Gibbs free energyEdexcel A-Level Chemistry: Subtopic test

1. 1
   The Haber process, N₂(g) + 3H₂(g) → 2NH₃(g), has ΔH = −92 kJ mol⁻¹ and ΔS system = −199 J K⁻¹ mol⁻¹. A student uses these values, assumed to be constant with temperature, to investigate how feasible the reaction is at different temperatures.

   (a)

   What is ΔG for the Haber process at 298 K?

   [1 mark]

   • A+32.7 kJ mol⁻¹
   • B−151.3 kJ mol⁻¹
   • C+59.3 kJ mol⁻¹
   • D−32.7 kJ mol⁻¹

   (b)

   Above which temperature does ΔG for the Haber process become positive?

   [1 mark]

   • A298 K
   • B2.16 K
   • C462 K
   • D−462 K

   (c)

   The Haber process has a negative ΔG at 298 K but a mixture of nitrogen and hydrogen does not form ammonia at an observable rate at this temperature. Explain why.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Magnesium carbonate decomposes on heating: MgCO₃(s) → MgO(s) + CO₂(g). The reaction has ΔH = +117 kJ mol⁻¹ and ΔS system = +176 J K⁻¹ mol⁻¹. Both values are assumed to be constant with temperature.

   (a)

   What is ΔG for the decomposition at 298 K?

   [1 mark]

   • A−64.6 kJ mol⁻¹
   • B+64.6 kJ mol⁻¹
   • C+117 kJ mol⁻¹
   • D−117 kJ mol⁻¹

   (b)

   What is the minimum temperature at which the decomposition becomes feasible?

   [1 mark]

   • A665 K
   • B298 K
   • C1.50 K
   • D0.665 K

   (c)

   Explain why the decomposition becomes feasible at high temperatures.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A student relates the Gibbs free energy change of a reaction to its equilibrium constant, K, using ΔG = −RT ln K. She uses R = 8.31 J K⁻¹ mol⁻¹ and T = 298 K.

   (a)

   A reaction has ΔG = −32.7 kJ mol⁻¹ at 298 K. Calculate the value of K for this reaction.

   [3 marks]

   (b)

   A second reaction has K = 1.0 × 10⁻⁵ at 298 K. Calculate ΔG for this reaction and state what it shows about the feasibility of the reaction.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A teacher uses ΔG to discuss whether reactions occur. The conversion of diamond into graphite has ΔG = −2.9 kJ mol⁻¹ at 298 K, yet diamond jewellery lasts for millions of years without changing. Two other reactions, X and Y, both at 298 K, have ΔG values of −40 kJ mol⁻¹ and +40 kJ mol⁻¹ respectively. Use R = 8.31 J K⁻¹ mol⁻¹.

   (a)

   Explain why diamond does not change into graphite at 298 K even though ΔG for the change is negative.

   [6 marks]

   (b)

   Calculate the equilibrium constant for each of reactions X and Y at 298 K, and explain what the values show about the equilibrium composition of each reaction.

   [6 marks]

   Total for question 4: 12 marks

End of questions