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5.3.3Additional lessons

The Equilibrium Constant — IB Chemistry HL Revision Notes & Topical Questions

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📚 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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Section 1 of 5Calculating K

Equilibrium constants are found by substituting measured concentrations into the expression.

Step one
  • Step one Write the balanced equation and the Kc expression for the equilibrium in question.
Step two
  • Step two Insert the equilibrium concentrations in mol dm⁻³, converting from moles by dividing by the volume where necessary.
Step three
  • Step three Evaluate the expression and work out the units by cancelling those of the concentrations.
Check
  • Check If the numbers are the same for powers in the top and bottom, K has no units.

Section 2 of 5What the Value Tells You

What the Value Tells You
K much greater than 1
  • K much greater than 1 The equilibrium mixture contains mostly products, so the reaction goes almost to completion.
K about 1
  • K about 1 Significant amounts of both reactants and products are present at equilibrium.
K much less than 1
  • K much less than 1 The mixture contains mostly reactants, so the reaction hardly proceeds.
Not about rate
  • Not about rate K says nothing about how fast equilibrium is reached, only about where the position lies.
Size of KPositionDominant species
K ≫ 1 (e.g. 10⁵)Far to the rightProducts
K ≈ 1Neither sideBoth
K ≪ 1 (e.g. 10⁻⁵)Far to the leftReactants

Section 3 of 5Temperature and K

1
The value of Kc changes only when the temperature changes; concentration, pressure and catalysts do not alter it.
2
For an exothermic forward reaction, raising the temperature decreases K because the equilibrium shifts to the left.
3
For an endothermic forward reaction, raising the temperature increases K because the equilibrium shifts to the right.
4
A large K does not mean a fast reaction, because the activation energy may still be very high.
5
The numerical value must always be quoted at a stated temperature.

Section 4 of 5Worked Example

Question

  • In a 2.00 dm³ flask at equilibrium there are 0.0800 mol N₂O₄ and 0.0240 mol NO₂.
  • Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g).

  1. [N₂O₄] = 0.0800 / 2.00 = 0.0400 mol dm⁻³ and [NO₂] = 0.0240 / 2.00 = 0.0120 mol dm⁻³.
  2. Kc = [NO₂]² / [N₂O₄] = (0.0120)² / 0.0400.
    What the symbols mean: Kc = equilibrium constant in terms of concentration · [ ] = concentration at equilibrium (mol dm−3) · each power is the number in front of that substance in the balanced equation
  3. Kc = 1.44 × 10⁻⁴ / 0.0400 = 3.6 × 10⁻³ mol dm⁻³, so the position lies well to the left.

Answer Kc = 3.6 × 10⁻³ mol dm⁻³

Section 5 of 5Exam Checklist & Tips

5.3.3 – Key Points for the Exam

  • Equilibrium constant A number that is constant at a given temperature and expresses the position of equilibrium.
  • Extent of reaction How far a reaction proceeds towards products before equilibrium is reached.
  • Position to the right An equilibrium mixture containing mostly products.
  • Temperature dependence K changes only with temperature.

Common Mistakes

  • Using the initial concentrations in the expression instead of the equilibrium concentrations.
  • Stating that a large K means a fast reaction.
  • Writing that increasing the pressure changes the value of K.
  • Giving K without the units, or with units that do not match the expression.

Last updated:

3 & IGCSE —

Kc

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