Topic 12 · page 5 of 17

Entropy change of the system

Add up the standard entropies of the products, subtract those of the reactants, and multiply each by its coefficient. The answer is in J K⁻¹ mol⁻¹, and its sign should match your prediction.

10 min read12.7

Read first: Why entropy changes

The calculation 12.7

Standard entropies are measured, and tables list them for each substance. For a reaction, the entropy change of the system is the entropy of the products minus the entropy of the reactants.

ΔSsystem⊖=ΣS⊖(products)−ΣS⊖(reactants) \Delta S^\ominus_\text{system} = \Sigma S^\ominus(\text{products}) - \Sigma S^\ominus(\text{reactants})

This looks like the enthalpy calculation of Topic 6, with products minus reactants. The difference is that nothing is zero: every substance, including every element, has a positive S⊖S^\ominus.

MethodCalculating the entropy change of the system

  1. Write the balanced equation, with state symbols.
  2. Predict the sign from the moles of gas (page 4).
  3. Multiply each S⊖S^\ominus by its coefficient, and add up the products and the reactants separately.
  4. Subtract: products minus reactants.
  5. Give the answer in J K⁻¹ mol⁻¹, and check that its sign matches your prediction.

Worked exampleAmmonia synthesis

Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)}. The entropies in J K⁻¹ mol⁻¹ are: N₂ 191.6, H₂ 130.6, NH₃ 192.3.

  1. Prediction: 4 mol of gas become 2 mol, so the sign should be negative.
  2. Products: 2 × 192.3 = 384.6
  3. Reactants: 191.6 + (3 × 130.6) = 191.6 + 391.8 = 583.4
  4. ΔSsystem⊖\Delta S^\ominus_\text{system} = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹. This is negative, as predicted.

Worked exampleBurning magnesium

Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for 2 Mg(s)+OX2(g)→2 MgO(s)\ce{2Mg(s) + O2(g) -> 2MgO(s)}. The entropies in J K⁻¹ mol⁻¹ are: Mg 32.7, O₂ 205.0, MgO 26.9.

  1. Prediction: 1 mol of gas becomes none, so negative.
  2. Products: 2 × 26.9 = 53.8
  3. Reactants: (2 × 32.7) + 205.0 = 65.4 + 205.0 = 270.4
  4. ΔSsystem⊖\Delta S^\ominus_\text{system} = 53.8 − 270.4 = −216.6 J K⁻¹ mol⁻¹

Worked exampleDecomposing calcium carbonate

Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for CaCOX3(s)→CaO(s)+COX2(g)\ce{CaCO3(s) -> CaO(s) + CO2(g)}. The entropies in J K⁻¹ mol⁻¹ are: CaCO₃ 92.9, CaO 39.7, CO₂ 213.6.

  1. Prediction: no gas becomes 1 mol of gas, so positive.
  2. Products: 39.7 + 213.6 = 253.3
  3. Reactants: 92.9
  4. ΔSsystem⊖\Delta S^\ominus_\text{system} = 253.3 − 92.9 = +160.4 J K⁻¹ mol⁻¹

The per-mole in the unit means "per mole of reaction as written". Double all the coefficients and you double ΔS\Delta S.

Check yourself

Use these values in J K⁻¹ mol⁻¹: H₂ 130.6, O₂ 205.0, H₂O(l) 70, CH₄(g) 186.3, CO₂(g) 213.6, NO₂(g) 240.1, N₂O₄(g) 304.3, N₂ 191.6, NH₃ 192.3.

  1. Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for NX2OX4(g)→2 NOX2(g)\ce{N2O4(g) -> 2NO2(g)}.
  2. Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for 2 HX2(g)+OX2(g)→2 HX2O(l)\ce{2H2(g) + O2(g) -> 2H2O(l)}.
  3. Calculate ΔSsystem⊖\Delta S^\ominus_\text{system} for CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l)\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}.
  4. Predict the sign of ΔSsystem⊖\Delta S^\ominus_\text{system} for 2 NOX2(g)→NX2OX4(g)\ce{2NO2(g) -> N2O4(g)}, then say what its value is.
  5. A student calculates the value for NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)} as +62.4 J K⁻¹ mol⁻¹. Find the mistake.

Answers cover these until you have tried the questions

  1. Products: 2 × 240.1 = 480.2. Reactants: 304.3. ΔSsystem⊖\Delta S^\ominus_\text{system} = 480.2 − 304.3 = +175.9 J K⁻¹ mol⁻¹ (positive: 1 mol of gas becomes 2 mol).
  2. Products: 2 × 70 = 140. Reactants: (2 × 130.6) + 205.0 = 261.2 + 205.0 = 466.2. ΔSsystem⊖\Delta S^\ominus_\text{system} = 140 − 466.2 = −326.2 J K⁻¹ mol⁻¹ (negative: 3 mol of gas become liquid).
  3. Products: 213.6 + (2 × 70) = 353.6. Reactants: 186.3 + (2 × 205.0) = 596.3. ΔSsystem⊖\Delta S^\ominus_\text{system} = 353.6 − 596.3 = −242.7 J K⁻¹ mol⁻¹. (3 mol of gas become 1 mol of gas plus liquid water.)
  4. 2 mol of gas become 1 mol, so the sign is negative. It is the reverse of question 1, so the value is −175.9 J K⁻¹ mol⁻¹.
  5. The student did not multiply the entropy of H₂ by 3. Reactants should be 191.6 + (3 × 130.6) = 583.4. The student used 191.6 + 130.6 = 322.2, giving 384.6 − 322.2 = +62.4. The correct answer is 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹, and the sign should have been negative because 4 mol of gas become 2 mol.