Lesson 3.1.8.2

3.1.8.2 Gibbs free-energy change, ∆G, and entropy change, ∆S Quiz: AQA Chemistry, Unit 1

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 3.1.8.2, Gibbs free-energy change, ∆G, and entropy change, ∆S: 20 multiple choice questions for the AQA Chemistry (7405), Unit 1: Physical chemistry, written with Revision Ninja.

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The 20 questions

  1. For a reaction to be feasible, what must be true of the Gibbs free-energy change, delta G?

    • It must be zero or negative
    • It must be exactly equal to the enthalpy change
    • It must be greater than the entropy change in value
    • It must be positive

  2. Which relationship links the Gibbs free-energy change, enthalpy change and entropy change?

    • delta G = delta H / T + delta S
    • delta G = T delta H - delta S
    • delta G = delta H + T delta S
    • delta G = delta H - T delta S

  3. For NH3 synthesis, delta H = -92 kJ/mol and delta S = -198 J K^-1 mol^-1. What is delta G at 298 K?

    • About +33 kJ/mol
    • About -59 kJ/mol
    • About -92 kJ/mol
    • About -33 kJ/mol

  4. For NH3 synthesis with delta H = -92 kJ/mol and delta S = -198 J K^-1 mol^-1, above what temperature does the reaction become non-feasible?

    • About 298 K
    • About 192 K
    • About 465 K
    • About 930 K

  5. Using standard entropy values for CaCO3 = 93, CaO = 40 and CO2 = 214 J K^-1 mol^-1, what is the entropy change for CaCO3 -> CaO + CO2?

    • -161 J K^-1 mol^-1
    • +347 J K^-1 mol^-1
    • +161 J K^-1 mol^-1
    • +54 J K^-1 mol^-1

  6. For CaCO3 -> CaO + CO2 with delta H = +178 kJ/mol and delta S = +161 J K^-1 mol^-1, what is the minimum temperature at which the decomposition is feasible?

    • About 161 K
    • About 1106 K
    • About 2 K
    • About 178 K

  7. Which process has a positive entropy change?

    • Water evaporating to form steam at 100 C
    • Water freezing to form ice at 0 C
    • Gaseous nitrogen and hydrogen combining to form solid ammonium chloride
    • Gaseous ammonia dissolving to form a more ordered solid

  8. Why is the enthalpy change alone not sufficient to explain whether a change is feasible?

    • The enthalpy change applies only to reactions in aqueous solution, so it cannot be used for gases or solids in the cycle
    • Entropy is always negative for feasible reactions, so the enthalpy change must be ignored in every calculation at all times
    • Feasibility also depends on the entropy change and the temperature, through delta G = delta H - T delta S
    • The enthalpy change is always zero for any feasible reaction at all temperatures, so it cannot be used in any calculation

  9. For a reaction with delta H = -90 kJ/mol and delta S = -150 J K^-1 mol^-1, the reaction is feasible below about 600 K. What is the reason?

    • delta G is zero at T = delta S/delta H = 0.0017 K, so the reaction is feasible at almost all temperatures
    • delta G is zero at T = delta H/delta S = 600 K, and delta G is negative below this temperature
    • delta G is positive below 600 K, so the reaction is feasible only above this temperature
    • delta G is always negative for any reaction with a negative entropy change at all temperatures

  10. For a reaction with delta H = -50 kJ/mol and delta S = -100 J K^-1 mol^-1 at 300 K, what is delta G?

    • -50 kJ/mol
    • -20 kJ/mol
    • +20 kJ/mol
    • -80 kJ/mol

  11. Which pair of signs for delta H and delta S gives a reaction that is feasible only at high temperatures?

    • delta H positive and delta S positive
    • delta H negative and delta S positive
    • delta H positive and delta S negative
    • delta H negative and delta S negative

  12. Which pair of signs gives a reaction that is feasible only at low temperatures?

    • delta H negative and delta S negative
    • delta H negative and delta S positive
    • delta H positive and delta S positive
    • delta H positive and delta S negative

  13. Estimate the entropy change for vaporising water at its boiling point of 373 K, given delta Hvap = +40.7 kJ/mol.

    • About +0.109 J K^-1 mol^-1
    • About +273 J K^-1 mol^-1
    • About +109 J K^-1 mol^-1
    • About +40.7 J K^-1 mol^-1

  14. On a graph of delta G against temperature T, what does the gradient of the straight line represent?

    • delta H
    • delta S
    • delta G
    • -delta S

  15. A reaction has delta H = -40 kJ/mol and delta S = -100 J K^-1 mol^-1. What is delta G at 400 K?

    • -80 kJ/mol
    • 0 kJ/mol
    • +80 kJ/mol
    • -40 kJ/mol

  16. A reaction has delta H = +50 kJ/mol and delta S = +0.1 kJ K^-1 mol^-1. At what temperature does delta G = 0?

    • 500 K
    • 1000 K
    • 50 K
    • 5 K

  17. Which expression correctly rearranges delta G = delta H - T delta S to give the entropy change?

    • delta S = T (delta H - delta G)
    • delta S = (delta G - delta H)/T
    • delta S = (delta H - delta G)/T
    • delta S = (delta H + delta G)/T

  18. A reaction has delta H = -90 kJ/mol and delta S = -150 J K^-1 mol^-1. Which statement is correct?

    • It is feasible below about 600 K, because delta G is zero or negative at those temperatures
    • It is feasible at all temperatures, because both delta H and delta S are negative
    • It is never feasible, because a negative entropy change always makes delta G positive
    • It is feasible above about 600 K, because delta G is zero or negative at those temperatures

  19. When delta H is in kJ/mol and delta S is in J K^-1 mol^-1, what must be done before using delta G = delta H - T delta S?

    • Convert the temperature to Celsius only, leaving delta S in J K^-1 mol^-1
    • Nothing, because the units cancel automatically in the equation
    • Convert delta S to kJ K^-1 mol^-1 by dividing by 1000
    • Multiply delta S by 1000 and convert the temperature to Celsius

  20. A reaction has delta H = -120 kJ/mol and delta S = +200 J K^-1 mol^-1. Which statement is correct?

    • The reaction is feasible only above about 600 K, because delta G is zero at that temperature
    • The reaction is never feasible, because a positive entropy change always makes delta G positive
    • The reaction is feasible only below about 0.6 K, because delta G is zero at that temperature
    • The reaction is feasible at all positive temperatures, because delta G is negative at every such temperature

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