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AP Unit 6 Solutions

Worked solutions to all 16 Unit 6 problems. These answers and reasons were added for the AP edition. Check the setup, units, assumptions and explanation alongside the final result.

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Problem 1

A supplied dissolution experiment in an insulated cup lowers the solution temperature. Identify energy-flow direction for the dissolution and for the solution’s thermal change. Explain the net interaction balance and why a separate melting experiment can absorb heat without a temperature rise.

  1. Cooling means qsolution <0, so the dissolution process absorbs heat: qdissolution >0.
  2. Solute separation and solvent reorganization require more energy than solvation formation releases.
  3. During fixed-pressure melting of a pure substance, absorbed energy changes particle interactions/phase rather than raising average kinetic energy.

Final answer

Endothermic dissolution; solution loses thermal energy; a melting plateau still permits energy absorption.

What earns credit
  • Credit boundary-specific signs, competing interactions and plateau explanation.

Watch out: Assigning endothermic/exothermic from thermometer direction without a boundary.

Return to problem 1 → · Review Topic 6.1

Problem 2

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 both axes, draw and label the initial and final horizontal levels, and add a vertical arrow showing the forward enthalpy change. Give ΔH for the forward and reverse processes. Explain the effect of adding 100 kJ mol−1 to both reference levels.

Worked enthalpy diagramVertical axis enthalpy in kilojoules per mole, horizontal axis transformation progress. Initial level 8.0, final level 30.0. An upward arrow between the levels shows forward delta H positive 22.0 kilojoules per mole.Enthalpy (kJ mol⁻¹)Transformation progressInitial: 8.0Final: 30.0ΔH = +22.0
  1. Place final level above initial and label vertical axis enthalpy (kJ mol−1); horizontal direction is transformation progress.
  2. Forward ΔH =30.0−8.0 =+22.0; reverse ΔH =−22.0 kJ mol−1.
  3. Both shifted levels retain the same difference, so the enthalpy changes are unchanged.

Final answer

Forward +22.0 and reverse −22.0 kJ mol−1; reference shift has no effect.

What earns credit
  • Award separate representation credit for enthalpy axis with units, transformation-progress axis, labeled initial/final levels in the correct relative positions, and upward forward ΔH arrow. Award calculation credit for +22.0/−22.0 kJ mol⁻¹ and explanation that a common reference shift leaves the difference unchanged.

Watch out: Using final level alone as ΔH.

Return to problem 2 → · Review Topic 6.2

Problem 3

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 phase/reaction change. Explain net particle-level energy transfer and find Tf.

  1. Initially the warmer body has greater average thermal kinetic energy, so interactions give net transfer toward the cooler body.
  2. 30.0(Tf−70.0)+90.0(Tf−10.0) =0.
  3. Tf =(30×70+90×10)/120 =25.0 °C. At this temperature microscopic exchange continues without net heat flow.

Final answer

Tf =25.0 °C.

What earns credit
  • Credit temperature-directed transfer, weighted balance and equilibrium meaning.

Watch out: Taking the midpoint 40.0 °C.

Return to problem 3 → · Review Topic 6.3

Problem 4

Read the initial and corrected-final temperatures from these two thermometer scales. Record each value with its unit and a sensible estimated digit. The smallest marked division is1 °C; estimate to0.1 °C, and treat each recorded reading as having a maximum uncertainty±0.1 °C.

Initial thermometer scaleThe scale is marked every one degree Celsius from20 to30. The top of the fluid lies between22 and23, approximately three tenths of that interval above22.InitialTemperature(°C)2021222324252627282930Smallest marked interval:1 °C
Corrected final thermometer scaleThe scale is marked every one degree Celsius from20 to30. The top of the fluid lies between27 and28, approximately nine tenths of that interval above27.Corrected finalTemperature(°C)2021222324252627282930Smallest marked interval:1 °C

Supplied dissolution data:80.0 g final solution, c=4.00 J g⁻¹ °C⁻¹, Ccup=15.0 J °C⁻¹, and0.0300 mol dissolved solute. Assume no remaining external heat loss. Use your recorded readings to calculate qdissolution and molar ΔH. Explain the bias from neglecting the cup, and give the largest ΔT uncertainty from the two reading uncertainties. These are supplied instruments/data; no experiment is required.

