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4.4.2Additional lessons

Calculating Standard Entropy Changes (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 86 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5The Calculation Formula

Equation
  • Equation ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants), with each value multiplied by its coefficient.
Data
  • Data Standard molar entropies are given in the data booklet or the question in J K⁻¹ mol⁻¹.
Elements
  • Elements Unlike ΔHf⦵, the standard molar entropy of an element is not zero, so all species are included.
Result
  • Result The answer is in J K⁻¹ and refers to the amounts in the balanced equation as written.

Part 2 of 5Step-by-Step Method

A sign check against gas moles catches most arithmetic slips.

1
Write a balanced equation including state symbols, since gases and solids have very different entropies.
2
List the S⦵ value for every species, including elements, and multiply by the coefficients.
3
Add the values for the products and add the values for the reactants separately.
4
Subtract the reactant total from the product total, keeping units in J K⁻¹ mol⁻¹.
5
Check the sign against the change in the number of moles of gas to confirm it is reasonable.

Part 3 of 5Judging the Answer

Judging the Answer
Positive ΔS⦵
  • Positive ΔS⦵ More gas moles or more dispersal in the products, such as in decomposition reactions.
Negative ΔS⦵
  • Negative ΔS⦵ Fewer gas moles, as in synthesis of ammonia or formation of a solid from gases.
Near zero
  • Near zero Equal moles of gas on both sides give a small ΔS⦵ whose sign is harder to predict.
Linking to ΔG
  • Linking to ΔG ΔS⦵ feeds into ΔG⦵ = ΔH⦵ − TΔS⦵ after being converted into kJ K⁻¹.

Part 4 of 5Worked Example

Question Calculate ΔS⦵ for N₂(g) + 3H₂(g) → 2NH₃(g) using S⦵ values (J K⁻¹ mol⁻¹): N₂ = 191.6, H₂ = 130.7, NH₃ = 192.5.

  1. Products2 × 192.5 = 385.0 J K⁻¹.
  2. Reactants191.6 + 3 × 130.7 = 191.6 + 392.1 = 583.7 J K⁻¹.
  3. ΔS⦵ = 385.0 − 583.7 = −198.7 J K⁻¹.

Answer ΔS⦵ = −198.7 J K⁻¹ (four moles of gas become two)

Part 5 of 5Exam Checklist & Tips

4.4.2 – Key Points for the Exam

  • ΔS⦵ The standard entropy change of a reaction, in J K⁻¹.
  • Σ The sum of the values for all products or all reactants, each multiplied by its coefficient.
  • Standard molar entropy Entropy of one mole of a substance at standard conditions.
  • Dispersal The spreading out of matter and energy among more particles and arrangements.

Common Mistakes

  • Forgetting to multiply each S⦵ value by its coefficient.
  • Setting S⦵ of elements to zero by analogy with enthalpy of formation.
  • Subtracting in the wrong order and getting the opposite sign.
  • Quoting the result in kJ without converting from J.

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