reactions dominate.
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- Check understanding: Compare adding acid to saturated solutions containing M(OH)2(s), MCO3(s), and MX(s), where X− is the extremely weak conjugate base of a strong… · Worked answer for Compare adding acid to saturated solutions containing M(OH)2(s), MCO3(s), and MX(s), where X− is the extremely weak conjugate base of a strong…
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Supports part of this goal.
Check understanding: Why can acid dissolve more carbonate salt? · Worked answer for Why can acid dissolve more carbonate salt?
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Required knowledge for 8.11.A
8.11.A.1 · The solubility of a salt is pH sensitive when one of the constituent ions is a weak acid, a weak base, or the hydroxide ion. These effects can be understood qualitatively using Le Châtelier’s principle.
Unit 9: Thermodynamics and Electrochemistry
9.1 · Introduction to Entropy
9.1.A · Identify the sign and relative magnitude of the entropy change associated with chemical or physical processes.
Lesson provided · Practice provided
Course and exam description p. 166
- Teach: Introduction to Entropy
More teaching passages in this lesson
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- Coverage describes an available learner task and scoring criteria, not student mastery. All stated task/model conditions apply.
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Supports part of this goal.
Check understanding: Can an isothermal expansion raise entropy? · Worked answer for Can an isothermal expansion raise entropy?
Required knowledge for 9.1.A
9.1.A.1 · Entropy increases when matter becomes more dispersed. For example, the phase change from solid to liquid or from liquid to gas results in a dispersal of matter as the individual particles become freer to move and generally occupy a larger volume. Similarly, for a gas, the entropy increases when there is an increase in volume (at constant temperature), and the gas molecules are able to move within a larger space. For reactions involving gas-phase reactants or products, the entropy generally increases when the total number of moles of gas-phase products is greater than the total number of moles of gas-phase reactants.
9.1.A.2 · Entropy increases when energy is dispersed. According to kinetic molecular theory (KMT), the distribution of kinetic energy among the particles of a gas broadens as the temperature increases. As a result, the entropy of the system increases with an increase in temperature.
9.2 · Absolute Entropy and Entropy Change
9.2.A · Calculate the standard entropy change for a chemical or physical process based on the absolute entropies (standard molar entropies) of the species involved in the process.
Lesson provided · Practice provided
Course and exam description p. 167
- Teach: Absolute Entropy and Entropy Change
More teaching passages in this lesson
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Supports part of this goal.
Worked example: For H2O(l) → H2O(g), suppose the stated-temperature S° values are 69.9 and 188.8 J mol−1 K−1.
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- Check understanding: At 298 K the standard molar entropies are S°(N2(g)) = 191.6, S°(H2(g)) = 130.7, and S°(NH3(g)) = 192.8 J mol−1 K−1. Calculate ΔS° for N2(g) +… · Worked answer for At 298 K the standard molar entropies are S°(N2(g)) = 191.6, S°(H2(g)) = 130.7, and S°(NH3(g)) = 192.8 J mol−1 K−1. Calculate ΔS° for N2(g) +…
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Supports part of this goal.
Check understanding: Why do coefficients appear in an entropy sum? · Worked answer for Why do coefficients appear in an entropy sum?
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Supports part of this goal.
Check understanding: Are elemental gas S° values zero at 298 K? · Worked answer for Are elemental gas S° values zero at 298 K?
Required knowledge for 9.2.A
9.2.A.1 · The entropy change for a process can be calculated from the absolute entropies of the species involved before and after the process occurs. EQN: ΔS° reaction = ΣSo products − ΣSo reactants
9.3 · Gibbs Free Energy and Thermodynamic Favorability
9.3.A · Explain whether a physical or chemical process is thermodynamically favored based on an evaluation of ΔG°.
Lesson provided · Practice provided
Course and exam description p. 168, 169
- Teach: Gibbs Free Energy and Thermodynamic Favorability
More teaching passages in this lesson
- Teaching passage: You can obtain ΔG° from standard formation free energies: ΔG°rxn = ΣνΔGf°products − ΣνΔGf°reactants. Use coefficients and phases just as in a…
- Teaching passage: The other route is ΔG° = ΔH° − TΔS°. Use kelvin and matching energy units. An endothermic process can be favored if its positive entropy term…
- Teaching passage: The four sign cases are useful: negative ΔH°/positive ΔS° is favored at all positive T in the constant-parameter model; positive ΔH°/negative ΔS°…
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Supports part of this goal.
