Indian Forest Service (Mains), Chemistry optional · 2025 · Paper I
Q1(c). Formation free energy of uranium hydride
Question
The equilibrium pressure of H₂(g) over solid uranium and UH₃ at 500 K is 139 Pa. Calculate the standard Gibbs energy of formation of UH₃(s) at 500 K and 1 atm.
Answer in brief
Using a 1 atm standard pressure, ΔfG°(UH₃, 500 K) = −41.10 kJ/mol.
Key points
- The formation reaction is U(s) + 3/2 H₂(g) → UH₃(s); activities of the pure solids are unity.
- The dimensionless equilibrium constant is K = (pH₂/p°)−3/2 = (139/101325)−3/2.
- ΔfG° = −RT ln K = (3/2)RT ln(139/101325) = (3/2)(8.314)(500) ln(0.00137182).
- This equals −41102 J/mol, or −41.10 kJ/mol; the low hydrogen pressure is consistent with favourable hydride formation.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversstandard Gibbs energy of formationuranium hydride UH₃equilibrium constantΔG° = −RT ln Kheterogeneous equilibrium
All parts of Question 1
- 1(a) Equivalent bonds in nitrate8 marks
- 1(b) Formula from occupied crystal voids8 marks
- 1(c) Formation free energy of uranium hydride8 marks
- 1(d) Pressure dependence of solid Gibbs energy8 marks
- 1(e) Fast and slow electrode systems8 marks