The scales give initial22.3 °C and corrected final27.9 °C. Record the estimated tenths digit because the markings are1 °C apart. These reading estimates are distinct from the stated±0.1 °C uncertainty bound.

  1. ΔT =5.6 °C. qsolution =80.0×4.00×5.6 =1792 J; qcup =15.0×5.6 =84 J.
  2. qdissolution =−1876 J ≈−1.9 kJ; ΔH =−1.876/0.0300 ≈−63 kJ mol−1 using the two significant figures of ΔT.
  3. Ignoring cup gives too small a released-heat magnitude. Worst-case ΔT uncertainty is 0.1+0.1 =0.2 °C because opposite reading errors can add.

Final answer

q ≈−1.9 kJ; ΔH ≈−63 kJ mol−1; ΔT uncertainty up to ±0.2 °C.

What earns credit
  • Award explicit reading credit for22.3 °C and27.9 °C, each with°C and the estimated tenths digit. Credit explaining one estimated digit beyond the1 °C scale divisions.
  • Credit reading difference, both heat-recipient terms, negative dissolution sign, reported precision and cup-neglect bias.
  • Credit worst-case uncertainty±0.2 °C from opposite reading errors; do not equate scale division with the stated uncertainty automatically.

Watch out: Using only solute mass or reporting excessive precision.

Return to problem 4 → · Review Topic 6.4

Problem 5

An ideal supplied thermal experiment gives 250 J to each of two 10.0 g nonreacting samples. Sample A has c =0.500 J g−1 °C−1; sample B has c =2.00 J g−1 °C−1. Find their temperature rises. If A’s molar mass is 40.0 g mol−1, find its molar heat capacity and show that the molar equation gives the same rise.

  1. A: ΔT =250/(10.0×0.500) =50.0 °C. B: 250/(10.0×2.00) =12.5 °C.
  2. Cm,A =0.500×40.0 =20.0 J mol−1 °C−1; n =10.0/40.0 =0.250 mol.
  3. ΔT =250/(0.250×20.0) =50.0 °C. Equal q does not imply equal ΔT. Each sample gains energy.

Final answer

ΔTA =50.0 °C; ΔTB =12.5 °C; Cm,A =20.0 J mol−1 °C−1.

What earns credit
  • Credit both rises, capacity conversion/units and equivalent molar calculation.

Watch out: Multiplying by molar mass twice.

Return to problem 5 → · Review Topic 6.4

Problem 6

A supplied exothermic calorimetry trial loses heat to room air during the temperature reading. A student uses the solution’s uncorrected rise alone to infer ΔH. Identify the missing heat-transfer term, predict the bias, and describe a method using an existing temperature-time record that can improve the estimate. Explain why q =mcΔT would not by itself account for any simultaneous melting.

  1. The heat balance must include qroom >0 as a recipient, along with solution/cup heat.
  2. Ignoring external loss gives too small a magnitude of qreaction, so calculated ΔH is less negative.
  3. Use pre/post-transfer temperature trends and extrapolation toward the mixing time, with stated heat-loss-model assumptions, to estimate the rise before loss.
  4. A melting segment absorbs phase enthalpy nΔHfus even at constant temperature; add that separate term.

Final answer

Missing external heat sink biases ΔH toward zero; extrapolation can estimate a corrected rise; include phase energy separately.

What earns credit
  • Credit heat sink, causal sign, an aligned correction method and distinct phase term.

Watch out: Saying all heat lost is visible in the solution temperature rise.

Return to problem 6 → · Review Topic 6.4

Problem 7

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 =4.18 J g−1 °C−1, molar mass 18.0 g mol−1, and ΔHfus =6.01 kJ mol−1. Calculate total heat and explain the plateau at the particle level.

  1. Warm ice: 36.0×2.10×10.0 =756 J =0.756 kJ.
  2. Melt: (36.0/18.0)×6.01 =12.02 kJ. Warm liquid: 36.0×4.18×20.0 =3009.6 J =3.01 kJ.
  3. Total =15.8 kJ. Plateau energy changes interactions/phase fractions with intact H2O molecules, not O–H bond cleavage.

Final answer

Total heat ≈15.8 kJ.

