Calculating Entropy Changes
A concise revision guide to calculating the entropy change of a reaction from standard entropies, the entropy change of the surroundings from ΔSsurroundings = −ΔH/T, and the total entropy change that decides whether a reaction is feasible.
- 12.6-a
- 12.6-b
- 12.7
- 12.8-a
- 12.8-b
What these spec points say
- 12.6-a understand that total entropy change of any reaction is the sum of entropy change of system and entropy change of surroundings
- 12.6-b know the expression ΔStotal = ΔSsystem + ΔSsurroundings
- 12.7 be able to calculate the entropy change of the system for a reaction, ΔSsystem, given the entropies of the reactants and products
- 12.8-a be able to calculate entropy change in the surroundings using ΔSsurroundings = −ΔH/T
- 12.8-b be able to calculate ΔStotal using ΔSsurroundings = −ΔH/T
ΔS of the System From Standard Entropies
The entropy change of the system, ΔS_system, 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_system = 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).
| Substance | S⦵ / J K⁻¹ mol⁻¹ |
|---|---|
| CaCO₃(s) | 92.9 |
| CaO(s) | 39.7 |
| CO₂(g) | 213.6 |
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_system = (39.7 + 213.6) − 92.9.
Answer. ΔS_system = +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_system = 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.
Worked example: CaCO₃(s) → CaO(s) + CO₂(g): ΔS_system = ΣS(products) − ΣS(reactants) = (39.7 + 213.6) − 92.9 = +160.4 J K⁻¹ mol⁻¹. Multiply each S by its balancing number, and include the elements.
Check: ΔS From Standard Entropies
Calculate the entropy change of the system with its sign and units for reactions not used on this page.
ΔS of the Surroundings
The reaction also changes the entropy of everything around it.
- An exothermic reaction gives out heat. That heat spreads among the particles of the surroundings and raises their entropy.
- An endothermic reaction takes heat in and lowers the entropy of the surroundings.
The size of the effect depends on how much heat is transferred and on the temperature at which it happens. The two are combined in a simple expression:
ΔS_surroundings = −ΔH ÷ T
| Part of the expression | What to do | Why |
|---|---|---|
| The minus sign | keep it | it makes the signs come out right: a negative ΔH (exothermic) gives a positive ΔS_surroundings |
| ΔH | convert from kJ to J before dividing | entropies are in J K⁻¹ mol⁻¹ |
| T | use kelvin | T is in the denominator because a given amount of heat makes a bigger difference to cold surroundings, which have little energy already, than to hot ones |
Worked example: calcium carbonate at 298 K
Step 1. Convert the enthalpy change: ΔH = +178 kJ mol⁻¹ = +178 000 J mol⁻¹.
Step 2. Substitute: ΔS_surroundings = −178 000 ÷ 298.
Answer. ΔS_surroundings = −597 J K⁻¹ mol⁻¹.
The reaction is endothermic, so it takes heat from the surroundings and lowers their entropy, by an amount that far outweighs the entropy gained by the system at this temperature.
Worked example: the Haber process reaction at 298 K
Step 1. Convert the enthalpy change: ΔH = −92 kJ mol⁻¹ = −92 000 J mol⁻¹.
Step 2. Substitute: ΔS_surroundings = −(−92 000) ÷ 298.
Answer. ΔS_surroundings = +309 J K⁻¹ mol⁻¹.
The heat given out raises the entropy of the surroundings by more than the entropy of the system falls.
Exam focus: ΔS_surroundings = −ΔH/T. Convert ΔH to joules (multiply by 1000), use T in kelvin, and keep the minus sign: exothermic reactions give a positive ΔS_surroundings, endothermic reactions a negative one.
Check: ΔS of the Surroundings
Calculate the entropy change of the surroundings from ΔH and T, with the right units and sign, for reactions not used on this page.
Total Entropy Change
The total entropy change of a reaction is the sum of the entropy change of the system and the entropy change of the surroundings:
ΔS_total = ΔS_system + ΔS_surroundings
A reaction is feasible (it can happen of its own accord) when ΔS_total is positive. It does not matter if one of the two terms is negative, as long as the other outweighs it.