Worked example: For a process, ΔH° = −24.0 kJ mol−1 and ΔS° = −80.0 J mol−1 K−1 = −0.0800 kJ mol−1 K−1.
- Check understanding: A hypothetical process has ΔH° = +40.0 kJ mol−1 and ΔS° = +100. J mol−1 K−1. Treat these as constant over the stated range. Calculate ΔG° at 298 K… · Worked answer for A hypothetical process has ΔH° = +40.0 kJ mol−1 and ΔS° = +100. J mol−1 K−1. Treat these as constant over the stated range. Calculate ΔG° at 298 K…
- Check understanding: At 298 K, ΔGf°(H2O(l)) = −237.1 kJ mol−1 and ΔGf°(H2O(g)) = −228.6 kJ mol−1. Use these formation values to find ΔG° for H2O(g) → H2O(l). Explain… · Worked answer for At 298 K, ΔGf°(H2O(l)) = −237.1 kJ mol−1 and ΔGf°(H2O(g)) = −228.6 kJ mol−1. Use these formation values to find ΔG° for H2O(g) → H2O(l). Explain…
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Supports part of this goal.
Check understanding: Does the superscript ° mean the temperature must be 25 °C? · Worked answer for Does the superscript ° mean the temperature must be 25 °C? -
Supports part of this goal.
Check understanding: Can positive ΔH° still give negative ΔG°? · Worked answer for Can positive ΔH° still give negative ΔG°?
Required knowledge for 9.3.A
9.3.A.1 · The Gibbs free energy change for a chemical process in which all the reactants and products are present in a standard state (as pure substances, as solutions of 1.0 M concentration, or as gases at a pressure of 1.0 atm (or 1.0 bar)) is given the symbol ΔG°.
9.3.A.2 · The standard Gibbs free energy change for a chemical or physical process is a measure of thermodynamic favorability. Historically, the term “spontaneous” has been used to describe processes for which ΔG° < 0. The phrase “thermodynamically favored” is preferred instead so that common misunderstandings (equating “spontaneous” with “suddenly” or “without cause”) can be avoided. When ΔG° < 0 for the process, it is said to be thermodynamically favored.
9.3.A.3 · The standard Gibbs free energy change for a physical or chemical process may also be determined from the standard Gibbs free energy of formation of the reactants and products. EQN: ΔG° reaction = ΣΔGf ° products − ΣΔGf ° reactants
9.3.A.4 · In some cases, it is necessary to consider both enthalpy and entropy to determine if a process will be thermodynamically favored. The freezing of water and the dissolution of sodium nitrate are examples of such phenomena.
9.3.A.5 · Knowing the values of ΔH° and ΔS° for a process at a given temperature allows ΔG° to be calculated directly. EQN: ΔG° = ΔH° − T ΔS°
9.3.A.6 · In general, the temperature conditions for a process to be thermodynamically favored (ΔG° < 0) can be predicted from the signs of ΔH° and ΔS° as shown in the table below: ΔH° < 0, ΔS° > 0, symbols < >, ΔG° < 0 favored at all T; ΔH° > 0, ΔS° < 0, symbols > <, ΔG° < 0 favored at no T; ΔH° > 0, ΔS° > 0, symbols > >, ΔG° < 0 favored at high T; ΔH° < 0, ΔS° < 0, symbols < <, ΔG° < 0 favored at low T. In cases where ΔH° < 0 and ΔS° > 0, no calculation of ΔG° is necessary to determine that the process is thermodynamically favored (ΔG° < 0). In cases where ΔH° > 0 and ΔS° < 0, no calculation of ΔG° is necessary to determine that the process is thermodynamically unfavored (ΔG° > 0).
9.4 · Thermodynamic and Kinetic Control
9.4.A · Explain, in terms of kinetics, why a thermodynamically favored reaction might not occur at a measurable rate.