What earns credit
  • Credit three separate terms, unit conversion and particle account.

Watch out: Applying water c to the ice segment or dropping fusion.

Return to problem 7 → · Review Topic 6.5

Problem 8

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 cooling. Find q for the sample, compare it with vaporizing 0.500 mol, and state whether temperature changes during the ideal transition.

  1. Condensation reverses vaporization, so q =0.250(−30.0) =−7.50 kJ.
  2. Vaporizing twice as many moles absorbs +15.0 kJ.
  3. During pure-substance two-phase equilibrium at fixed pressure, temperature stays constant.

Final answer

Condensation −7.50 kJ; vaporization +15.0 kJ; ideal plateau temperature fixed.

What earns credit
  • Credit sign reversal, amount scaling and conditions.

Watch out: Changing sign without retaining the same per-mole magnitude.

Return to problem 8 → · Review Topic 6.5

Problem 9

For 2CO(g) + O2(g) → 2CO2(g), ΔH = −566.0 kJ. A supplied sample contains 0.300 mol CO and 0.100 mol O2.

  1. Find the heat as the reaction proceeds to the limiting-reagent maximum. Explain warming if the released energy is transferred to a surrounding liquid.
  2. In a separate supplied post-reaction observation, reaction is complete: the products are at 40.0 °C and a surrounding liquid is at 20.0 °C. They can exchange heat through a conductive wall. The combined assembly is insulated from the room; no further chemical reaction, phase change or work occurs, and both parts have positive heat capacities. Sketch or describe how both temperatures change over time, the direction of heat transfer, and the condition reached at the end. Is the final common temperature necessarily 20.0 °C?
  3. Would reaching that final temperature establish chemical equilibrium between CO, O2 and CO2? Explain.
  1. CO supports reaction extent 0.300/2 = 0.150 mol; oxygen supports 0.100/1 = 0.100 mol. Oxygen limits.
  2. q = 0.100(−566.0) = −56.6 kJ. Only 0.200 mol CO is used, so 0.100 mol CO remains. Net bond formation lowers chemical potential energy; energy transferred to a surrounding liquid increases its thermal motion and temperature.
  3. In the separate post-reaction observation, energy transfers as heat from hotter products to cooler liquid. The products cool from 40.0 °C and the liquid warms from 20.0 °C. Under the stated insulated, heat-only conditions, their curves approach one common temperature between those initial temperatures. The exact temperature needs their heat capacities; it is not necessarily 20.0 °C because the finite liquid also warms.
  4. Once their temperatures are equal, there is no net heat transfer between them. Thermal equilibrium describes this temperature equality; it does not require that the earlier chemical energy release be reversed.
  5. Thermal equilibrium does not by itself establish reaction equilibrium. Chemical equilibrium would require equal forward and reverse reaction rates and an appropriate equilibrium composition. A common temperature alone supplies neither piece of evidence.

Final answer

qreaction = −56.6 kJ; 0.100 mol CO remains. After the separate hot-products/cool-liquid observation, heat flows outward from the products until both reach the same temperature between 20.0 and 40.0 °C. That is thermal equilibrium, not proof of chemical equilibrium.

What earns credit
  • Find the limiting reaction extent, scale ΔH with coefficients, and explain surrounding-liquid warming.
  • Describe products cooling and finite surroundings warming, with heat moving from hotter to cooler material.
  • Give a common final temperature between the two initial values and no net heat flow; do not assume an infinite 20.0 °C reservoir.
  • Distinguish temperature equality from equal forward/reverse reaction rates.

Watch out: Using all initial CO despite insufficient O₂.

Return to problem 9 → · Review Topic 6.6

Problem 10

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 equation B → 2A.

  1. B coefficient is one, so extent is 0.250 mol.
  2. q =0.250×80.0 =+20.0 kJ.
  3. Reversing switches sign without changing magnitude: −80.0 kJ for B →2A.

Final answer

q =+20.0 kJ; reversed ΔH =−80.0 kJ.

What earns credit
  • Credit mole basis and reverse sign.

Watch out: Dividing by two merely because A’s coefficient is two.

Return to problem 10 → · Review Topic 6.6

Problem 11

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 436, C–C 347, C–H 413 kJ mol−1. Show cancellation of unchanged bonds and discuss the estimate’s limitation.

