Calculating Entropy Changes

The entropy change of the system, ΔS, is the difference between the entropies of the products and the reactants.

It is found from tabulated standard entropies, in the same way that an enthalpy change is found from enthalpies of formation:

ΔS = ΣS⦵(products) − ΣS⦵(reactants)

• Each entropy is multiplied by the number of moles of that substance in the equation.
• The answer has the same units as the data, J K⁻¹ mol⁻¹.
• There is one difference from enthalpy calculations: elements have non-zero entropies, so an element in the equation must be included with its tabulated value, not treated as zero.

Worked example 1: a gas is made

The thermal decomposition of calcium carbonate is CaCO₃(s) → CaO(s) + CO₂(g).

Step 1. Add the entropies of the products: 39.7 + 213.6 = 253.3 J K⁻¹ mol⁻¹.

Step 2. Subtract the entropy of the reactant: ΔS = (39.7 + 213.6) − 92.9.

Answer. ΔS = +160.4 J K⁻¹ mol⁻¹.

• The sign is positive, as page 1 predicted for a reaction that makes a gas from a solid.
• The size is typical of a reaction that produces one mole of gas: roughly 150 to 200 J K⁻¹ mol⁻¹ per mole of gas gained.

Worked example 2: gas moles fall

For N₂(g) + 3H₂(g) → 2NH₃(g), S⦵ = 192 (N₂), 131 (H₂) and 193 (NH₃) J K⁻¹ mol⁻¹.

Step 1. Products: 2 × 193 = 386 J K⁻¹ mol⁻¹.

Step 2. Reactants: 192 + (3 × 131) = 585 J K⁻¹ mol⁻¹.

Answer. ΔS = 386 − 585 = −199 J K⁻¹ mol⁻¹. Four moles of gas become two, so the entropy of the system falls.

The calcium carbonate calculation laid out in full: standard entropies, ΔS of the system, then the surroundings and total entropy at 298 K and at 1200 K.

For the decomposition of calcium carbonate ΔH = +178 kJ mol⁻¹ and ΔS = +160.4 J K⁻¹ mol⁻¹.

The enthalpy change is unfavourable and the entropy change favourable, so whether the reaction is feasible depends on the temperature.

The Gibbs energy change, ΔG = ΔH − TΔS, weighs the two against each other. The reaction is feasible when ΔG is negative.

Worked example: calcium carbonate at two temperatures

Step 1. Convert the entropy term to kJ first: 160.4 J K⁻¹ mol⁻¹ = 0.1604 kJ K⁻¹ mol⁻¹.

Step 2. At 298 K: ΔG = 178 − (298 × 0.1604) = 178 − 47.8 = +130 kJ mol⁻¹.

Step 3. At 1200 K: ΔG = 178 − (1200 × 0.1604) = 178 − 192.5 = −15 kJ mol⁻¹.

Answer. ΔG is positive at 298 K, so limestone does not decompose at room temperature. ΔG is negative at 1200 K, so in a lime kiln the reaction goes.

The method as a flow chart: ΔS of the system first, then the total entropy route or the Gibbs energy route, with the unit traps marked and both routes giving the same verdict for calcium carbonate.

• The TΔS term grows with temperature until it outweighs ΔH.
• This is the reason an endothermic reaction with a positive entropy change becomes feasible when hot.
• Page 3 deals with the full equation, its sign combinations and the temperature at which ΔG changes sign.

Say

• “ΔS = ΣS(products) − ΣS(reactants), including the elements.”
• “Convert ΔS to kJ K⁻¹ mol⁻¹ before using ΔG = ΔH − TΔS.”
• “ΔG is negative, so the reaction is feasible.”

Do not say

• “Elements have an entropy of zero” (only their enthalpy of formation is zero).
• “T = 25” (use kelvin: 298 K).
• “ΔG = 178 − 298 × 160.4” (units mixed: use 0.1604 kJ K⁻¹ mol⁻¹).

Watch for

• The most common slip in this topic is units: entropies are in J K⁻¹ mol⁻¹ and enthalpies in kJ mol⁻¹, so one of them must be converted before they are combined.
• Check that the sign of ΔS matches the change in moles of gas.
• Check that a temperature given in °C has had 273 added.