Lesson provided · Practice provided
Course and exam description p. 170
- Teach: Thermodynamic and Kinetic Control
More teaching passages in this lesson
- Check understanding: A reaction has ΔG° < 0 but produces no detectable change over a class period at the stated composition, for which forward reaction is… · Worked answer for A reaction has ΔG° < 0 but produces no detectable change over a class period at the stated composition, for which forward reaction is…
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Supports part of this goal.
Check understanding: What does a catalyst change on an energy profile? · Worked answer for What does a catalyst change on an energy profile?
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Required knowledge for 9.4.A
9.4.A.1 · Many processes that are thermodynamically favored do not occur to any measurable extent, or they occur at extremely slow rates.
9.4.A.2 · Processes that are thermodynamically favored, but do not proceed at a measurable rate, are under “kinetic control.” High activation energy is a common reason for a process to be under kinetic control. The fact that a process does not proceed at a noticeable rate does not mean that the chemical system is at equilibrium. If a process is known to be thermodynamically favored, and yet does not occur at a measurable rate, it is reasonable to conclude that the process is under kinetic control.
9.5 · Free Energy and Equilibrium
9.5.A · Explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T.
Lesson provided · Practice provided
Course and exam description p. 171
- Teach: Free Energy and Equilibrium
More teaching passages in this lesson
- Check understanding: At 298 K, a reaction has ΔG° = −11.4 kJ mol−1. Calculate K using R = 8.314 J mol−1 K−1. Estimate qualitatively whether K is close to one before… · Worked answer for At 298 K, a reaction has ΔG° = −11.4 kJ mol−1. Calculate K using R = 8.314 J mol−1 K−1. Estimate qualitatively whether K is close to one before…
- Check understanding: Two reactions at the same temperature have ΔG° values −0.10RT and −10RT. Compare their equilibrium constants without inserting a numerical… · Worked answer for Two reactions at the same temperature have ΔG° values −0.10RT and −10RT. Compare their equilibrium constants without inserting a numerical…
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Supports part of this goal.
Check understanding: What does ΔG° = 0 imply for K? · Worked answer for What does ΔG° = 0 imply for K? -
Supports part of this goal.
Check understanding: At equilibrium must ΔG° be zero? · Worked answer for At equilibrium must ΔG° be zero?
Required knowledge for 9.5.A
9.5.A.1 · The phrase “thermodynamically favored” (ΔG° < 0) means that the products are favored at equilibrium (K > 1) under standard conditions.
9.5.A.2 · The equilibrium constant is related to free energy by the equations EQN: K = e−ΔG°/RT and EQN: ΔG° = −RT ln K.
9.5.A.3 · Connections between K and ΔG° can be made qualitatively through estimation. When ΔG° is near zero, the equilibrium constant will be close to 1. When ΔG° is much larger or much smaller than RT, the value of K deviates strongly from 1.
9.5.A.4 · Processes with ΔG° < 0 favor products (i.e., K > 1) and those with ΔG° > 0 favor reactants (i.e., K < 1).
9.6 · Free Energy of Dissolution
9.6.A · Explain the relationship between the solubility of a salt and changes in the enthalpy and entropy that occur in the dissolution process.
Lesson provided · Practice provided
Course and exam description p. 172
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Supports part of this goal.
Teach: Free Energy of DissolutionMore teaching passages in this lesson
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Supports part of this goal.
Teaching passage: Entropy also has competing contributions. Dispersing ions can increase accessible arrangements, but ordered hydration shells can restrict nearby… -
Supports part of this goal.
Worked example: A model salt has net ΔH°diss = +8.00 kJ mol−1 and ΔS°diss = +40.0 J mol−1 K−1 at 298 K.
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- Check understanding: Two hypothetical salts have standard dissolution values at 298 K: salt A, ΔH° = +12.0 kJ mol−1 and ΔS° = +60.0 J mol−1 K−1; salt B, ΔH° = −10.0 kJ… · Worked answer for Two hypothetical salts have standard dissolution values at 298 K: salt A, ΔH° = +12.0 kJ mol−1 and ΔS° = +60.0 J mol−1 K−1; salt B, ΔH° = −10.0 kJ…
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Supports part of this goal.
Check understanding: Why can hydration lower entropy? · Worked answer for Why can hydration lower entropy?