  1. Four C–H bonds appear in both inventories and cancel. Remaining breaking cost: 614+436 =1050 kJ.
  2. Remaining formation release: 347+2×413 =1173 kJ.
  3. ΔH ≈1050−1173 =−123 kJ per balanced reaction. Average bond values depend on environment, so the prediction is approximate.

Final answer

ΔH ≈−123 kJ.

What earns credit
  • Credit inventories/cancellation, sign and average-value limitation.

Watch out: Counting the C=C double bond as two tabulated C–C single bonds.

Return to problem 11 → · Review Topic 6.7

Problem 12

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 formation release.

  1. Break one N≡N and three H–H bonds: 945+3×436 =2253 kJ input.
  2. Two NH3 contain six N–H bonds: formation releases 6×391 =2346 kJ.
  3. Net ΔH ≈2253−2346 =−93 kJ per equation as written.

Final answer

Breaking +2253 kJ; formation −2346 kJ; net ≈−93 kJ.

What earns credit
  • Credit coefficients in bond counts and separate signs.

Watch out: Counting three N–H bonds despite forming two ammonia molecules.

Return to problem 12 → · Review Topic 6.7

Problem 13

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°. Recalculate if H2O(g), −241.8 kJ mol−1, is the product and explain the difference.

  1. Liquid-water product sum =−393.5+2(−285.8) =−965.1; reactant sum =−74.8. ΔH° =−890.3 kJ.
  2. Vapor-water product sum =−393.5+2(−241.8) =−877.1; subtract −74.8 to get −802.3 kJ.
  3. Producing vapor releases 88.0 kJ less per reaction because the two moles of water retain 2×44.0 kJ of phase enthalpy compared with liquid.

Final answer

Liquid products: −890.3 kJ; vapor products: −802.3 kJ.

What earns credit
  • Credit signed weighted sums and the two-mole phase difference.

Watch out: Ignoring the product phase or forgetting the water coefficient.

Return to problem 13 → · Review Topic 6.8

Problem 14

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 data for O(g) cannot be replaced by zero when computing another reaction.

  1. Use 2C(graphite,s) +3H2(g) +½O2(g) → C2H5OH(l).
  2. Count two C, six H and one O on each side; the product coefficient is one by definition.
  3. Zero applies to the reference elemental form O2(g), not isolated O(g), whose formation requires bond dissociation.

Final answer

2C(graphite,s) +3H2(g) +½O2(g) → C2H5OH(l).

What earns credit
  • Credit reference states, one-mole product and O/O₂ distinction.

Watch out: Assigning zero to any species containing only one element.

Return to problem 14 → · Review Topic 6.8

Problem 15

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. Show operations and cancellation, then justify addition of enthalpies.

  1. Double the first two equations; their heats become −787.0 and −571.6 kJ. Reverse the third; its heat becomes +1411.0 kJ.
  2. Adding cancels 2CO2, 2H2O(l) and 3O2, leaving the target.
  3. ΔH =−787.0−571.6+1411.0 =+52.4 kJ. Conservation makes the same-endpoint constant-pressure heat equal to the sum of step heats.

Final answer

ΔH =+52.4 kJ for the target as written.

What earns credit
  • Credit each scaling/reversal, actual cancellation and energy-conservation reason.

Watch out: Adding original equations without reversing ethene combustion.

Return to problem 15 → · Review Topic 6.9

Problem 16

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 ΔH for 0.300 mol A converted to B(l). Explain why stopping at B(g) would answer a different question and whether these two steps prove the actual mechanism.

  1. For the specified liquid endpoint, add the two steps: +25.0−40.0 =−15.0 kJ mol−1.
  2. For 0.300 mol, q =0.300(−15.0) =−4.50 kJ.
  3. B(g) and B(l) differ in phase enthalpy. The accounting path establishes the endpoint energy change, not a unique kinetic sequence.

Final answer

ΔH =−15.0 kJ mol−1; q =−4.50 kJ.

What earns credit
  • Credit phase step, amount scaling and mechanism limitation.

Watch out: Ignoring condensation because B has the same chemical formula in both phases.

Return to problem 16 → · Review Topic 6.9

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