AP Unit 9 Solutions

1. Melting usually increases accessible particle arrangements and freedom of motion, so ΔSsystem > 0. Isothermal gas expansion gives particles more spatial arrangements, so ΔSsystem > 0 even without a temperature increase.
2. Ammonia formation decreases total gaseous moles from four to two for the equation as written, so its entropy change is expected to be negative.
3. Heating increases the range of accessible particle energies and broadens the energy distribution, increasing entropy.
4. Thermodynamic favorability also involves the surroundings; the enthalpy contribution to free energy accounts for heat transfer at fixed T and P. A system entropy decrease can be outweighed by a surroundings entropy increase.

Magnitude comparison

1. B has the same doubling of available volume per particle as A, but twice the amount: ΔSB = 2.00 × 5.76 = +11.5 J K−1 to three significant figures. A has +5.76 J K−1; C has +11.5 J K−1. Thus |ΔSA| < |ΔSB| ≈ |ΔSC|.
2. C offers each particle four times its initial spatial volume instead of twice, giving a larger increase in accessible arrangements than A. B doubles the number of particles undergoing the same per-particle change, so the extensive entropy change doubles. The equal B/C values follow from the supplied ideal-gas data; they are not a universal volume-only rule.
3. A positive sign does not specify a magnitude. To compare unrelated processes, specify amounts, states, temperatures and enough entropy data or a justified quantitative model.

1. Multiply each standard molar entropy by its equation coefficient: products = 2(192.8) = 385.6; reactants = 191.6 + 3(130.7) = 583.7.
2. ΔS° = 385.6 − 583.7 = −198.1 J K−1 per mole of reaction as written.
3. Elements in their reference forms have zero standard formation enthalpy/free energy by convention, but not zero standard molar entropy at 298 K. A perfect crystal’s entropy approaches zero at 0 K, which is a different statement.
4. Four gaseous moles become two, reducing accessible arrangements and supporting the negative sign.

1. Convert entropy to matching units: +0.100 kJ mol−1 K−1. At 298 K, ΔG° = 40.0 − 298(0.100) = +10.2 kJ mol−1; the forward standard-state process is unfavored.
2. At 600 K, ΔG° = 40.0 − 600(0.100) = −20.0 kJ mol−1; it is favored.
3. Set ΔG° = 0: T = ΔH°/ΔS° = 400 K. It is favored for T > 400 K under the constant-parameter approximation. At the boundary there is no standard-state directional driving force.
4. ΔH° < 0 and ΔS° > 0 gives negative ΔG° at every positive T in this model. The activation barrier and pathway still determine speed; favorability is an energy comparison.

Reference state and all sign cases

1. The AP solution reference is 1.0 M; 0.100 M is not that reference concentration. More precisely the reference is unit activity, with concentration used in the ideal approximation. Temperature is independently specified, so ° does not require 298 K. The actual driving force also depends on composition, through Q; it cannot be obtained from ΔG° alone for this mixture.
2. For ΔH° < 0, ΔS° > 0, both terms make ΔG° negative: favored at all positive T within the model. For ΔH° > 0, ΔS° < 0, both terms are positive: favored at no positive T. For both positive, −TΔS° wins above the boundary; for both negative, the positive −TΔS° penalty wins at high T and the process is favored below the boundary.
3. For the second process, ΔG° = −40.0 + 0.100T in kJ mol−1, so the boundary is 400 K. It is favored for 0 < T < 400 K within the stated constant-H/S, unchanged-phase approximation. At 400 K the standard-state comparison has zero directional driving force; this is not a rate statement.
4. For the new reference-state comparison, a solid or liquid reference is the pure substance in its stated stable phase at the specified temperature and reference pressure (unit activity). The AP gas-pressure approximation uses 1 atm; modern thermodynamic tabulations often use 1 bar, so use the convention associated with the supplied data consistently. Purity defines the condensed-phase reference, whereas the gas reference specifies pressure. The ° symbol alone does not set 298 K or guarantee that a chosen phase is stable at all temperatures.

1. Formation-value sums give ΔG° = (−237.1) − (−228.6) = −8.5 kJ mol−1. Both species have coefficient one but different phases.
2. Negative ΔG° favors condensation for the written standard-state comparison at 298 K: pure liquid and water vapor at its standard pressure.
3. Actual vapor in air may have a far lower water partial pressure. Its composition/pressure changes the actual free-energy driving force; condensation also has a nucleation/pathway issue.
4. The standard-state result is a reference comparison, not a universal statement about every nonstandard humidity and rate.

1. Draw or describe a reactant-to-product free-energy profile with products below reactants and a substantial activation barrier separating them. A low product energy does not remove the barrier.
2. Few particles can reach the necessary transition configuration, so the net reaction can be too slow to measure during the observation window. This is kinetic control.
3. At equilibrium the forward and reverse rates balance and the driving force vanishes. A stalled but thermodynamically favored mixture need not be at equilibrium.
4. A catalyst provides an alternative lower-barrier pathway for both directions; it changes speed, not the reactant/product free-energy difference or fixed-temperature K.