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Required knowledge for 9.6.A
9.6.A.1 · The free energy change (ΔG°) for dissolution of a substance reflects a number of factors: the breaking of the intermolecular interactions that hold the solid together, the reorganization of the solvent around the dissolved species, and the interaction of the dissolved species with the solvent. It is possible to estimate the sign and relative magnitude of the enthalpic and entropic contributions to each of these factors. However, making predictions for the total change in free energy of dissolution can be challenging due to the cancellations among the free energies associated with the three factors cited.
9.7 · Coupled Reactions
9.7.A · Explain the relationship between external sources of energy or coupled reactions and their ability to drive thermodynamically unfavorable processes.
Lesson provided · Practice provided
Course and exam description p. 173
- Teach: Coupled Reactions
More teaching passages in this lesson
- Teaching passage: Chemical coupling links an unfavorable target step to a sufficiently favorable step through a shared intermediate and mechanism. Add the balanced…
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Supports part of this goal.
Worked example: A target transformation X + I → Y has ΔG° = +25 kJ mol−1. A linked step Z → I has ΔG° = −40 kJ mol−1.
- Check understanding: A proposed coupled pathway has A + I → P with ΔG° = +18.0 kJ mol−1 and D → I with ΔG° = −30.0 kJ mol−1. Write the overall equation, identify the… · Worked answer for A proposed coupled pathway has A + I → P with ΔG° = +18.0 kJ mol−1 and D → I with ΔG° = −30.0 kJ mol−1. Write the overall equation, identify the…
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Supports part of this goal.
Check understanding: How can electrolysis drive an unfavored reaction? · Worked answer for How can electrolysis drive an unfavored reaction?
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Required knowledge for 9.7.A
9.7.A.1 · An external source of energy can be used to make a thermodynamically unfavorable process occur. Examples include: i. Electrical energy to drive an electrolytic cell or charge a battery. ii. Light to drive the overall conversion of carbon dioxide to glucose in photosynthesis.
9.7.A.2 · A desired product can be formed by coupling a thermodynamically unfavorable reaction that produces that product to a favorable reaction (e.g., the conversion of ATP to ADP in biological systems). In the coupled system, the individual reactions share one or more common intermediates. The sum of the individual reactions produces an overall reaction that achieves the desired outcome and has ΔG° < 0.
9.8 · Galvanic (Voltaic) and Electrolytic Cells
9.8.A · Explain the relationship between the physical components of an electrochemical cell and the overall operational principles of the cell.
Lesson provided · Practice provided
Course and exam description p. 174
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Supports part of this goal.
Teach: Galvanic (Voltaic) and Electrolytic CellsMore teaching passages in this lesson
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Supports part of this goal.
Teaching passage: For a separated Zn/Cu galvanic model, Zn → Zn2+ + 2e− supplies electrons, and Cu2+ + 2e− → Cu consumes them. A salt bridge allows ions to move and… -
Supports part of this goal.
Teaching passage: A galvanic/voltaic cell uses a favored redox reaction to deliver electrical energy. An electrolytic cell uses an external source to drive an… -
Supports part of this goal.
Worked example: Write the Zn oxidation and Cu2+ reduction equations, each involving two electrons. Add them to obtain Zn + Cu2+ → Zn2+ + Cu.
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- Check understanding: An idealized galvanic cell operates by Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Its half-cells are joined by a wire, a conducting load, and an… · Worked answer for An idealized galvanic cell operates by Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Its half-cells are joined by a wire, a conducting load, and an…
- Check understanding: A supplied electrolytic model decomposes molten MX into M metal and X2 gas, with M+ and X− as its only ions and inert electrodes. Write each… · Worked answer for A supplied electrolytic model decomposes molten MX into M metal and X2 gas, with M+ and X− as its only ions and inert electrodes. Write each…
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Supports part of this goal.
Check understanding: Do electrons travel through a salt bridge? · Worked answer for Do electrons travel through a salt bridge? -
Supports part of this goal.
Check understanding: Does electrolysis reverse the anode definition? · Worked answer for Does electrolysis reverse the anode definition?
