UPSC Combined Geo-Scientist, Mains · 2024 · Mains Paper II
Q1(f). Relation between Kp, Kc and Kx
Question
Obtain a relation between equilibrium constants Kp, Kc and Kx. Under what conditions is Kp = Kc = Kx?
Answer in brief
Kp = Kc(RT)^Δn = Kx P^Δn; all three are equal when Δn = 0 (no change in the number of moles of gas).
Key points
- Ideal gas: pi = (ni/V)RT = ciRT ⇒ Kp = Kc(RT)^Δn, Δn = (c + d) − (a + b).
- Dalton's law: pi = xiP ⇒ Kp = Kx P^Δn.
- Condition: Δn = 0 ⇒ Kp = Kc = Kx, dimensionless and independent of pressure, e.g. H₂ + I₂ ⇌ 2HI.
- Δn ≠ 0, e.g. N₂ + 3H₂ ⇌ 2NH₃ (Δn = −2): Kx = KpP² rises with P, so more NH₃ at high pressure.
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.
CoversKp Kc Kx relationequilibrium constantΔn gasDalton's lawLe Chatelier principle
All parts of Question 1
- 1(a) Volume and critical temperature of two van der Waals gases5 marks
- 1(b) Second virial coefficient A₂ and the Boyle temperature5 marks
- 1(c) Conductors, semiconductors and insulators: band picture5 marks
- 1(d) Cubic and tetragonal Bravais lattices5 marks
- 1(e) Chemical potential and its change with temperature5 marks
- 1(f) Relation between Kp, Kc and Kx5 marks
- 1(g) Rate constant in dm³ mol⁻¹ s⁻¹5 marks
- 1(h) BET equation and its terms5 marks
- 1(i) Zinc–copper cell from two half-reactions5 marks
- 1(j) Alkaline (sodium) error of the glass electrode5 marks
- 1(k) Commutator of position and d/dx5 marks
- 1(l) Meaning and dimensions of |Ψ(x)|²5 marks
- 1(m) Raman effect against fluorescence and phosphorescence5 marks
- 1(n) H₂: no IR or microwave spectrum, but Raman active5 marks
- 1(o) Fluorescence rate constant of tryptophan5 marks
- 1(p) Why 511 nm light dissociates Br₂ but not H₂5 marks