JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
The variation of equilibrium constant with temperature is given below :
The values of ΔHᵒ, ΔGᵒ at
T₁ and ΔGᵒ at T₂ (in kJ mol –1 ) respectively, are close to :
[Use R = 8.314 J K –1 mol –1 ]
| Temperature | Equilibrium Constant |
|---|---|
| T₁ = 25ᵒC | K₁ = 10 |
| T₂ = 100ᵒC | K₂ = 100 |
- (a) 28.4, –5.71 and –14.29
- (b) 0.64, –7.14 and –5.71
- (c) 28.4, –7.14 and –5.71
- (d) 0.64, –5.71 and –14.29
Answer and solution
Answer (a) 28.4, –5.71 and –14.29
Solution
Chemistry ma’am · 4 steps
Change both temperatures to kelvin first, then use van't Hoff for ΔH and -RT ln K for each ΔG.
- Kelvin first. T₁ = 25 + 273 = 298 K and T₂ = 100 + 273 = 373 K. Every thermodynamic formula here needs absolute temperature.
ln K₂/K₁ = (Δ H°)/R(1/T₁-1/T₂)
- Find ΔH. ln 100/10 = ln 10 = 2.303, and 1/298-1/373 = 75/111154 = 6.75 × 10⁻⁴ K⁻¹. So Δ H° = (8.314 × 2.303)/(6.75 × 10⁻⁴) = 2.84 × 10⁴ J mol⁻¹ ≈ 28.4 kJ mol⁻¹.
- Find each ΔG. Use Δ G° = -RT ln K with the K that belongs to that temperature. At T₁: -8.314 × 298 × 2.303 = -5.70 × 10³ J mol⁻¹ ≈ -5.71 kJ mol⁻¹. At T₂: -8.314 × 373 × 4.605 = -1.43 × 10⁴ J mol⁻¹ ≈ -14.3 kJ mol⁻¹.
- So. The values are about 28.4, −5.71 and −14.29 kJ mol⁻¹, which is option (a). This makes sense: a bigger K at T₂ gives a more negative Δ G°.
The idea to keep. K rising with temperature means the reaction is endothermic. Always use kelvin in van't Hoff, and pair each Δ G° = -RT ln K with its own T and its own K.
Answer from the NTA final answer key. Solution by PrepWiser.