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5.3.6Additional lessons

Equilibrium Law Problem Solving (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 114 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5Setting Up a Table

5.3.6 Equilibrium Law Problem Solving (HL)

A table keeps the stoichiometry organised when the concentrations at equilibrium are unknown.

Initial
  • List the starting concentration of every species, using zero for any product that is not yet present.
Change
  • Use a variable x for the change in one species and write the others in proportion to the coefficients.
Equilibrium
  • Add the change to the initial value for each species to give its equilibrium concentration.
Substitute
  • Place the equilibrium terms into the Kc expression and solve for x.

Part 2 of 5Solving the Algebra

Solving the Algebra
Perfect square
  • If both sides of the equation are perfect squares, take the square root of both sides to avoid a quadratic.
Quadratic
  • Otherwise rearrange to the form ax² + bx + c = 0 and use the quadratic formula, rejecting any root that gives a negative concentration.
Small K
  • When K is very small, x is tiny compared with the initial value, so (initial − x) can be approximated as the initial value.
Check
  • Substitute the final concentrations back into the expression to confirm that it gives the original K.

Part 3 of 5Method for Any Problem

1
Write the balanced equation and the Kc expression.
2
Convert amounts in mol to concentrations in mol dm⁻³ before anything else.
3
Construct the table and express every equilibrium concentration using x.
4
Solve for x, ignoring physically impossible values, and calculate each equilibrium concentration.
5
Verify your answer by recalculating K, and give the result with appropriate significant figures.

Part 4 of 5Worked Example

Question

  • 1.00 mol H₂ and 1.00 mol I₂ are placed in a 1.00 dm³ vessel.
  • At equilibrium H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 50.
  • Calculate the equilibrium concentrations.

  1. Let x mol dm⁻³ of H₂ react: [H₂] = [I₂] = 1.00 − x and [HI] = 2x.
  2. Kc = (2x)² / (1.00 − x)² = 50, a perfect square, so 2x / (1.00 − x) = √50 = 7.07.
  3. 2x = 7.07 − 7.07x, so x = 7.07 / 9.07 = 0.780 mol dm⁻³.
  4. [HI] = 1.56 mol dm⁻³ and [H₂] = [I₂] = 0.22 mol dm⁻³. Check: 1.56² / 0.22² ≈ 50.

Answer

  • [HI] = 1.56 mol dm⁻³;
    • [H₂] = [I₂] = 0.22 mol dm⁻³

Part 5 of 5Exam Checklist & Tips

5.3.6 – Key Points for the Exam

  • ICE table A table of initial, change and equilibrium concentrations used to solve equilibrium problems.
  • Quadratic formula x = (−b ± √(b² − 4ac)) / 2a, used when an equation of the form ax² + bx + c = 0 appears.
  • Approximation Ignoring x compared with a much larger initial concentration when K is very small.
  • Stoichiometric ratio The ratio of coefficients that sets the proportion of the changes in the table.

Common Mistakes

  • Putting x only on the species that is used up and forgetting the coefficients for the others.
  • Substituting amounts in moles into Kc when the volume is not 1 dm³.
  • Keeping a negative root of a quadratic equation.
  • Using the approximation when K is not small.

Last updated:

3 & IGCSE —

Kc

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