Lesson 3.1.6.2

3.1.6.2 Equilibrium constant Kc for homogeneous systems Quiz: AQA Chemistry, Unit 1

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

In partnership with Revision Ninja

Lesson 3.1.6.2, Equilibrium constant Kc for homogeneous systems: 20 multiple choice questions for the AQA Chemistry (7405), Unit 1: Physical chemistry, written with Revision Ninja.

Host it live on the board and students join with a game code on their own devices, or revise alone with Free Play. The answers are revealed in the game.

Host this setFree Play

The 20 questions

  1. Which expression is the correct Kc for the homogeneous equilibrium 2A(g) + B(g) <=> C(g)?

    • Kc = [C]^2/([A] [B])
    • Kc = [A][B]/[C]
    • Kc = [C]/([A]^2 [B])
    • Kc = [A]^2 [B]/[C]

  2. Which expression is the correct Kc for the equilibrium N2O4(g) <=> 2NO2(g)?

    • Kc = [NO2]/[N2O4]
    • Kc = [N2O4]/[NO2]^2
    • Kc = [NO2]^2/[N2O4]
    • Kc = [NO2]^2 [N2O4]

  3. At equilibrium for H2(g) + I2(g) <=> 2HI(g), [H2] = 0.10, [I2] = 0.10 and [HI] = 0.80 mol dm^-3. What is Kc?

    • 16
    • 0.016
    • 64
    • 8

  4. For 2SO2(g) + O2(g) <=> 2SO3(g), at equilibrium [SO2] = 0.20, [O2] = 0.10 and [SO3] = 0.50 mol dm^-3. What is Kc?

    • 62.5 dm^3 mol^-1
    • 250 dm^3 mol^-1
    • 0.4 dm^3 mol^-1
    • 2.5 dm^3 mol^-1

  5. For A + B <=> C + D, the equilibrium concentrations are [A] = 0.50, [B] = 0.20, [C] = 0.40 and [D] = 0.10 mol dm^-3. What is Kc?

    • 2.5
    • 0.4
    • 4.0
    • 0.04

  6. The equilibrium PCl5(g) <=> PCl3(g) + Cl2(g) has [PCl5] = 0.20, [PCl3] = 0.40 and [Cl2] = 0.40 mol dm^-3. What is Kc?

    • 0.8 mol dm^-3
    • 1.6 mol dm^-3
    • 0.4 mol dm^-3
    • 0.2 mol dm^-3

  7. In an equilibrium 2HI <=> H2 + I2 in a 2 dm^3 vessel, the amounts at equilibrium are 0.4 mol HI, 0.1 mol H2 and 0.1 mol I2. What is Kc?

    • 0.0625
    • 0.0156
    • 0.0313
    • 0.2500

  8. For N2(g) + 3H2(g) <=> 2NH3(g), what are the units of Kc?

    • mol dm^-3
    • No units, because Kc is a pure number
    • dm^6 mol^-2
    • dm^3 mol^-1

  9. For the equilibrium A + B <=> 2C, which of the following is the units of Kc?

    • dm^3 mol^-1, because the product concentration is squared while each reactant concentration appears only once
    • mol^2 dm^-6, because the square of the product concentration is divided by the product of the reactant concentrations
    • No units, because the number of moles of C equals the sum of A and B
    • mol dm^-3, because the expression has one net mole of gas more on the product side than on the reactant side

  10. Which change will change the numerical value of Kc for a reversible reaction?

    • A change in temperature, because the value of Kc depends on temperature
    • A change in the concentration of one of the reactants, at constant temperature
    • A change in the total volume of the mixture at constant temperature
    • The addition of a catalyst, at constant temperature and pressure

  11. For an exothermic reversible reaction, what happens to the value of Kc when the temperature is increased?

    • Kc stays the same, because only the rate of reaction changes with temperature
    • Kc decreases, because the equilibrium shifts towards the reactants
    • Kc becomes zero, because the reactants are fully consumed at high temperature
    • Kc increases, because the equilibrium shifts towards the products

  12. A student says that adding more reactant to an equilibrium mixture increases the value of Kc. Which response is correct?

    • Correct, because a higher reactant concentration always increases the value of Kc at all temperatures
    • Incorrect, because Kc depends only on temperature, and the equilibrium shifts to restore the same Kc
    • Partly correct, because Kc rises with reactant concentration only in gas-phase equilibria
    • Correct, because Kc is the ratio of reactants to products and so rises with more reactant

  13. In the Kc expression for a reaction, what happens to the value of Kc if the equilibrium position moves further to the right?

    • Kc increases, because the numerator product concentrations grow relative to the reactants at the same temperature
    • Kc becomes negative, because the reverse reaction has started to dominate the system and reverses the sign of the constant
    • Kc decreases, because the reactant concentration in the denominator becomes smaller as the equilibrium moves in the reverse direction
    • Kc is unchanged, because movement along the equilibrium position does not change the value of the equilibrium constant at all

  14. A reaction is catalysed. Which statement about the equilibrium constant Kc is correct?

    • Kc increases, because the catalyst makes the forward reaction faster than the reverse
    • Kc is unchanged, because the catalyst speeds up the forward and reverse reactions equally
    • Kc decreases, because the catalyst lowers the number of molecules available to react
    • Kc increases, because a catalyst always shifts the equilibrium towards the products

  15. The equilibrium concentration of ester in the esterification of ethanol and ethanoic acid is found using Kc. Why must the Kc expression be written with the correct stoichiometric powers?

    • The powers are needed only when the reaction is at a high temperature, not in aqueous solution
    • The powers are used to convert concentration into partial pressure before the Kc expression is written
    • Each concentration term must be raised to its coefficient, so the expression reflects the balanced equation
    • Each concentration term must be raised to the power of the total moles, so all species are counted equally

  16. For the equilibrium 2A(g) + B(g) <=> 2C(g) + D(g), which is the correct expression for Kc?

    • Kc = [C]^2 [D]^2/([A]^2 [B]^2)
    • Kc = [C][D]/([A][B])
    • Kc = [C]^2 [D]/([A]^2 [B])
    • Kc = [A]^2 [B]/([C]^2 [D])

  17. A solution of ethanol and ethanoic acid reaches equilibrium with Kc about 4 at a fixed temperature. If the initial moles of reactants are doubled, what happens to Kc?

    • Kc becomes 8, because the number of moles of each reactant is multiplied
    • Kc remains about 4, because Kc depends only on temperature for this equilibrium
    • Kc doubles, because the amount of each reactant has doubled
    • Kc halves, because the extra reactant reduces the equilibrium constant

  18. For N2O4(g) <=> 2NO2(g), a vessel has [N2O4] = 0.040 and [NO2] = 0.040 mol dm^-3 at equilibrium. What is Kc?

    • 25 mol dm^-3
    • 0.0016 mol dm^-3
    • 1.0 mol dm^-3
    • 0.040 mol dm^-3

  19. A mixture contains 0.30 mol of A and 0.30 mol of B in 1.0 dm^3 and reaches equilibrium with 0.10 mol of C for A + B <=> C. What is Kc?

    • 0.10 dm^3 mol^-1
    • 0.033 dm^3 mol^-1
    • 2.5 dm^3 mol^-1
    • 10 dm^3 mol^-1

  20. For N2O4(g) <=> 2NO2(g), Kc = 0.50 mol dm^-3 at a fixed temperature. If [N2O4] = 0.20 mol dm^-3 at equilibrium, what is [NO2]?

    • 0.32 mol dm^-3
    • 2.5 mol dm^-3
    • 0.40 mol dm^-3
    • 0.10 mol dm^-3

All AQA Chemistry quizzes