Calculate reaction entropy changes
Calculate standard reaction entropy from tabulated molar entropies of products and reactants.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 23. Use the syllabus sign convention for lattice enthalpy (gaseous ions forming the ionic solid); state phases, signs, units and temperature. Written practice is self-checked
Before you begin
- Use AS enthalpy changes and Hess's law, formulae for ions, balanced equations, and temperature in kelvins.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which order is used in the standard reaction entropy calculation?
What you will learn
- Use ΔS° = ΣS°(products) − ΣS°(reactants)
- Multiply each molar entropy by its balanced coefficient
- Use consistent units and check the sign qualitatively
Products minus reactants
For a reaction, ΔS° = ΣS°(products) − ΣS°(reactants). Multiply each standard molar entropy by its equation coefficient before summing. The result is the standard entropy change for the balanced reaction as written.
Standard molar entropies are commonly given in J K⁻¹ mol⁻¹. Keep all values in the same units, include phases, and use the exact balanced coefficients.
Use a qualitative check after the arithmetic
Compare the computed sign with the physical changes. A reaction that makes fewer gas molecules often has a negative ΔS, but phase changes also influence the result.
Do not subtract products from reactants. If the result disagrees with a strong qualitative pattern, recheck coefficients, phase data, arithmetic and the direction of the equation.
Pause and explain
In ΔS° for 2A(g) → B(g), how is S°(A) included?
Worked example
Calculate ΔS° for 2H₂(g) + O₂(g) → 2H₂O(l), given S° values H₂ = 131, O₂ = 205 and H₂O(l) = 70 J K⁻¹ mol⁻¹.
Show the worked solution
- Products: 2 × 70 = 140.
- Reactants: 2 × 131 + 205 = 467.
- Subtract reactants from products.
Answer ΔS° = 140 − 467 = −327 J K⁻¹ mol⁻¹ for the reaction as written.
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Calculate ΔS° for A(g) + B(g) → AB(g), with S°(A)=150, S°(B)=120 and S°(AB)=210 J K⁻¹ mol⁻¹.
Give me a hint
Use the product entropy minus both reactant entropies.
Compare my reasoning
- Product sum = 210.
- Reactant sum = 150 + 120 = 270.
- Subtract 270 from 210.
ΔS° = 210 − 270 = −60 J K⁻¹ mol⁻¹.
Look for these in your work
- Correct order
- Both reactants included
- Units and sign stated
Calculate ΔS° for 2NO₂(g) → N₂O₄(g), given S°(NO₂)=240 and S°(N₂O₄)=305 J K⁻¹ mol⁻¹.
Give me a hint
Multiply the NO₂ entropy by two, then subtract from the product value.
Compare my reasoning
- Products: 305.
- Reactants: 2×240 = 480.
- ΔS° = 305−480.
ΔS° = −175 J K⁻¹ mol⁻¹.
Look for these in your work
- Coefficient applied
- Products-minus-reactants used
- Result matches reduced gas count
A calculation gives ΔS° = +327 J K⁻¹ mol⁻¹ for 2H₂(g) + O₂(g) → 2H₂O(l) using the example values. Identify the likely sign error.
Give me a hint
Check the order of the two sums.
Compare my reasoning
- Products total 140.
- Reactants total 467.
- The correct subtraction is 140−467, not 467−140.
The subtraction order was reversed. ΔS° = 140−467 = −327 J K⁻¹ mol⁻¹.
Look for these in your work
- Sums checked
- Error located
- Correct sign and calculation supplied
Common mistakes
- Subtracting products from reactants
- Omitting stoichiometric coefficients
- Combining J and kJ without conversion
- Dropping state labels or the per-mole basis
What can you explain now?
Which order is used in the standard reaction entropy calculation?
Compare with the explanation
Products minus reactants
ΔS° = ΣS°(products) − ΣS°(reactants).
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Make it stick
How do coefficients, phases and the products-minus-reactants rule work together in a ΔS° calculation?
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