AP Thermodynamics and Thermochemistry 2

Question 1

Calculate ΔH°rxn for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Given ΔH°f values: CH₄(g) = −74.8 kJ/mol, CO₂(g) = −393.5 kJ/mol, H₂O(l) = −285.8 kJ/mol.

Accepted Answer

−890.3 kJ/mol

Answer

ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants) = [−393.5 + 2(−285.8)] − [−74.8 + 0] = [−393.5 − 571.6] − [−74.8] = −965.1 + 74.8 = −890.3 kJ/mol.

Question 2

For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g). ΔH° = −92 kJ and ΔS° = −198 J/K. At what temperature does this reaction change from spontaneous to non-spontaneous?

Accepted Answer

464 K

Answer

At the crossover temperature, ΔG = 0: T = ΔH/ΔS = −92,000 J/(−198 J/K) = 464 K. Below 464 K the reaction is spontaneous (negative ΔG); above 464 K it is non-spontaneous.

Question 3

A bomb calorimeter experiment burns 1.00 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) and releases 15.57 kJ of heat, raising the temperature by 1.40°C. What is the heat capacity of the calorimeter?

Accepted Answer

11.1 kJ/°C

Answer

Heat capacity = q / ΔT = 15.57 kJ / 1.40°C = 11.1 kJ/°C. The bomb calorimeter absorbs all the heat released by combustion.

Question 4

Calculate ΔS° for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l). S° values: H₂(g) = 130.7 J/mol·K, O₂(g) = 205.1 J/mol·K, H₂O(l) = 70.0 J/mol·K.

Accepted Answer

−326.5 J/K

Answer

ΔS° = ΣS°(products) − ΣS°(reactants) = 2(70.0) − [2(130.7) + 205.1] = 140.0 − 466.5 = −326.5 J/K. Decrease in entropy expected: 3 mol gas → 2 mol liquid.

Question 5

For a reaction at 298 K with ΔG° = −25.0 kJ/mol, what is the equilibrium constant K?

Accepted Answer

K ≈ 2.48 × 10⁴

Answer

ΔG° = −RT ln K. ln K = −ΔG°/RT = −(−25,000)/(8.314 × 298) = 25,000/2477.6 = 10.09. K = e^10.09 = 2.41 × 10⁴ ≈ 2.48 × 10⁴.

AP Practice Test

AP Practice Test

AP Thermodynamics and Thermochemistry 2