Thermodynamics Solved Examples (11 Chemistry)
Chemical thermodynamics studies enthalpy, entropy, and Gibbs free energy. These examples cover heat of reaction, entropy changes, spontaneity, and equilibr
TL;DR: Chemical thermodynamics studies enthalpy, entropy, and Gibbs free energy. These examples cover heat of reaction, entropy changes, spontaneity, and equ…
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Chemical thermodynamics studies enthalpy, entropy, and Gibbs free energy. These examples cover heat of reaction, entropy changes, spontaneity, and equilibr
Thermodynamics — Solved Numerical Examples (Step by Step)
Example 1: Calculate the enthalpy change for the reaction 2H2 + O2 → 2H2O. Bond energies: H-H = 436 kJ/mol, O=O = 498 kJ/mol, O-H = 467 kJ/mol.
Solution: Enthalpy change ΔH = Energy required to break bonds - Energy released when forming bonds
Step 1: Bonds broken (energy required).
2 mol H-H bonds: 2 × 436 = 872 kJ
1 mol O=O bond: 1 × 498 = 498 kJ
Total energy required = 872 + 498 = 1370 kJ
Step 2: Bonds formed (energy released).
2 mol H2O has 4 O-H bonds: 4 × 467 = 1868 kJ
Total energy released = 1868 kJ
Step 3: Calculate ΔH.
ΔH = 1370 - 1868 = -498 kJ/mol
The reaction is exothermic (releases 498 kJ).
Example 2: For a reaction, ΔH = -60 kJ/mol and ΔS = 150 J/mol⋅K. At what temperature does this reaction become spontaneous? (ΔG = ΔH - TΔS)
Solution: For spontaneity, ΔG < 0
ΔG = ΔH - TΔS < 0
Given: ΔH = -60 kJ = -60,000 J, ΔS = 150 J/mol⋅K
-60,000 - T(150) < 0
-60,000 < 150T
T > -60,000 / 150
T > -400 K
This means at all positive temperatures, ΔG < 0.
The reaction is spontaneous at all temperatures (since ΔH is negative and ΔS is positive).
Example 3: The equilibrium constant Kc = 4 at 300 K for the reaction A + B ⇌ C. Find ΔG°. (R = 8.314 J/mol⋅K)
Solution: Use the relationship: ΔG° = -RT ln(Kc)
Given: Kc = 4, T = 300 K, R = 8.314 J/mol⋅K
ΔG° = -8.314 × 300 × ln(4)
ΔG° = -2494.2 × 1.386
ΔG° = -3460 J/mol ≈ -3.46 kJ/mol
The negative value indicates the forward reaction is spontaneous under standard conditions.
Example 4: Calculate the entropy change when 1 mole of ice at 0°C melts to water at 0°C. Latent heat of fusion = 334 kJ/mol.
Solution: For a phase change at constant temperature:
ΔS = q_rev / T = ΔH_fusion / T
Given: ΔH_fusion = 334 kJ = 334,000 J, T = 273.15 K (0°C)
ΔS = 334,000 / 273.15 = 1223 J/mol⋅K ≈ 1.22 kJ/mol⋅K
The entropy increases during melting (disorder increases as solid becomes liquid).
Example 5: At what temperature will the reaction N2O4 ⇌ 2NO2 have ΔG = 0? Given ΔH = 58 kJ/mol and ΔS = 176 J/mol⋅K.
Solution: At equilibrium, ΔG = 0
0 = ΔH - TΔS
T = ΔH / ΔS
Given: ΔH = 58 kJ = 58,000 J, ΔS = 176 J/mol⋅K
T = 58,000 / 176 = 329.5 K ≈ 330 K or 57°C
Above this temperature, the forward reaction becomes spontaneous (ΔG < 0).
Example 6: For the reaction 2SO2 + O2 ⇌ 2SO3, ΔH = -198 kJ. Is this reaction exothermic or endothermic? What does this tell us about the forward and reverse reactions?
Solution: Since ΔH = -198 kJ is negative, the forward reaction is exothermic.
The reverse reaction (2SO3 → 2SO2 + O2) is endothermic and requires 198 kJ of energy.
This means:
- Forward reaction releases 198 kJ of heat
- Reverse reaction absorbs 198 kJ of heat
- The products (SO3) are more stable (lower energy) than reactants
- Cooling shifts equilibrium toward products (Le Chatelier's principle)
Example 7: Calculate the standard free energy change for a reaction where Kp = 1. What does this mean?
Solution: Use: ΔG° = -RT ln(Kp)
When Kp = 1:
ΔG° = -RT ln(1)
ΔG° = -RT × 0 = 0
This means:
- The reaction is at equilibrium under standard conditions
- Forward and reverse reactions occur at equal rates
- The system has no net tendency to form products or reactants
- Reactants and products are equally favored
Tips
- ΔH negative = exothermic (releases heat), ΔH positive = endothermic (absorbs heat).
- ΔS positive = disorder increases; ΔS negative = disorder decreases.
- ΔG = ΔH - TΔS; ΔG < 0 means spontaneous; ΔG > 0 means non-spontaneous.
- ln(Kc) > 0 when Kc > 1 (products favored); ln(Kc) < 0 when Kc < 1 (reactants favored).
Frequently Asked Questions
Can a non-spontaneous reaction become spontaneous if conditions change?
Yes. A non-spontaneous reaction (ΔG > 0) can become spontaneous at higher temperatures if ΔS > 0. This is because ΔG = ΔH - TΔS becomes negative when the TΔS term dominates.
Why is entropy important in thermodynamics?
Entropy measures disorder and randomness in a system. The second law of thermodynamics states that entropy of the universe always increases in spontaneous processes, which is why spontaneity depends on both enthalpy and entropy changes.
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