Find out the value of the equilibrium constant for the following reaction at $298 \, K$.
$2 NH_{3(g)} + CO_{2(g)} \leftrightharpoons NH_{2}CONH_{2(aq)} + H_{2}O_{(l)}$
Standard Gibbs energy change,$\Delta_{r} G^{\ominus}$ at the given temperature is $-13.6 \, kJ \, mol^{-1}$.
$2 NH_{3(g)} + CO_{2(g)} \leftrightharpoons NH_{2}CONH_{2(aq)} + H_{2}O_{(l)}$
Standard Gibbs energy change,$\Delta_{r} G^{\ominus}$ at the given temperature is $-13.6 \, kJ \, mol^{-1}$.
(N/A) We know the relationship between the equilibrium constant $K$ and the standard Gibbs energy change $\Delta_{r} G^{\ominus}$ is given by:
$\Delta_{r} G^{\ominus} = -RT \ln K = -2.303 \, RT \log K$
Rearranging for $\log K$:
$\log K = \frac{-\Delta_{r} G^{\ominus}}{2.303 \, RT}$
Given:
$\Delta_{r} G^{\ominus} = -13.6 \, kJ \, mol^{-1} = -13.6 \times 10^{3} \, J \, mol^{-1}$
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
$T = 298 \, K$
Substituting the values:
$\log K = \frac{-(-13.6 \times 10^{3})}{2.303 \times 8.314 \times 298}$
$\log K = \frac{13600}{5705.84} \approx 2.3835$
$K = \text{antilog}(2.3835) \approx 2.42 \times 10^{2}$.
$\Delta_{r} G^{\ominus} = -RT \ln K = -2.303 \, RT \log K$
Rearranging for $\log K$:
$\log K = \frac{-\Delta_{r} G^{\ominus}}{2.303 \, RT}$
Given:
$\Delta_{r} G^{\ominus} = -13.6 \, kJ \, mol^{-1} = -13.6 \times 10^{3} \, J \, mol^{-1}$
$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
$T = 298 \, K$
Substituting the values:
$\log K = \frac{-(-13.6 \times 10^{3})}{2.303 \times 8.314 \times 298}$
$\log K = \frac{13600}{5705.84} \approx 2.3835$
$K = \text{antilog}(2.3835) \approx 2.42 \times 10^{2}$.
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DifficultJEE MAIN 2021
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