JEE Main 2021 · 27 August · evening · Numerical answer · +4 / −0
Data given for the following reaction is as follows :
FeO₍ₛ₎ + C (graphite) → Fe₍ₛ₎ + CO (g)
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
| Substance | Δ H° (kJ mol⁻¹) | Δ S° (J mol⁻¹ K⁻¹) |
|---|---|---|
| FeO₍ₛ₎ | - 266.3 | 57.49 |
| C_((graphite)) | 0 | 5.74 |
| Fe₍ₛ₎ | 0 | 27.28 |
| CO_((g)) | - 110.5 | 197.6 |
Answer and solution
Answer 964
Solution
Chemistry ma’am · 4 steps
Both ΔH and ΔS are positive here, so the reaction turns spontaneous only above T = ΔH/ΔS.
- Find ΔH. Take products minus reactants: Δ H° = (0 + (-110.5)) - ((-266.3) + 0) = +155.8 kJ mol⁻¹. The reaction is endothermic.
- Find ΔS. Products: 27.28 + 197.6 = 224.88. Reactants: 57.49 + 5.74 = 63.23. So Δ S° = 224.88 - 63.23 = +161.65 J K⁻¹ mol⁻¹, mostly from the gaseous CO formed.
- Set ΔG to zero. Since Δ G° = Δ H° - TΔ S°, the reaction becomes spontaneous once TΔ S° outweighs Δ H°. The crossover is where Δ G° = 0. Convert Δ H° to joules first.
T = (Δ H°)/(Δ S°) = (155800 J mol⁻¹)/(161.65 J K⁻¹ mol⁻¹) = 963.8 K
- So. The reaction must be above 963.8 K, so the minimum integer temperature is 964 K.
The common slip: 1. Dividing 155.8 (kJ) by 161.65 (J) without converting gives about 0.96, which rounds to 1, not 964.
The idea to keep. When ΔH and ΔS are both positive, the reaction becomes spontaneous above T = ΔH/ΔS. Always put ΔH in joules before dividing by ΔS.
Answer from the NTA final answer key. Solution by PrepWiser.