Lesson 13B.3

13B.3 Free energy, feasibility and equilibrium Quiz: Pearson Edexcel Chemistry, Unit 13

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 13B.3, Free energy, feasibility and equilibrium: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 13: Energetics II, written with Revision Ninja.

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The 20 questions

  1. Which equation defines the Gibbs free energy change?

    • dG = dH divided by T dSsystem
    • dG = T dSsystem - dH
    • dG = dH - T dSsystem
    • dG = dH + T dSsystem

  2. What value of dG indicates that a reaction is feasible?

    • A dG equal to dH
    • A positive value of dG
    • A dG of exactly zero at all temperatures
    • A negative value of dG

  3. What does a large value of the equilibrium constant K indicate?

    • The equilibrium lies far towards the reactants, consistent with a positive dG
    • The reaction must be endothermic by definition
    • K is unaffected by dG
    • The equilibrium lies far towards the products, consistent with a negative dG

  4. At what temperature is a reaction at equilibrium, with dG = 0?

    • T = dS / dH
    • T = dH / dS
    • T = dH x dS
    • T = dG / dH

  5. Why can some endothermic reactions occur at room temperature?

    • Endothermic reactions absorb entropy from the surroundings only
    • Endothermic reactions always have negative entropy changes
    • The entropy term T dS can make dG negative even though dH is positive
    • Enthalpy is ignored when the temperature is low

  6. Why might a reaction with a negative dG not occur in practice?

    • The entropy of the reactants is always zero
    • The reaction is at equilibrium, so no net change happens
    • The dG value is wrongly signed and must always be positive
    • Kinetic factors, such as a high activation energy, can prevent the reaction from proceeding at a measurable rate

  7. What units must dH and T dS have in the equation dG = dH - T dS?

    • dH in K and T dS in kJ, because temperature is in the enthalpy term
    • Both must be in kJ mol-1, so T dS must be converted from J to kJ
    • Both must be in J K-1, so dH is converted to kelvin
    • dG must be dimensionless, so the units are ignored

  8. For dH = +180 kJ mol-1 and dS = +160 J K-1 mol-1, at what temperature does dG become zero?

    • 340 K
    • 0.889 K
    • 1125 K
    • 28800 K

  9. For dH = -92 kJ mol-1 and dS = -198 J K-1 mol-1 at 298 K, what is dG and is the reaction feasible?

    • +151 kJ mol-1, so the reaction is not feasible at 298 K
    • +33.0 kJ mol-1, so the reaction is not feasible at 298 K
    • -151 kJ mol-1, so the reaction is feasible at 298 K
    • -33.0 kJ mol-1, so the reaction is feasible at 298 K

  10. For dH = +50 kJ mol-1 and dS = +100 J K-1 mol-1 at 298 K, what is dG?

    • -20.2 kJ mol-1, so the reaction is feasible at 298 K
    • +50.0 kJ mol-1, so the reaction is not feasible at 298 K
    • -50.0 kJ mol-1, so the reaction is feasible at 298 K
    • +20.2 kJ mol-1, so the reaction is not feasible at 298 K

  11. For dH = +50 kJ mol-1 and dS = +100 J K-1 mol-1, above what temperature is the reaction feasible?

    • Below 500 K, because the entropy term is then larger
    • Above 298 K only, because the temperature is then fixed
    • Above 0.5 K, because dS is in J and dH in kJ
    • Above 500 K, because dG becomes negative only when T dS exceeds dH

  12. For dG = -10 kJ mol-1 at 298 K, what is the equilibrium constant K approximately?

    • About 2.5 x 10^3
    • About 57
    • About 0.018
    • About 4.0

  13. For dG = +5 kJ mol-1 at 298 K, what is K approximately?

    • About 0.13
    • About 0.0067
    • About 1.0
    • About 7.5

  14. For a reaction with dH negative and dS negative, at what temperatures is it feasible?

    • At low temperatures, because the T dS term is small so dH dominates dG
    • At all temperatures, because dH is negative
    • At high temperatures only, because dS is negative
    • At no temperature, since both terms are negative

  15. For dH = -92 kJ mol-1 and dS = -198 J K-1 mol-1, what is dG at 500 K?

    • +7.0 kJ mol-1, so the reaction is not feasible at 500 K
    • +191 kJ mol-1, so the reaction is not feasible at 500 K
    • -191 kJ mol-1, so the reaction is feasible at 500 K
    • -7.0 kJ mol-1, so the reaction is feasible at 500 K

  16. For dH = -92 kJ mol-1 and dS = -198 J K-1 mol-1, what is the temperature at which dG = 0?

    • About 0.46 K
    • About 2150 K
    • About 465 K
    • About 298 K

  17. A student says a feasible reaction will occur quickly. Which evaluation is best?

    • Correct, because feasibility and rate are the same measure
    • Correct, because a negative dG means the reaction proceeds quickly
    • Incorrect, because feasible reactions are always at equilibrium and never proceed
    • Incorrect, because feasibility from dG says nothing about rate, which depends on kinetic factors such as activation energy

  18. At 298 K, which value of dG corresponds to an equilibrium constant K of 1.0 x 10^5?

    • dG = -2.9 kJ mol-1 corresponds to K = 10^5 at 298 K
    • dG = -0.23 kJ mol-1 corresponds to K = 10^5 at 298 K
    • dG = -28.5 kJ mol-1 corresponds to K = 10^5 at 298 K
    • dG = +28.5 kJ mol-1 corresponds to K = 10^5 at 298 K

  19. When dG = 0 for a reaction, what is the value of the equilibrium constant?

    • K = 0, because no products form
    • K is infinite, so the reaction goes to completion
    • K = 298, the temperature in kelvin
    • K = 1, so reactants and products are equally favoured at equilibrium

  20. Which type of reaction becomes feasible only at high temperature?

    • An exothermic reaction with positive dS, which is feasible at all temperatures
    • An exothermic reaction with negative dS, which is feasible at high temperature
    • An endothermic reaction with negative dS, which is feasible at high temperature
    • An endothermic reaction with positive dS, since T dS grows with temperature until it outweighs dH

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