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4.4.5Unit 4. Rates, Equilibria and Further Organic Chemistry

Equilibrium Constant and Entropy — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 33 of 56 Unit quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
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Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
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Section 1 of 5Total Entropy Change

A reaction is feasible when the total entropy of the system plus surroundings increases.

System entropy
  • System entropy ΔS(system) is the difference between the entropy of the products and that of the reactants in the equation.
Surroundings entropy
  • Surroundings entropy ΔS(surroundings) = −ΔH / T, so an exothermic reaction increases the entropy of the surroundings.
Total
  • Total ΔS(total) = ΔS(system) + ΔS(surroundings), and a reaction is thermodynamically feasible if this is positive.
Equilibrium
  • Equilibrium At equilibrium the system is balanced between forward and reverse reactions, and K describes how far the reaction goes.

Section 2 of 5Linking to K

The two ideas are connected by one equation.

Linking to K
Equation
  • Equation ΔS(total) = R ln K, where R is 8.31 J K⁻¹ mol⁻¹ and K is Kc or Kp as appropriate.
K greater than 1
  • K greater than 1 ln K is positive, so ΔS(total) is positive and the reaction lies mainly toward the products.
K less than 1
  • K less than 1 ln K is negative, so ΔS(total) is negative and the equilibrium mixture contains mostly reactants.
K close to 1
  • K close to 1 ΔS(total) is close to zero, so appreciable amounts of reactants and products are present.

Section 3 of 5Calculating

1
Calculate ΔS(total) in J K⁻¹ mol⁻¹, converting ΔH from kJ to J when finding ΔS(surroundings).
2
To find K, rearrange to ln K = ΔS(total) / R and then take the exponential of both sides.
3
To find ΔS(total) from a known K, take the natural logarithm of K and multiply by R.
4
Give K with units if appropriate and state the answer to three significant figures.

Section 4 of 5Worked Example

Question

  • For a reaction at 298 K, K = 2.0 × 10³.
  • Calculate ΔS(total) and then calculate K for a different reaction with ΔS(total) = +25.0 J K⁻¹ mol⁻¹.
  • Use R = 8.31 J K⁻¹ mol⁻¹.

  1. First reactionΔS(total) = R ln K = 8.31 × ln(2.0 × 10³) = 8.31 × 7.601.
  2. Evaluate8.31 × 7.601 = 63.2 J K⁻¹ mol⁻¹.
  3. Second reactionln K = 25.0 / 8.31 = 3.008, so K = e3.008 = 20.3.

Answer

  • ΔS(total) = +63.2 J K⁻¹ mol⁻¹ for the first reaction;
    • K = 20.3 for the second.

Section 5 of 5Exam Checklist & Tips

4.4.5 – Key Points for the Exam

  • Entropy A measure of the dispersal of energy and the number of ways particles can be arranged.
  • ΔS(total) Sum of the entropy changes of system and surroundings.
  • Natural logarithm Logarithm to base e, written ln on a calculator.
  • Gas constant R 8.31 J K⁻¹ mol⁻¹.

Common Mistakes

  • Using kJ instead of J when calculating entropy values.
  • Using log₁₀ instead of ln in ΔS(total) = R ln K.
  • Forgetting the minus sign in ΔS(surroundings) = −ΔH / T.
  • Treating a positive ΔS(total) as proof that the reaction is fast.

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3 & IGCSE —

Kc

  • : Kc! (ΔH < 0) Kc. (ΔH > 0) Kc.
  • : ( ).
  • : Kc .
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