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Simple Equilibria and Acid-Base Reactions

The equilibrium constant Kc

  • 3 things to know
  • Components 1 and 2

Overview

Recall questions on writing a Kc expression from a balanced equation, deducing its units by cancelling, substituting given equilibrium concentrations, what a large or small value indicates about the extent of reaction, and the one condition that changes the value.

What you need to know

Answer each one before you reveal it.

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Key term

Write the expression for Kc for the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Key term

Deduce the units of Kc for the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Numbers

Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g) in a mixture where [N₂O₄] = 0.050 mol dm⁻³ and [NO₂] = 0.020 mol dm⁻³.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write the expression for Kc for the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

    Model answer

    Kc = [NH₃]² ÷ ([N₂][H₂]³).

  2. 2

    Deduce the units of Kc for the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

    Model answer

    Two concentration terms on the top and four on the bottom leave mol⁻² dm⁶.

  3. 3

    Why does Kc for H₂(g) + I₂(g) ⇌ 2HI(g) have no units?

    Model answer

    There are two concentration terms on the top and two on the bottom, so the units cancel completely.

  4. 4

    An equilibrium mixture contains [H₂] = 0.20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³ and [HI] = 1.60 mol dm⁻³. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

    Model answer

    Kc = 1.60² ÷ (0.20 × 0.20) = 2.56 ÷ 0.040 = 64, with no units.

  5. 5

    Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g) in a mixture where [N₂O₄] = 0.050 mol dm⁻³ and [NO₂] = 0.020 mol dm⁻³.

    Model answer

    Kc = 0.020² ÷ 0.050 = 8.0 × 10⁻³ mol dm⁻³.

  6. 6

    An esterification mixture at equilibrium contains 0.33 mol dm⁻³ of ethanoic acid, 0.33 mol dm⁻³ of ethanol, 0.67 mol dm⁻³ of ethyl ethanoate and 0.67 mol dm⁻³ of water. Calculate Kc.

    Model answer

    Kc = (0.67 × 0.67) ÷ (0.33 × 0.33) = 4.1, and it has no units.

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