AP - Advanced Placement Thermodynamics and Thermochemistry 1

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

Accepted Answer

−74.8 kJ

Answer

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

Accepted Answer

−141.8 kJ

Answer

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?

Accepted Answer

−57.7 kJ/mol

Answer

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

Accepted Answer

−185 kJ

Answer

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?

Accepted Answer

1079 K

Answer

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.

AP Practice Test

AP Practice Test

AP - Advanced Placement Thermodynamics and Thermochemistry 1