The problems are on the thermochemistry worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Thermochemistry measures heat changes in reactions. These worked answers correct the 20 problems of the AP Chemistry worksheet on thermochemistry, from exothermic or endothermic to reading a temperature-time graph: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result in bold. Pay attention to the sign of ΔH and to the number of significant figures, which follows the data given. Some estimates, such as those from bond enthalpies, are expected to differ from tabulated values.
Level: AP Chemistry · Curriculum: AP Chem Unit 6
Problem 1 Exothermic or endothermic
1. Exothermic (ΔH < 0).2. Endothermic (absorbs heat from the surroundings).
3. Endothermic (ΔH > 0).
4. Exothermic (heat released).
Problem 2 Heat absorbed by water
1. q = 250 × 4.18 × 15.0 = 1.57 × 10⁴ J (15 675 J).2. 15.7 kJ.
Problem 3 Specific heat of copper
1. ΔT = 45.0 − 20.0 = 25.0 °C.2. c = q/(mΔT) = 962/(100.0 × 25.0) = 0.385 J/(g °C).
Problem 4 Heat from burning methane
1. ΔH is negative (−890 kJ/mol).2. n = 4.00/16.04 = 0.249 mol; heat = 0.249 × 890 = 222 kJ released.
Problem 5 Adding two steps
1. A → B: +40 kJ; B → C: −95 kJ.2. A → C: 40 + (−95) = −55 kJ.
3. C → A: +55 kJ.
Problem 6 Reversing and scaling
1. −180 kJ.2. +90 kJ (halved).
3. +540 kJ (equation ×3).
Problem 7 Writing formation equations
1. C(s) + O₂(g) → CO₂(g).2. 2 C(s) + 3 H₂(g) + ½O₂(g) → C₂H₅OH(l).
3. Cl₂(g) is the element in its standard state, so no change is needed to form it.
Problem 8 Neutralisation in a cup
1. q = 100.0 × 4.18 × 6.5 = 2.72 × 10³ J.2. n = 0.0500 L × 1.00 = 0.0500 mol.
3. ΔH = −2.72 kJ/0.0500 mol = −54 kJ/mol.
Problem 9 A cold dissolving
1. q = 108.0 × 4.18 × (−5.6) = −2528 J = −2.53 kJ for the solution.2. n = 8.00/80.04 = 0.09995 mol; ΔH = +2.53/0.09995 = +25 kJ/mol (endothermic).
3. The dissolving needs energy to separate the ions, which it takes from the water, so the solution cools.
Problem 10 Hess’s law with combustion data
1. C₂H₄(g) + H₂(g) → C₂H₆(g).2. ΔH = ΔHc(C₂H₄) + ΔHc(H₂) − ΔHc(C₂H₆) = −1411 − 286 + 1560 = −137 kJ.
Problem 11 Formation enthalpies
1. 0 (element in standard state).2. ΔH° = 3(−393.5) − [−824.2 + 3(−110.5)] = −1180.5 + 1155.7 = −24.8 kJ.
3. Yes, exothermic.
Problem 12 Bond enthalpies for HCl
1. 436 + 243 = 679 kJ.2. 2 × 432 = 864 kJ.
3. ΔH = 679 − 864 = −185 kJ.
Problem 13 Spirit burner
1. q = 200.0 × 4.18 × 25.0 = 20.9 kJ.2. n = 1.20/46.07 = 0.02605 mol; ΔH = −20.9/0.02605 = −802 kJ/mol.
3. 802/1367 = 59 %; the rest was lost to the air, the can and incomplete combustion.
Problem 14 Melting and warming ice
1. 10.0 × 334 = 3340 J.2. 10.0 × 4.18 × 25.0 = 1045 J.
3. Total = 4385 J = 4.39 kJ.
Problem 15 Hydrogenation by bond enthalpy
1. Broken: 614 + 436 = 1050. Formed: 347 + 2(413) = 1173. ΔH = 1050 − 1173 = −123 kJ.2. The Hess value is −137 kJ. Bond enthalpies are averages for many molecules, so the estimate is about 14 kJ different.
Problem 16 Calorimeter with a heat capacity
1. ΔT = 4.9 °C. q(solution) = 100.0 × 4.18 × 4.9 = 2048 J; q(calorimeter) = 25.0 × 4.9 = 122.5 J. Total 2171 J.2. n(Cu²⁺) = 0.1000 × 0.100 = 0.0100 mol; ΔH = −2.171/0.0100 = −2.2 × 10² kJ/mol.
3. Excess zinc makes sure all the Cu²⁺ reacts, so the moles of Cu²⁺ is the limiting amount.
Problem 17 Formation enthalpy from combustion
1. CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l).2. −726 = [−393.5 + 2(−285.8)] − ΔH°f, so ΔH°f = −965.1 + 726 = −239.1 kJ/mol.
Problem 18 Ethyne from its elements
1. Target = 2 × (2) + (3) − (1): 2 C + 2 O₂ → 2 CO₂; H₂ + ½O₂ → H₂O; 2 CO₂ + H₂O → C₂H₂ + 5/2 O₂. Cancel to give 2 C(s) + H₂(g) → C₂H₂(g).2. ΔH = 2(−393.5) + (−285.8) − (−1300) = −787.0 − 285.8 + 1300 = +227.2 kJ.
3. ΔH is positive, so ethyne is less stable (higher enthalpy) than carbon and hydrogen.
Problem 19 Ammonia synthesis
1. Broken: 945 + 3(436) = 2253. Formed: 6(391) = 2346. ΔH = −93 kJ for 2 mol NH₃.2. 1000 g/17.03 = 58.7 mol; heat = 58.7 × 46.5 = 2.7 × 10³ kJ per kg.
3. The reaction is exothermic, and heat build-up would raise the temperature and shift the equilibrium back towards reactants, so heat must be removed.
Problem 20 Reading a temperature-time graph
1. During mixing the cup loses heat to the surroundings. The cooling trend is extended back to the time of mixing to find the temperature the mixture would have reached without heat loss.2. Extrapolated maximum = 35.6 °C; ΔT = 35.6 − 22.0 = 13.6 °C.
3. q = 100.0 × 4.18 × 13.6 = 5685 J; n = 0.100 mol; ΔH = −57 kJ/mol (−56.8).
4. With 35.4 °C, ΔT = 13.4 °C and ΔH = −56 kJ/mol (−56.0). The raw reading is lower because heat had already been lost, so the magnitude of ΔH is underestimated.