Required knowledge for 9.8.A
9.8.A.1 · Each component of an electrochemical cell (electrodes, solutions in the half-cells, salt bridge, voltage/current measuring device) plays a specific role in the overall functioning of the cell. The operational characteristics of the cell (galvanic vs. electrolytic, direction of electron flow, reactions occurring in each half-cell, change in electrode mass, evolution of a gas at an electrode, ion flow through the salt bridge) can be described at both the macroscopic and particulate levels.
9.8.A.2 · Galvanic, sometimes called voltaic, cells involve a thermodynamically favored reaction, whereas electrolytic cells involve a thermodynamically unfavored reaction. Visual representations of galvanic and electrolytic cells are tools of analysis to identify where half-reactions occur and in what direction current flows.
9.8.A.3 · For all electrochemical cells, oxidation occurs at the anode and reduction occurs at the cathode.
9.9 · Cell Potential and Free Energy
9.9.A · Explain whether an electrochemical cell is thermodynamically favored, based on its standard cell potential and the constituent half-reactions within the cell.
Lesson provided · Practice provided
Course and exam description p. 175
- Teach: Cell Potential and Free Energy
More teaching passages in this lesson
- Teaching passage: The energy connection is ΔG° = −nFE°cell. n is the positive number of moles of electrons transferred per mole of the balanced net reaction; F ≈…
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Supports part of this goal.
Worked example: For Zn2+ + 2e− → Zn, E°red = −0.76 V, and Cu2+ + 2e− → Cu, E°red = +0.34 V, Cu is the cathode and Zn oxidation is the anode process.
- Check understanding: Use the supplied rounded standard reduction potentials: Ti2+(aq) + 2e− → Ti(s), E° = −1.63 V; Zn2+(aq) + 2e− → Zn(s), E° = −0.76 V. In an… · Worked answer for Use the supplied rounded standard reduction potentials: Ti2+(aq) + 2e− → Ti(s), E° = −1.63 V; Zn2+(aq) + 2e− → Zn(s), E° = −0.76 V. In an…
- Check understanding: For a proposed written reaction, the assigned cathode has E°red = +0.20 V and the assigned anode has E°red = +0.65 V; the balanced reaction… · Worked answer for For a proposed written reaction, the assigned cathode has E°red = +0.20 V and the assigned anode has E°red = +0.65 V; the balanced reaction…
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Supports part of this goal.
Check understanding: Why do you not multiply E° by the balancing coefficient? · Worked answer for Why do you not multiply E° by the balancing coefficient? -
Supports part of this goal.
Check understanding: Why must n come from the balanced net equation? · Worked answer for Why must n come from the balanced net equation?
Required knowledge for 9.9.A
9.9.A.1 · Electrochemistry encompasses the study of redox reactions that occur within electrochemical cells. The reactions are either thermodynamically favored (resulting in a positive voltage) or thermodynamically unfavored (resulting in a negative voltage and requiring an externally applied potential for the reaction to proceed).
9.9.A.2 · The standard cell potential of electrochemical cells can be calculated by identifying the oxidation and reduction half-reactions and their respective standard reduction potentials.
9.9.A.3 · ΔG° (standard Gibbs free energy change) is proportional to the negative of the cell potential for the redox reaction from which it is constructed. Thus, a cell with a positive E° involves a thermodynamically favored reaction, and a cell with a negative E° involves a thermodynamically unfavored reaction. EQN: ΔG° = −nFE°
9.10 · Cell Potential Under Nonstandard Conditions
9.10.A · Explain the relationship between deviations from standard cell conditions and changes in the cell potential.
Lesson provided · Practice provided
Course and exam description p. 176, 177
- Teach: Cell Potential Under Nonstandard Conditions
More teaching passages in this lesson
- Teaching passage: The qualitative Nernst form is E = E° − (RT/nF) ln Q. If Q falls below one, the logarithm is negative and E for the written direction exceeds E°.…
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Supports part of this goal.
Worked example: For Zn + Cu2+ → Zn2+ + Cu, take Q = [Zn2+]/[Cu2+]. Doubling Cu2+ while leaving Zn2+ unchanged halves Q.
