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Thermodynamics question

2016 · 9 Apr · Shift 1 · Q12
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Thermodynamics question

2016 · 9 Apr · Shift 1 · Q12

JEE MainChemistryThermodynamicsMCQ+4 / −1
For the reaction, A(g) + B(g) →\to→ C(g) + D(g), Δ\DeltaΔ Ho and Δ\DeltaΔ So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
  1. A
    1.0 ×\times× 10 −-− 10
  2. B
    1.0 ×\times× 1010
  3. C
    10
  4. D
    1
View written solutionFree

Correct answer: D

  1. Use the Gibbs free energy relation

For a reaction at temperature TTT,

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circΔG∘=ΔH∘−TΔS∘

Given:

ΔH∘=−29.8 kJ mol−1\Delta H^\circ = -29.8\ \text{kJ mol}^{-1}ΔH∘=−29.8 kJ mol−1 ΔS∘=−0.100 kJ K−1mol−1\Delta S^\circ = -0.100\ \text{kJ K}^{-1}\text{mol}^{-1}ΔS∘=−0.100 kJ K−1mol−1 T=298 KT = 298\ \text{K}T=298 K
  1. Calculate ΔG∘\Delta G^\circΔG∘
ΔG∘=−29.8−298(−0.100)\Delta G^\circ = -29.8 - 298(-0.100)ΔG∘=−29.8−298(−0.100) ΔG∘=−29.8+29.8=0\Delta G^\circ = -29.8 + 29.8 = 0ΔG∘=−29.8+29.8=0

So,

ΔG∘=0\Delta G^\circ = 0ΔG∘=0
  1. Relate ΔG∘\Delta G^\circΔG∘ to equilibrium constant

We use:

ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln KΔG∘=−RTlnK

Since ΔG∘=0\Delta G^\circ = 0ΔG∘=0,

0=−RTln⁡K0 = -RT\ln K0=−RTlnK ln⁡K=0\ln K = 0lnK=0 K=e0=1K = e^0 = 1K=e0=1
  1. Match with the options

The equilibrium constant is:

K=1K = 1K=1

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

This matches the derived answer.

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