Calcium carbonate at two temperatures
All values in the table are in J K⁻¹ mol⁻¹.
| Quantity | At 298 K | At 1200 K |
|---|---|---|
| ΔS_system | +160.4 | +160.4 |
| ΔS_surroundings = −ΔH ÷ T | −178 000 ÷ 298 = −597 | −178 000 ÷ 1200 = −148 |
| ΔS_total | +160.4 + (−597) = −437 | +160.4 + (−148) = +12 |
| Feasible? | no: ΔS_total is negative | yes: ΔS_total is positive |
- At 298 K limestone does not decompose, however long you wait.
- Raising the temperature makes the surroundings term smaller, because the same ΔH is divided by a larger T.
- At 1200 K, in a lime kiln, the reaction goes.
- Page 3 shows how to find the exact temperature at which the sign changes.
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 link to Gibbs energy
The same numbers can be expressed through the Gibbs energy change, ΔG = ΔH − TΔS_system, which is −T × ΔS_total in kJ.
You may use it in answers, though the total entropy method is the one the specification asks for: a positive ΔS_total is the same statement as a negative ΔG.
Exam wording: “ΔS_total = ΔS_system + ΔS_surroundings = +160.4 + (−597) = −437 J K⁻¹ mol⁻¹. ΔS_total is negative, so the reaction is not feasible at 298 K.” Give both terms, the sum, its sign and the conclusion.
Common Exam Points
Say
- “ΔS_system = ΣS(products) − ΣS(reactants), including the elements.”
- “ΔS_surroundings = −ΔH/T, with ΔH in J and T in K.”
- “ΔS_total is positive, 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).
- “ΔS_surroundings = −178 ÷ 298” (ΔH must be in joules: −178 000 ÷ 298).
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_system matches the change in moles of gas.
- Check that a temperature given in °C has had 273 added.
Check: A Full Calculation
Carry out a complete calculation of ΔS_system, ΔS_surroundings and ΔS_total and decide whether a reaction not used on this page is feasible.
FAQs
Use these quick answers to check the entropy calculations.
Why must I convert J to kJ?
Because standard entropies are tabulated in J K⁻¹ mol⁻¹ while enthalpy changes are in kJ mol⁻¹. Whenever the two meet in one equation they must share a unit, so either divide the entropy term by 1000 or multiply ΔH by 1000. Mixing them is the most common reason for an answer that is out by a factor of a thousand.
Why do I have to multiply each standard entropy by the balancing number?
Because standard entropy is quoted per mole of substance, and the equation may involve two or three moles. In 2H₂(g) + O₂(g) → 2H₂O(l) the hydrogen contributes 2 × 131 J K⁻¹ mol⁻¹ and the water 2 × 70 J K⁻¹ mol⁻¹. Forgetting the multiplier changes both the size and, sometimes, the sign of ΔS.
Why do I divide by T in ΔSsurroundings = −ΔH/T?
Because the entropy change caused by adding a given amount of heat depends on how hot the surroundings already are. The same quantity of heat spreads energy among far more extra arrangements in cold surroundings than in hot ones, so the entropy gain is bigger at low temperature. Dividing by the temperature in kelvin builds that in, and the minus sign is there because heat given out by the system (negative ΔH) is heat gained by the surroundings.
Why does ΔSsurroundings have the opposite sign to ΔH?
Because an exothermic reaction (negative ΔH) gives heat to the surroundings, which increases their entropy, so ΔSsurroundings is positive. An endothermic reaction takes heat in and lowers the entropy of the surroundings. Sort out the sign first and then check that the number you get agrees with it.
What temperature do I use if the question does not say?
Standard conditions, which means 298 K. Always convert Celsius to kelvin by adding 273 before substituting; an entropy calculation at 25 K instead of 298 K gives nonsense. If the question sets a different temperature, use that one throughout.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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