1. RT = 8.314 × 298 ≈ 2.48 kJ mol−1. −11.4 kJ is several RT units below zero, so K should be appreciably greater than one, not close to one.
2. K = exp(−ΔG°/RT) = exp(11400/(8.314 × 298)) ≈ 99.6, or about 1.00 × 10² at the precision of the data.
3. For K = 0.250, ΔG° = −RT ln K = +3434.64 J mol−1 ≈ +3.43 kJ mol−1. Reactants are favored under the standard comparison.
4. ΔG° = 0 implies K = 1. An actual mixture is at equilibrium only when its current Q equals K; an arbitrary Q need not be one.

1. K = exp(−ΔG°/RT). The first gives e^0.10 ≈ 1.11, close to one; the second gives e^10 ≈ 2.20 × 10^4, strongly product-favored.
2. A negative sign determines K > 1, but the magnitude relative to RT determines how far K lies from one. Sign alone does not establish similar composition preference.
3. Reversing the second equation changes ΔG° to +10RT, giving K = e^−10 ≈ 4.54 × 10^−5, the reciprocal of its original value.

1. Use matching units: ΔG°A = 12.0 − 298(0.0600) = −5.88 kJ mol−1; ΔG°B = −10.0 − 298(−0.0600) = +7.88 kJ mol−1.
2. A is favored in the standard comparison because its entropy contribution outweighs the enthalpy cost. B is unfavored in that comparison despite being exothermic.
3. Separating the solid disperses its particles, but hydration reorganizes nearby water. Strong ordering of solvent can oppose or outweigh that entropy gain; interaction and solvent-reorganization contributions must be included.
4. At saturated dissolution equilibrium the actual driving force is zero, because Qsp = Ksp. Positive standard ΔG° does not mean literally no salt ever dissolves in initially ion-free water.

Component reasoning

1. Lattice separation has ΔH = +800. kJ mol−1 and ΔS = +100. J mol−1 K−1. Solvent separation has +15.0 and +20.0 in the corresponding units for this specified path. Hydration has −803 and −60.0. Breaking attractions costs enthalpy; forming them releases it. The assigned entropy signs follow the stipulated accessible-arrangement changes. The two lattice/hydration enthalpies dominate and nearly cancel. The positive entropy contributions total +120.; the hydration restriction contributes −60.0, leaving +60.0 J mol−1 K−1.
2. Using entropy in kJ mol−1 K−1, the component free energies are 800. − 298(0.100) = +770.2; 15.0 − 298(0.0200) = +9.04; and −803 − 298(−0.0600) = −785.12 kJ mol−1. Their sum is −5.88 kJ mol−1. The summed enthalpy is +12.0 and summed entropy +60.0, consistent with the first calculation. Extra intermediate digits are retained for the cancellation; the estimates do not justify extra final precision.
3. With hydration ΔH = −795 kJ mol−1, net ΔH becomes +20.0 kJ mol−1, while ΔS remains +60.0 J mol−1 K−1. ΔG° = 20.0 − 298(0.0600) = +2.12 kJ mol−1: the predicted sign reverses. The change is small compared with the large component terms, but large compared with their remainder. The path estimates and assumptions need validation, and actual equilibrium solubility also depends on activities and composition.

For the coarse component estimates, report the net predictions as approximately −6 and +2 kJ mol−1; −5.88 and +2.12 above are retained arithmetic values, not extra experimental precision. The earlier calculation using the separately supplied net dissolution data has its own stated precision.

1. Add the equations and cancel I: A + D → P. I is the shared intermediate connecting the steps.
2. ΔG°overall = +18.0 − 30.0 = −12.0 kJ mol−1, so the overall standard-state process is favored.
3. A mechanism that actually links these steps can route the favorable D conversion into P formation. Merely adding two unrelated energy numbers does not establish physical coupling; disconnected containers do not transmit that driving force to the first reaction.
4. An applied electrical source can drive an electrolytic process, and light can drive photochemical conversion such as the overall photosynthetic storage of energy. The source supplies energy while the full system obeys conservation.

1. Anode oxidation: Zn → Zn2+ + 2e−. Cathode reduction: Cu2+ + 2e− → Cu. Electrons travel through the external conductor from Zn to Cu; zinc mass falls and copper mass rises.
2. Zn2+ production tends to increase positive charge in the anode solution, so bridge anions migrate toward that half-cell. Consumption of Cu2+ at the cathode requires compensating positive-ion migration toward its solution.
3. A salt bridge closes the ionic part of the circuit and limits charge buildup without intentionally supplying reacting ions. Removing it prevents sustained macroscopic current as charge imbalance develops.
4. Ions transport charge in the liquid; electrons transport charge in the metal wire. A correct labeled drawing distinguishes both paths and conserves atoms and charge.

