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Q.

For the reaction $C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g),$ the $K_{p} = 0.05 atm$ . The value of $Δ G^{\circ}$ of the reaction at 627 °C would be $(take p^{\circ} = 1 atm)$

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a

$11.19 kJ {mol}^{- 1}$

b

$22.40 kJ {mol}^{- 1}$

c

$33.57 kJ {mol}^{- 1}$

d

$27.98 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

$Δ G^{\circ} = - R T \ln K_{p}^{o} = - (8.314 {JK}^{- 1} {mol}^{- 1}) (900 K) (2.303) \log (0.05) = 22400 J {mol}^{- 1}$

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For the reaction C2H6(g)⇌C2H4(g)+H2(g), the Kp=0.05atm. The value of ΔG∘ of the reaction at 627 °C would be take⁡p∘=1 atm