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Simple Equilibria and Acid-Base Reactions

The equilibrium constant Kc

  • 3 things to know
  • AS Unit 1

Overview

Sixteen recall questions on the equilibrium constant Kc.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

Why does adding more of a reactant increase the yield of product without changing the value of Kc?

Numbers

Write the expression for Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) and deduce its units.

Key term

Write the expression for Kc for the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g).

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write the expression for Kc for the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g).

    Model answer

    Kc = [HI]² / ([H₂][I₂]), where each square bracket is the equilibrium concentration in mol dm⁻³.

  2. 2

    Write the expression for Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) and deduce its units.

    Model answer

    Kc = [NH₃]² / ([N₂][H₂]³). The units are (mol dm⁻³)² divided by (mol dm⁻³)⁴, which simplifies to mol⁻² dm⁶.

  3. 3

    Why does the equilibrium constant 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 of mol dm⁻³ cancel completely and Kc is simply a number.

  4. 4

    An equilibrium mixture of hydrogen, iodine and hydrogen iodide contains [H₂] = 0.0200 mol dm⁻³, [I₂] = 0.0200 mol dm⁻³ and [HI] = 0.140 mol dm⁻³. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

    Model answer

    Kc = [HI]² / ([H₂][I₂]) = (0.140)² / (0.0200 × 0.0200) = 0.0196 / (4.00 × 10⁻⁴) = 49.0. The units cancel, so Kc = 49.0 with no units.

  5. 5

    At a certain temperature an equilibrium mixture contains [N₂] = 0.500 mol dm⁻³, [H₂] = 1.00 mol dm⁻³ and [NH₃] = 0.200 mol dm⁻³. Calculate Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), including units.

    Model answer

    Kc = [NH₃]² / ([N₂][H₂]³) = (0.200)² / (0.500 × (1.00)³) = 0.0400 / 0.500 = 0.0800 mol⁻² dm⁶.

  6. 6

    2.00 mol of hydrogen and 2.00 mol of iodine are sealed in a 5.00 dm³ vessel. At equilibrium 3.20 mol of hydrogen iodide is present. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

    Model answer

    Forming 3.20 mol of HI uses 1.60 mol of H₂ and 1.60 mol of I₂, so 0.40 mol of each remains. Dividing by 5.00 dm³ gives [H₂] = [I₂] = 0.080 mol dm⁻³ and [HI] = 0.640 mol dm⁻³. Kc = (0.640)² / (0.080 × 0.080) = 0.4096 / (6.4 × 10⁻³) = 64. Kc = 64, with no units.

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