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Q.

For the reaction at 25° C,
$C_{2} H_{4} (g) + H_{2} (g) ⇌ C_{2} H_{6} (g); K_{p} = 308.08$
What is $Δ G^{\circ}$ for this reaction in $kJ \times {mol}^{- 1}$ ?

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a

$- 14.2$

b

$- 19.2$

c

$- 0.436$

d

$- 3.71$

answer is D.

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Detailed Solution

Gibbs free energy,

$Δ G^{\circ} = - 2.303 R T \log (K_{p})$

$= \frac{- 2.303 \times 8.314 \times 298}{1000} \log (308.08)$

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