Entropy and free energy

Whether a reaction can happen by itself depends on two things: the enthalpy change, ΔH, and the entropy change, ΔS. Entropy measures disorder, the number of ways particles and their energy can be arranged. The Gibbs free energy change, ΔG = ΔH − TΔS, combines them: a reaction is feasible when ΔG is zero or negative.

Key facts

  • Entropy, S, is measured in J K⁻¹ mol⁻¹. Gases have far more entropy than liquids, and liquids more than solids. A perfect crystal at 0 K has S = 0.
  • ΔS = ΣS(products) − ΣS(reactants). ΔS is positive when a solid melts or dissolves, a liquid boils, or the number of moles of gas increases.
  • ΔG = ΔH − TΔS, with T in kelvin. Change ΔS to kJ K⁻¹ mol⁻¹ (divide by 1000) to match ΔH in kJ/mol.
  • A reaction is feasible when ΔG ≤ 0. Feasible does not mean fast: a high activation energy can make a feasible reaction too slow to see.
  • ΔH negative and ΔS positive: always feasible. ΔH positive and ΔS negative: never feasible. When the signs are the same, the reaction changes over at T = ΔH ÷ ΔS.
  • A graph of ΔG against T is a straight line with gradient −ΔS and intercept ΔH.

Worked examples

Example 1

Calculate ΔS for N₂(g) + 3H₂(g) → 2NH₃(g). S° in J K⁻¹ mol⁻¹: N₂ 192, H₂ 131, NH₃ 193.

  1. Products. 2 × 193 = 386
  2. Reactants. 192 + 3 × 131 = 585
  3. Subtract. ΔS = 386 − 585 = −199 J K⁻¹ mol⁻¹
  4. It is negative, as expected: 4 moles of gas become 2.

Answer: −199 J K⁻¹ mol⁻¹

Example 2

For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol and ΔS = +161 J K⁻¹ mol⁻¹. Is it feasible at 298 K? Above what temperature does it become feasible?

  1. Change ΔS to kJ. ΔS = 0.161 kJ K⁻¹ mol⁻¹
  2. At 298 K: ΔG = 178 − 298 × 0.161 = +130 kJ/mol
  3. ΔG is positive, so it is not feasible at room temperature.
  4. It becomes feasible when ΔG = 0. T = ΔH ÷ ΔS = 178 000 ÷ 161 = 1106 K

Answer: Not at 298 K (ΔG = +130 kJ/mol); feasible above 1106 K

Example 3

For ice melting, H₂O(s) → H₂O(l), ΔH = +6.01 kJ/mol and ΔS = +22.0 J K⁻¹ mol⁻¹. Show why ice melts above 0 °C but not below.

  1. Both ΔH and ΔS are positive, so melting is feasible above T = ΔH ÷ ΔS. T = 6010 ÷ 22.0 = 273 K
  2. At 263 K (−10 °C): ΔG = 6.01 − 263 × 0.0220 = +0.22 kJ/mol
  3. At 283 K (+10 °C): ΔG = 6.01 − 283 × 0.0220 = −0.22 kJ/mol
  4. So ice melts only above 273 K, which is 0 °C.

Answer: T = 273 K: below it ΔG > 0, above it ΔG < 0