Gibbs Free Energy - UNSOLVED PRACTICE SET

Q7. Define Gibbs free energy. Why is it called the 'thermodynamic potential'?

Q8. Derive the Gibbs-Helmholtz equation: ΔG = ΔH – TΔS. Explain the significance of each term.

Q9. Predict the sign of ΔG for a reaction where ΔH is positive and ΔS is negative. Is such a reaction ever spontaneous? Explain.

Q10. What is the significance of ΔG = 0? In what type of process does this condition occur?

Q11. A reaction has ΔH = +50 kJ/mol and ΔS = +120 J/K·mol at 298 K. Calculate ΔG and predict whether the reaction is spontaneous.

Q12. Explain why the decomposition of limestone (CaCO₃ → CaO + CO₂) is non-spontaneous at room temperature but becomes spontaneous at high temperatures.

Q13. (a) Define Gibbs free energy and derive the relationship ΔG = ΔH – TΔS.

(b) Explain how the sign of ΔG helps predict the spontaneity of a process under different conditions of ΔH and ΔS. Summarise your answer in a table.

(c) A reaction has ΔH = –85 kJ and ΔS = –120 J/K at 298 K. Calculate ΔG and predict spontaneity. What happens if the temperature is increased to 500 K?

Q14. (a) Derive the relationship between standard Gibbs free energy change (ΔG°) and the equilibrium constant (K): ΔG° = –RT ln K.

(b) For a reaction with K = 10 at 298 K, calculate ΔG°. Is the reaction spontaneous under standard conditions?

(c) If K < 1, what can you say about the spontaneity of the reaction under standard conditions? Explain.

Q15. (a) Explain the term 'standard Gibbs free energy of formation' (ΔG°f). Why is ΔG°f of elements in their standard states taken as zero?

(b) Calculate ΔG° for the following reaction at 298 K:

2NO(g) + O₂(g) → 2NO₂(g)

Given: ΔG°f [NO(g)] = 86.6 kJ/mol, ΔG°f [NO₂(g)] = 51.3 kJ/mol, ΔG°f [O₂(g)] = 0

(c) Is this reaction spontaneous under standard conditions? What does this imply about the stability of NO₂ compared to NO?

Q16. For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

Given: ΔH° = –92.4 kJ/mol, ΔS° = –198.3 J/K·mol at 298 K

(a) Calculate ΔG° at 298 K. Is the reaction spontaneous at this temperature?

(b) Calculate the temperature above which the reaction becomes non-spontaneous.

(c) The Haber process is carried out industrially at 450-500°C and 200 atm. Why is a higher temperature used despite the reaction being exothermic? Explain the industrial compromise.

Q17. For the reaction: 2SO₂(g) + O₂(g) → 2SO₃(g)

Given: ΔH° = –198.2 kJ, ΔS° = –187.9 J/K at 298 K

(a) Calculate ΔG° at 298 K and predict spontaneity.

(b) Calculate the equilibrium constant K at 298 K.

(c) In the contact process for manufacturing sulphuric acid, this reaction is carried out at 450°C with a catalyst. Calculate ΔG at 450°C (723 K) and comment on why the reaction is still feasible at this temperature.

(d) A student suggests carrying out the reaction at 1000 K to increase the rate. Would this be thermodynamically favourable? Calculate ΔG at 1000 K and explain.

(R = 8.314 J/mol·K)

Q18. In India, the production of urea (CO(NH₂)₂) is vital for agriculture. The synthesis of urea involves the reaction:

2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(l)

Given at 298 K:

ΔG°f [NH₃(g)] = –16.5 kJ/mol

ΔG°f [CO₂(g)] = –394.4 kJ/mol

ΔG°f [CO(NH₂)₂(s)] = –196.9 kJ/mol

ΔG°f [H₂O(l)] = –237.1 kJ/mol

(a) Calculate ΔG° for the synthesis of urea at 298 K.

(b) Is this reaction spontaneous under standard conditions? Explain.

(c) Urea is produced industrially at high pressure (150-250 atm). Explain how increasing pressure affects the spontaneity of this reaction using Le Chatelier's principle and Gibbs free energy concepts.

(d) Discuss why understanding ΔG is crucial for India's agricultural economy and food security.