Device roles

The voltmeter measures the potential difference between the electrode terminals and connects across them (in parallel). Its high resistance minimizes current draw; a nonzero near-open-circuit voltage can be measured with negligible current. The ammeter measures charge flow per time in an external branch and is inserted in series with the load, not placed directly across the cell as a low-resistance short. The galvanic chemical reaction supplies the driving energy. A load and complete external/ionic paths permit sustained operation; the meters measure quantities rather than act as an electrolytic power source.

1. Cathode reduction: M+ + e− → M. Anode oxidation: 2X− → X2 + 2e−. Balance electrons by doubling the cathode half-reaction; overall 2M+ + 2X− → 2M + X2.
2. Metal is deposited at the cathode and gas evolves at the anode. The applied source removes electrons from the anode and supplies them to the cathode, driving the otherwise unfavorable written chemical process.
3. The definitions are universal: oxidation at anode, reduction at cathode. A galvanic cell supplies electrical energy from a favored reaction; an electrolytic cell consumes externally supplied electrical energy.
4. Aqueous electrolysis introduces water and possible competing half-reactions, so products need additional information. This is an abstract molten model and no procedure is being prescribed.

1. Zn2+/Zn has the higher reduction potential, so Zn2+ is reduced at the cathode. Ti metal is oxidized at the anode. Write Ti → Ti2+ + 2e−, not the reduction equation for both electrodes.
2. Both half-reactions transfer two electrons, so they cancel directly: Ti(s) + Zn2+(aq) → Ti2+(aq) + Zn(s). n = 2 for this written reaction.
3. E°cell = E°red,cathode − E°red,anode = −0.76 − (−1.63) = +0.87 V. Do not multiply either tabulated voltage by its coefficient.
4. ΔG° = −nFE° = −2(96485)(0.87) = −167884 J ≈ −1.7 × 10² kJ per mole of the balanced reaction, limited by the voltage precision. Doubling gives n = 4 and doubles ΔG°, while E° remains 0.87 V.

1. E°written = +0.20 − (+0.65) = −0.45 V. ΔG°written = −2(96485)(−0.45) = +86836.5 J ≈ +87 kJ per mole of the written reaction.
2. The negative E° and positive ΔG° mean the assigned direction is not favored in the standard-state galvanic comparison.
3. Reversing swaps oxidation/reduction: E°reverse = +0.45 V and ΔG°reverse ≈ −87 kJ for the correspondingly reversed equation.
4. An electrical source must oppose the natural direction and supply sufficient driving potential; real cells may require additional voltage for resistance and kinetic losses. The standard sign comparison alone does not quantify those losses.

1. Q = 0.10 gives ln Q < 0, so subtracting the negative term makes E > E°. Q = 10 gives ln Q > 0, so E < E°; this comparison alone does not assert E is negative.
2. Forward reaction produces Zn2+ and consumes Cu2+, so Q rises toward K. The forward driving potential decreases and reaches zero at Q = K.
3. E° corresponds to Q = 1, not to equilibrium unless K = 1. A functioning cell is generally away from equilibrium; its free-energy difference supplies the electrical driving force.
4. Le Châtelier describes response of a system initially at equilibrium. For an operating cell use the actual Q, its relation to K, and the free-energy/Nernst driving-force relation rather than pretending the original cell is at equilibrium.

1. Oxidation M → M2+ + 2e− occurs on the dilute side, increasing its dissolved-ion concentration. Reduction M2+ + 2e− → M occurs on the concentrated side, decreasing its dissolved-ion concentration.
2. Electrons travel externally from the dilute-side anode to the concentrated-side cathode. This chemical direction reduces the concentration imbalance.
3. For the forward net transfer, Q is effectively concentration(dilute)/concentration(concentrated), less than one initially. With E° = 0, the negative ln Q contribution gives positive E.
4. As activities approach equality, Q approaches one and E approaches zero. The initial nonstandard concentration difference, not a difference in standard electrode identities, supplies the driving force.

1. Convert time: 20.0 min × 60 s min−1 = 1200 s. q = It = 1.50 C s−1 × 1200 s = 1800 C.
2. n(e−) = q/F = 1800/96485 = 0.0187 mol. The half-reaction needs two electrons per copper atom, so nCu = n(e−)/2 = 0.00933 mol.
3. Theoretical mass = (1800/96485)(63.55/2) = 0.593 g, assuming all charge produces adherent Cu.
4. Efficiency = measured/theoretical × 100 = 0.500/0.592786 × 100 = 84.3%. Competing reductions can consume charge, or deposited material can fail to remain on the measured electrode. State the chosen mechanism rather than inferring efficiency above 100%.

1. M needs three electrons per atom: n(e−) = 3 × 0.0100 = 0.0300 mol; q = 0.0300 × 96485 = 2894.55 C ≈ 2.89 × 10³ C.
2. At constant 2.00 C s−1, t = q/I = 1447.28 s ≈ 1.45 × 10³ s or 24.1 min.
3. The same 0.0300 mol electrons deposits 0.0300/2 = 0.0150 mol N. Charge fixes electron amount, and the half-reaction fixes metal/electron mole ratio.
4. Mass additionally depends on each metal’s molar mass. Without those values, equal or ordered deposited masses cannot be concluded just from the charge.