Lesson 11.1.1

11.1.1 Kp and partial pressures Quiz: Pearson Edexcel Chemistry, Unit 11

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 11.1.1, Kp and partial pressures: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 11: Equilibrium II, written with Revision Ninja.

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The 20 questions

  1. What is the partial pressure of a gas in a mixture?

    • The pressure of the gas at standard temperature only
    • The volume of the gas as a fraction of total volume
    • The total pressure of the mixture divided by the number of gases
    • The pressure that gas would exert if it alone occupied the container

  2. How is the partial pressure of a gas calculated from its mole fraction?

    • Partial pressure = mole fraction + total pressure
    • Partial pressure = mole fraction x molar mass
    • Partial pressure = total pressure / mole fraction
    • Partial pressure = mole fraction x total pressure

  3. For the equilibrium N2O4(g) ⇌ 2NO2(g), what is the expression for Kp?

    • Kp = p(NO2)^2 / p(N2O4)
    • Kp = p(NO2)^2 x p(N2O4)
    • Kp = p(NO2) / p(N2O4)
    • Kp = p(N2O4) / p(NO2)^2

  4. What are the usual units of Kp when partial pressures are given in atm?

    • Always mol dm-3, because partial pressures are converted to concentrations before Kp is calculated
    • Depends on the reaction, given as atm to the power of the change in gas moles
    • Always Pa, because pressure in the SI system is used throughout all gas equilibrium calculations
    • Always dimensionless, because the partial pressures are divided and the units cancel in every case

  5. In a mixture at total pressure 2.00 atm, NO2 has mole fraction 0.50. What is the partial pressure of NO2?

    • 1.00 atm
    • 4.00 atm
    • 0.50 atm
    • 2.00 atm

  6. For the equilibrium N2O4 ⇌ 2NO2, partial pressures at equilibrium are p(N2O4) = 0.50 atm and p(NO2) = 1.00 atm. What is Kp?

    • 0.50 atm
    • 0.67 atm
    • 2.00 atm
    • 4.00 atm

  7. For the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), partial pressures are p(SO2) = 0.40 atm, p(O2) = 0.20 atm and p(SO3) = 0.80 atm. What is Kp?

    • 20 atm-1
    • 2.5 atm
    • 0.05 atm-1
    • 0.40 atm-1

  8. Which statement about Kp in a heterogeneous equilibrium is correct?

    • Solids are included as their molar mass, which converts the expression into a mass ratio
    • Solids are included with their partial pressures, since they also exert a vapour pressure in the vessel
    • Solids and liquids are omitted from Kp because their activities are constant
    • Pure liquids are included as their mole fractions, which are multiplied together in the expression

  9. For the equilibrium CaCO3(s) ⇌ CaO(s) + CO2(g), what is the expression for Kp?

    • Kp = p(CaO) / p(CaCO3)
    • Kp = p(CaO) x p(CO2) / p(CaCO3)
    • Kp = p(CO2)
    • Kp = p(CaCO3) / p(CO2)

  10. A gaseous equilibrium has Kp = 4.0 atm for A(g) ⇌ 2B(g). If p(A) = 0.50 atm, what is p(B) at equilibrium?

    • 1.41 atm
    • 2.00 atm
    • 0.25 atm
    • 8.0 atm

  11. Which statement about Kp is correct?

    • Kp changes whenever the total pressure changes, so its value depends on the pressure of the mixture
    • Kp changes with the amount of catalyst added, so a larger catalyst mass gives a larger value
    • Kp is always equal to one for gases, because the partial pressures cancel in every equilibrium
    • Kp is a value for a given equilibrium at a fixed temperature

  12. What is the relationship between Kp and Kc for a gaseous reaction?

    • Kp = 1 / Kc always, because partial pressure is the reciprocal of concentration in every case
    • Kp = Kc only when the reaction has no gas, since the gas terms then cancel in the expression
    • Kp = Kc always, regardless of the change in gas moles, because the two constants are identical
    • Kp = Kc(RT)^change in moles of gas, with R and T in the appropriate units

  13. For the equilibrium 2NO(g) + O2(g) ⇌ 2NO2(g), what is the expression for Kp?

    • Kp = p(NO)^2 x p(O2) / p(NO2)^2
    • Kp = p(NO2)^2 / (p(NO)^2 x p(O2))
    • Kp = p(NO2) / (p(NO) x p(O2))
    • Kp = p(NO2)^2 / (p(NO) + p(O2))

  14. If the total pressure of an equilibrium mixture is 5.0 atm and the mole fraction of a gas is 0.20, what is its partial pressure?

    • 5.2 atm
    • 25 atm
    • 1.0 atm
    • 0.20 atm

  15. A gas mixture contains 0.40 mol of A and 0.60 mol of B in total. What is the mole fraction of A?

    • 0.67
    • 0.40
    • 0.60
    • 1.50

  16. For the equilibrium H2(g) + I2(g) ⇌ 2HI(g), what is the expression for Kp?

    • Kp = p(H2) x p(I2) / p(HI)^2
    • Kp = p(HI)^2 / (p(H2) x p(I2))
    • Kp = p(HI) / (p(H2) + p(I2))
    • Kp = p(HI)^2 / (p(H2) + p(I2))

  17. Why are partial pressures usually used for gaseous equilibria in industry?

    • They remove the need for temperature control, because partial pressures are unaffected by changes in heat
    • They directly relate the pressure of each gas to the position of equilibrium at fixed conditions
    • They allow the enthalpy change to be measured directly, without any need for a calorimeter
    • They show the mass of catalyst required, since the catalyst amount is calculated from the pressure

  18. A mixture at equilibrium at 3.0 atm contains equal mole fractions of two gases. What is the partial pressure of each gas?

    • 6.0 atm
    • 1.5 atm
    • 0.50 atm
    • 3.0 atm

  19. For a reaction where Kp has units of atm-1, which statement is consistent?

    • The number of moles of gaseous products is one more than reactants, so the units become atm to a positive power
    • The reaction has no gaseous species, so the Kp value cannot be defined for the equilibrium at all
    • The number of moles of gaseous products is one less than the number of moles of gaseous reactants
    • The reaction is at zero temperature, which is required for the units of atm-1 to arise

  20. What is the expression for Kp for the equilibrium 2SO3(g) ⇌ 2SO2(g) + O2(g)?

    • Kp = p(SO3) / (p(SO2) + p(O2))
    • Kp = p(SO2)^2 x p(O2) / p(SO3)^2
    • Kp = p(SO2) x p(O2) / p(SO3)
    • Kp = p(SO3)^2 / (p(SO2)^2 x p(O2))

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