Find out the value of equilibrium constant for the following reaction at 298 K. $2 N H_{3 (g)} + C O_{2 (g)} ⇌ N H_{2} C O N H_{2 (a q)} + H_{2} O_{(l)}$
Standard Gibbs energy change $Δ_{r} G^{\circ}$ at the given temperature is $- 13.6 K J m o l^{- 1}$

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240

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2400

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Solution

The correct option is C 240
We know logK
$R = 8.314 J / (K m o l)$
$= \frac{- Δ_{r} G^{\circ}}{2.303 R T} = \frac{(- 13.6 \times 10^{3} J m o l^{- 1})}{2.303 (8.314 J K^{- 1} m o l^{- 1})} = \frac{2.38}{(298 K)}$
$H e n c e K = a n t i l o g 2.38$
$= 10^{2.38} = 2.4 \times 10^{2}$
$= 240$

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