For the equilibrium: NiO(s) + CO(g) ⇌ Ni(s) + CO₂(g), ΔG°(cal/mol⁻¹) = -5320 - 5.6T (K). The temperature (in Kelvin) at which the gaseous mixture at equilibrium contains 400 ppm of CO, by mole, is [ln

Question

For the equilibrium: NiO(s) + CO(g) ⇌ Ni(s) + CO₂(g), ΔG°(cal/mol⁻¹) = -5320 - 5.6T (K). The temperature (in Kelvin) at which the gaseous mixture at equilibrium contains 400 ppm of CO, by mole, is [ln10 = 2.3, ln2 = 0.7]

Prashant Kotian · Accepted Answer

Correct answer: 0532. At equilibrium, [CO] = 400 ppm = 4 × 10⁻⁴. Since CO + CO₂ ≈ 1 (solids do not contribute to gaseous mole fraction), [CO₂] ≈ 1. K = P_CO₂ / P_CO ≈ 1 / (4 × 10⁻⁴) = 2500. ln K = ln(2500) = ln(10000/4) = 4 ln 10 - 2 ln 2 = 4(2.3) - 2(0.7) = 9.2 - 1.4 = 7.8. We know ΔG° = -RT ln K. Since ΔG° is given in cal, use R ≈ 2 cal/(mol·K). -5320 - 5.6T = -2 × T × 7.8 = -15.6T. 10.0T = 5320 => T = 532 K. Therefore, correct answer is 0532.

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