AP - Advanced Placement Thermodynamics and Thermochemistry 1 — Questions and Answers
Question 1: Using Hess's Law, calculate ΔH° for: C(s) + 2H₂(g) → CH₄(g) Given: (1) C(s) + O₂(g) → CO₂(g), ΔH° = −393.5 kJ (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH° = −285.8 kJ (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH° = −890.3 kJ
- −74.8 kJ (Correct answer)
- +74.8 kJ
- −965.1 kJ
- −1869.4 kJ
Correct answer: −74.8 kJ
Apply Hess's Law: use reaction (1) as written, reaction (2) × 2, and reverse reaction (3). ΔH° = −393.5 + 2(−285.8) + (+890.3) = −393.5 − 571.6 + 890.3 = −74.8 kJ. This equals the standard enthalpy of formation of CH₄, which is a useful check.
Question 2: Calculate ΔG° at 298 K for: 2SO₂(g) + O₂(g) → 2SO₃(g) Given: ΔH° = −197.8 kJ, ΔS° = −187.9 J/K
- −197.8 kJ
- −141.8 kJ (Correct answer)
- −253.8 kJ
- +55.0 kJ
Correct answer: −141.8 kJ
Using ΔG° = ΔH° − TΔS°: convert ΔS° to kJ/K (−0.1879 kJ/K), then ΔG° = −197.8 − (298)(−0.1879) = −197.8 + 56.0 = −141.8 kJ. The negative ΔG° confirms the reaction is spontaneous at 298 K.
Question 3: 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH, both initially at 22.0°C. The final temperature is 28.9°C. Assuming the solution density is 1.0 g/mL and specific heat is 4.184 J/g·°C, what is ΔH per mole of water formed?
- −2.89 kJ/mol
- −57.7 kJ/mol (Correct answer)
- −28.9 kJ/mol
- −115.4 kJ/mol
Correct answer: −57.7 kJ/mol
q = mcΔT = (100.0 g)(4.184 J/g·°C)(6.9°C) = 2887 J released. Moles of water = (0.0500 L)(1.0 mol/L) = 0.0500 mol. ΔH = −2887 J ÷ 0.0500 mol = −57,740 J/mol ≈ −57.7 kJ/mol. The sign is negative because heat is released by the reaction.
Question 4: Estimate ΔH for: H₂(g) + Cl₂(g) → 2HCl(g) Bond energies: H−H = 436 kJ/mol, Cl−Cl = 243 kJ/mol, H−Cl = 432 kJ/mol
- −185 kJ (Correct answer)
- +185 kJ
- −864 kJ
- +679 kJ
Correct answer: −185 kJ
Bonds broken (endothermic): H−H (436) + Cl−Cl (243) = +679 kJ. Bonds formed (exothermic): 2 × H−Cl = 2 × 432 = −864 kJ. ΔH = 679 − 864 = −185 kJ. The reaction is exothermic because more energy is released forming H−Cl bonds than is required to break H−H and Cl−Cl bonds.
Question 5: For the decomposition: CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ and ΔS° = +165 J/K. At what minimum temperature does this reaction become spontaneous?
- 465 K
- 806 K
- 1079 K (Correct answer)
- 1257 K
Correct answer: 1079 K
The crossover from non-spontaneous to spontaneous occurs when ΔG° = 0: T = ΔH°/ΔS° = 178,000 J ÷ 165 J/K ≈ 1079 K. Below this temperature ΔG° > 0 (non-spontaneous); above it ΔG° < 0 (spontaneous), consistent with a reaction driven by entropy at high temperature.
Question 6: Calculate ΔS° for: C(s, graphite) + O₂(g) → CO₂(g) S° values: C(s, graphite) = 5.7 J/mol·K, O₂(g) = 205.1 J/mol·K, CO₂(g) = 213.8 J/mol·K
- −3.0 J/mol·K
- +3.0 J/mol·K (Correct answer)
- +424.6 J/mol·K
- −196.0 J/mol·K
Correct answer: +3.0 J/mol·K
ΔS° = ΣS°(products) − ΣS°(reactants) = 213.8 − (5.7 + 205.1) = 213.8 − 210.8 = +3.0 J/mol·K. The very small positive value reflects that one mole of gas is consumed and one mole of gas is produced, so the net entropy change is near zero but slightly positive because CO₂ has higher absolute entropy than O₂.
AP - Advanced Placement Thermodynamics and Thermochemistry 1
Using Hess's Law, calculate ΔH° for: C(s) + 2H₂(g) → CH₄(g)
Given:
(1) C(s) + O₂(g) → CO₂(g), ΔH° = −393.5 kJ
(2) H₂(g) + ½O₂(g) → H₂O(l), ΔH° = −285.8 kJ
(3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH° = −890.3 kJ