AP Thermodynamics and Thermochemistry 2 — Questions and Answers

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.

−890.3 kJ/mol (Correct answer)
−504.0 kJ/mol
+890.3 kJ/mol
−815.5 kJ/mol

Correct answer: −890.3 kJ/mol

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.

Using standard enthalpies of formation: ΔH°rxn = Σ[n × ΔH°f(products)] − Σ[n × ΔH°f(reactants)] Products: - CO₂(g): 1 mol × (−393.5 kJ/mol) = −393.5 kJ - H₂O(l): 2 mol × (−285.8 kJ/mol) = −571.6 kJ Sum of products = −393.5 + (−571.6) = −965.1 kJ Reactants: - CH₄(g): 1 mol × (−74.8 kJ/mol) = −74.8 kJ - O₂(g): 2 mol × 0 kJ/mol = 0 kJ (elements in standard state) Sum of reactants = −74.8 kJ ΔH°rxn = −965.1 − (−74.8) = −965.1 + 74.8 = −890.3 kJ/mol The large negative value confirms combustion of methane is highly exothermic — about 890 kJ of heat released per mole. This is why natural gas (methane) is an effective fuel. The standard enthalpy of combustion (−890 kJ/mol) is tabulated and used in energy calculations for home heating, power generation, and chemical engineering.

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?

464 K (Correct answer)
232 K
198 K
925 K

Correct answer: 464 K

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.

Spontaneity is governed by ΔG = ΔH − TΔS. For the Haber process: ΔH° = −92 kJ (exothermic, favorable), ΔS° = −198 J/K (unfavorable — 4 mol gas → 2 mol gas, decrease in disorder). The signs predict: - At low T: −TΔS term is small; ΔH dominates → ΔG < 0 → spontaneous - At high T: −TΔS = −T(−198) = +198T becomes large positive → ΔG becomes positive → non-spontaneous Crossover at ΔG = 0: 0 = ΔH − TΔS T = ΔH/ΔS = (−92,000 J) / (−198 J/K) = 464.6 K ≈ 464 K This has real industrial significance: the Haber process operates at 400–500°C (~700–800 K) despite crossing into non-spontaneous territory thermodynamically, because: 1. Above 464 K: thermodynamically unfavorable but kinetically faster 2. The equilibrium yield of NH₃ decreases at higher temperatures 3. A compromise temperature (~450°C) and high pressure (150–300 atm) optimize practical NH₃ yield

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?

11.1 kJ/°C (Correct answer)
15.57 kJ/°C
7.79 kJ/°C
21.8 kJ/°C

Correct answer: 11.1 kJ/°C

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

A bomb calorimeter is a constant-volume calorimeter that directly measures qᵥ (heat at constant volume = ΔU, not ΔH, but the difference is small for most combustions). The key principle: all heat released by the combustion reaction is absorbed by the calorimeter: q_rxn = C_cal × ΔT (in magnitude) C_cal = q_rxn / ΔT = 15.57 kJ / 1.40°C = 11.1 kJ/°C To find the standard enthalpy of combustion per mole of glucose: mol glucose = 1.00 g / 180.2 g/mol = 0.00555 mol qᵥ per mol = −15.57 kJ / 0.00555 mol = −2,805 kJ/mol (The actual ΔH°_combustion of glucose is −2,803 kJ/mol — excellent agreement.) Bomb calorimeters are used in: (1) nutritional calorie measurement (food energy content), (2) fuel energy analysis (coal, oil), (3) thermodynamic property determination. The calorimeter constant (11.1 kJ/°C) is determined once by calibration with a standard such as benzoic acid, then used for subsequent measurements.

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.

−326.5 J/K (Correct answer)
+326.5 J/K
−257.5 J/K
+257.5 J/K

Correct answer: −326.5 J/K

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.

Standard entropy change: ΔS° = Σ[n × S°(products)] − Σ[n × S°(reactants)] Products: - H₂O(l): 2 mol × 70.0 J/mol·K = 140.0 J/K Reactants: - H₂(g): 2 mol × 130.7 J/mol·K = 261.4 J/K - O₂(g): 1 mol × 205.1 J/mol·K = 205.1 J/K - Total reactants: 261.4 + 205.1 = 466.5 J/K ΔS° = 140.0 − 466.5 = −326.5 J/K The large negative ΔS° makes physical sense: 3 moles of gas (high entropy) are converted to 2 moles of liquid (low entropy). The entropy decrease per mole gas → liquid is typically ~100–120 J/mol·K (Trouton's rule), and here 3 moles of gas are lost with only 2 moles of liquid gained. Despite the unfavorable ΔS°, the reaction of H₂ and O₂ is spontaneous because ΔH° = −572 kJ is strongly negative — enthalpy dominates at room temperature. This is the reaction in hydrogen fuel cells, which generate electricity from this spontaneous but controlled reaction.

Question 5: For a reaction at 298 K with ΔG° = −25.0 kJ/mol, what is the equilibrium constant K?

K ≈ 2.48 × 10⁴ (Correct answer)
K ≈ 5.47 × 10⁻³
K ≈ 1.00
K ≈ 1.61 × 10²

Correct answer: K ≈ 2.48 × 10⁴

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.

The relationship between standard free energy and equilibrium constant: ΔG° = −RT ln K → K = e^(−ΔG°/RT) Given: ΔG° = −25,000 J/mol (note conversion from kJ), T = 298 K, R = 8.314 J/mol·K. −ΔG°/RT = −(−25,000)/(8.314 × 298) = 25,000/2477.6 = 10.09 K = e^10.09 = 2.41 × 10⁴ ≈ 2.48 × 10⁴ Interpretation: K >> 1 means products are strongly favored at equilibrium, consistent with ΔG° << 0 (spontaneous in forward direction). The relationship between ΔG°, K, and spontaneity: - ΔG° < 0 → K > 1 → products favored at equilibrium - ΔG° = 0 → K = 1 → equal amounts of products and reactants - ΔG° > 0 → K < 1 → reactants favored at equilibrium Using base 10: log K = −ΔG°/(2.303RT) = 25,000/(2.303 × 8.314 × 298) = 4.38. K = 10^4.38 = 2.40 × 10⁴ ✓

Question 6: The specific heat capacity of iron is 0.449 J/(g·°C). How much heat is required to raise the temperature of a 250. g iron bar from 20.0°C to 150.°C?

14,600 J (Correct answer)
7,300 J
29,200 J
3,650 J

Correct answer: 14,600 J

q = m × c × ΔT = 250. g × 0.449 J/(g·°C) × (150.0 − 20.0)°C = 250 × 0.449 × 130 = 14,593 J ≈ 14,600 J.

The heat required to change the temperature of a substance is: q = m × c × ΔT Where: - m = 250. g (mass) - c = 0.449 J/(g·°C) (specific heat of iron) - ΔT = T_final − T_initial = 150.0°C − 20.0°C = 130.°C q = 250. g × 0.449 J/(g·°C) × 130.°C q = 250. × 0.449 × 130. q = 250. × 58.37 q = 14,593 J ≈ 14,600 J For comparison: q(water) = 250. × 4.184 × 130 = 135,980 J — about 9.3× more heat to warm the same mass of water by 130°C. This reflects iron's much lower specific heat than water, which is why metals heat up and cool down much faster than water. Specific heat capacity is an intensive property related to molecular structure: metallic solids with heavy atoms and simple crystal structures (like iron) have low specific heats per gram. Water's high specific heat (4.184 J/g·°C) is due to H-bonding and the molecule's ability to absorb thermal energy in many vibrational modes.

AP Thermodynamics and Thermochemistry 2

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.