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Difference between Internal Energy and Enthalpy

For the react...

Question

For the reaction at $25^{\circ} C N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$ , if $Δ G_{f}^{\circ}$ for $N_{2} O_{4}$ and $N O_{2}$ are 23.49 and 12.39 $K C a l . m o l^{- 1}$ , then $l o g (K_{p})$ for the reaction is:

A

-113.332

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B

-113.332

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C

-113.332

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D

-0.9393

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Solution

The correct option is D
-0.9393

$N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$

$△ G_{R e a c t i o n}^{\circ} = 2 \times △ G_{f o r m a t i o n o f N O_{2}}^{\circ} - △ G_{f o r m a t i o n o f N_{2} O_{4}}^{\circ}$

$△ G_{R e a c t i o n}^{\circ} = 2 \times (12.39) - (23.49) = 1.29 K . C a l . m o l^{- 1}$

$△ G_{R e a c t i o n}^{\circ} = - 2.303 \times R T \times l o g (K_{P})$

$1.29 = - 2.303 \times (2 \times 10^{- 3} \times 298 \times l o g (K_{p})$

$l o g (K_{P}) = - 0.9398$

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