- Check understanding: For the written reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), E° > 0 at the stated temperature and Q = [Zn2+]/[Cu2+] in the ideal dilute model.… · Worked answer for For the written reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), E° > 0 at the stated temperature and Q = [Zn2+]/[Cu2+] in the ideal dilute model.…
- Check understanding: A concentration cell has identical M(s)/M2+(aq) electrodes, with one side dilute and the other concentrated. Standard reduction potentials cancel,… · Worked answer for A concentration cell has identical M(s)/M2+(aq) electrodes, with one side dilute and the other concentrated. Standard reduction potentials cancel,…
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Supports part of this goal.
Check understanding: Is Q = 1 the same as equilibrium? · Worked answer for Is Q = 1 the same as equilibrium? -
Supports part of this goal.
Check understanding: In an M/M2+ concentration cell, where is oxidation favored? · Worked answer for In an M/M2+ concentration cell, where is oxidation favored?
Required knowledge for 9.10.A
9.10.A.1 · In a real system under nonstandard conditions, the cell potential will vary depending on the concentrations of the active species. The cell potential is a driving force toward equilibrium; the farther the reaction is from equilibrium, the greater the magnitude of the cell potential.
9.10.A.2 · Equilibrium arguments such as Le Châtelier’s principle do not apply to electrochemical systems, because the systems are not in equilibrium.
9.10.A.3 · The standard cell potential E° corresponds to the standard conditions of Q = 1. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when Q = K). Deviations from standard conditions that take the cell further from equilibrium than Q = 1 will increase the magnitude of the cell potential relative to E°. Deviations from standard conditions that take the cell closer to equilibrium than Q = 1 will decrease the magnitude of the cell potential relative to E°. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.
9.10.A.4 · Algorithmic calculations using the Nernst equation are insufficient to demonstrate an understanding of electrochemical cells under nonstandard conditions. However, students should qualitatively understand the effects of concentration on cell potential and use conceptual reasoning, including the qualitative use of the Nernst equation: EQN: E = E° − (RT/nF) ln Q to solve problems.
9.11 · Electrolysis and Faraday’s Law
9.11.A · Calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.
Lesson provided · Practice provided
Course and exam description p. 178
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Supports part of this goal.
Teach: Electrolysis and Faraday’s LawMore teaching passages in this lesson
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Teaching passage: Next use the balanced electrode half-reaction. Mz+ + z e− → M deposits one mole of metal per z moles of electrons, so nM = q/(zF) and mass = nM ×… -
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Teaching passage: A theoretical result usually assumes 100% current efficiency, a known current/time record, no competing electrode reactions, enough reactant, and… -
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Worked example: A model Zn anode loses 0.654 g by Zn → Zn2+ + 2e−. With MZn = 65.4 g mol−1, nZn = 0.654/65.4 = 0.0100 mol.
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- Check understanding: In a supplied-data copper electroplating model, Cu2+ + 2e− → Cu occurs with a constant current of 1.50 A for 20.0 min. Calculate charge, moles of… · Worked answer for In a supplied-data copper electroplating model, Cu2+ + 2e− → Cu occurs with a constant current of 1.50 A for 20.0 min. Calculate charge, moles of…
- Check understanding: An idealized M3+ + 3e− → M deposition transfers enough charge to deposit 0.0100 mol metal. Calculate the required charge and time at constant 2.00… · Worked answer for An idealized M3+ + 3e− → M deposition transfers enough charge to deposit 0.0100 mol metal. Calculate the required charge and time at constant 2.00…
More checks for this goal
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Supports part of this goal.
Check understanding: What changes if the current varies with time? · Worked answer for What changes if the current varies with time? -
Supports part of this goal.
Check understanding: Why divide electron moles by two for Cu2+ plating? · Worked answer for Why divide electron moles by two for Cu2+ plating?
Required knowledge for 9.11.A
9.11.A.1 · Faraday’s laws can be used to determine the stoichiometry of the redox reaction occurring in an electrochemical cell with respect to the following: i. Number of electrons transferred ii. Mass of material deposited on or removed from an electrode (as in electroplating) iii. Current iv. Time elapsed v. Charge of ionic species EQN: I = q/t
About this outline
AP Chemistry · Fall 2024. Official source: Course and exam description (College Board). The current outline uses the accepted teaching and practice mappings for this edition. Individual resources retain their own scope.
Activities and teaching plans support classroom preparation. They do not establish students’ mastery or actual classroom laboratory time.
Detailed mappings and prior review records remain available at their existing edition-specific addresses.