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Thermodynamics question

2019 · 11 Jan · Shift 2 · Q17
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Thermodynamics question

2019 · 11 Jan · Shift 2 · Q17

JEE MainChemistryThermodynamicsMCQ+4 / −1
The reaction, MgO(s) + C(s) →\to→ Mg(s) + CO(g), for which Δ\DeltaΔ rHo + 491.1 kJ mol–1 and Δ\DeltaΔ rSo = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaciton will be feasible is :
  1. A
    2480.3 K
  2. B
    2040.5 K
  3. C
    2380.5 K
  4. D
    1890.0 K
View written solutionFree

Correct answer: A

  1. Feasibility condition

A reaction becomes feasible when

ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0ΔG=ΔH−TΔS<0

The threshold temperature is obtained by setting

ΔG=0\Delta G = 0ΔG=0

So,

T=ΔHΔST = \frac{\Delta H}{\Delta S}T=ΔSΔH​
  1. Given data
ΔrH∘=+491.1 kJ mol−1\Delta_r H^\circ = +491.1\ \text{kJ mol}^{-1}Δr​H∘=+491.1 kJ mol−1 ΔrS∘=198.0 J K−1mol−1\Delta_r S^\circ = 198.0\ \text{J K}^{-1}\text{mol}^{-1}Δr​S∘=198.0 J K−1mol−1

Convert enthalpy to joules:

491.1 kJ mol−1=491100 J mol−1491.1\ \text{kJ mol}^{-1} = 491100\ \text{J mol}^{-1}491.1 kJ mol−1=491100 J mol−1
  1. Calculate threshold temperature
T=491100198.0T = \frac{491100}{198.0}T=198.0491100​ T=2480.3 KT = 2480.3\ \text{K}T=2480.3 K
  1. Interpretation

Since both ΔH>0\Delta H > 0ΔH>0 and ΔS>0\Delta S > 0ΔS>0, the reaction is feasible only at sufficiently high temperature, i.e.

T>2480.3 KT > 2480.3\ \text{K}T>2480.3 K

Hence, the reaction becomes feasible above 2480.3 K.

  1. Option check
  • A: 2480.3 K ✅
  • B: 2040.5 K ❌
  • C: 2380.5 K ❌
  • D: 1890.0 K ❌

Therefore, the correct option is A.

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