(C) Given, $\Delta G^{\circ} = -25 \ kJ \ mol^{-1} = -25000 \ J \ mol^{-1}$.Temperature $T = 300 \ K$.Gas constant $R = 8.33 \ J \ mol^{-1} \ K^{-1}$.

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(C) Given, $\Delta G^{\circ} = -25 \ kJ \ mol^{-1} = -25000 \ J \ mol^{-1}$.Temperature $T = 300 \ K$.Gas constant $R = 8.33 \ J \ mol^{-1} \ K^{-1}$.The relationship between standard Gibbs free energy and equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K$.Substituting the values: $-25000 = -(8.33 \times 300) \ln K$.$-25000 = -2499 \ln K$.$\ln K = \frac{25000}{2499} \approx 10.004$.Therefore, $K = e^{10.004} \approx e^{10}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Standard free energy

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Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

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$e^{8}$
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$e^{9}$
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$e^{10}$
D
$e^{11}$

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Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

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For a system in equilibrium,$\Delta G = 0$ under conditions of constant:

The correct relationship between the standard free energy change in a reaction and the corresponding equilibrium constant $K_c$ is:

The $INCORRECT$ match in the following is

If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$

For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

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