4.2.3Unit 4. Rates, Equilibria and Further Organic Chemistry
Total Entropy — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 21 of 56 Unit quiz
Section 1 of 6Entropy of the Surroundings
The surroundings gain or lose entropy as energy is transferred to or from them.
Formula
- ΔS°surroundings = −ΔH ÷ T, where T is the temperature in kelvin.
Exothermic reaction
- Heat released to the surroundings makes their particles move more, so ΔS of the surroundings is positive.
Endothermic reaction
- Heat taken in from the surroundings reduces their disorder, so ΔS of the surroundings is negative.
Units
- ΔH must be converted from kJ mol⁻¹ to J mol⁻¹ so that the answer is in J K⁻¹ mol⁻¹.
Section 2 of 6Total Entropy
Total Entropy
Equation
- The total entropy change is the sum ΔS°total = ΔS°system + ΔS°surroundings.
Meaning
- This measures the overall change in disorder of the universe for the reaction.
Second law
- A reaction can occur spontaneously only if ΔS°total is positive, or zero at equilibrium.
Temperature
- T affects the surroundings term, so the sign of the total can change with temperature.
Section 3 of 6Calculation Steps
1
Find ΔS°system from the standard entropies of the products and reactants.
2
Convert ΔH from kJ to J by multiplying by 1000.
3
Calculate ΔS°surroundings = −ΔH ÷ T using T in kelvin, usually 298 K.
4
Add the two values together to obtain the overall value of ΔS°total.
5
Interpret the sign: a positive total shows that the reaction is thermodynamically feasible at that temperature.
Section 4 of 6Interpreting Results
Both positive
- The reaction is feasible at every temperature, because the total is positive throughout.
Both negative
- The reaction is not feasible at any temperature.
Opposing signs
- The temperature decides which term dominates, so the reaction may be feasible only at high or low temperature.
Rate
- A positive total shows only that a reaction can happen, not that it will be fast.
Section 5 of 6Worked Example
Question
- For N₂(g) + 3H₂(g) → 2NH₃(g), ΔS°system = −198.8 J K⁻¹ mol⁻¹ and ΔH = −92.0 kJ mol⁻¹ at 298 K.
- Find ΔS°total.
- ΔH = −92.0 × 1000 = −92 000 J mol⁻¹.
- ΔS°surroundings = −(−92 000) ÷ 298 = +308.7 J K⁻¹ mol⁻¹.
- ΔS°total = −198.8 + 308.7 = +109.9 J K⁻¹ mol⁻¹.
Answer ΔS°total = +109.9 J K⁻¹ mol⁻¹, so the reaction is feasible at 298 K
Section 6 of 6Exam Checklist & Tips
4.2.3 – Key Points for the Exam
- ΔS°surroundings The entropy change of the surroundings, equal to −ΔH ÷ T.
- ΔS°total The sum of the entropy changes of the system and the surroundings.
- Second law of thermodynamics Total entropy increases in any spontaneous change.
- Exothermic A reaction that releases heat to the surroundings.
- Feasible Able to occur in terms of energy and entropy, regardless of rate.
Common Mistakes
- Using ΔH in kJ and entropy in J without converting.
- Dropping the minus sign in ΔS°surroundings = −ΔH ÷ T.
- Using the temperature in °C rather than kelvin.
- Concluding that a reaction with positive ΔS°total will be fast.
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