(C) The standard free energy change is given by the formula: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$. Given: $R = 8.314 \ J \ mol^{-1} \ K^{

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(C) The standard free energy change is given by the formula: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$. Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$, $T = 25 + 273 = 298 \ K$, and $K = 3.8 \times 10^{-3}$. Substituting the values: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log (3.8 \times 10^{-3})$. Since $\log (3.8 \times 10^{-3}) = -2.42$, we have: $\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times (-2.42)$. $\Delta G^{\circ} \approx 13817 \ J \ mol^{-1} \approx 13.8 \ kJ \ mol^{-1}$.

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

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If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

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A
$5.7 \ kJ \ mol^{-1}$
B
$9.9 \ kJ \ mol^{-1}$
C
$13.8 \ kJ \ mol^{-1}$
D
$15.6 \ kJ \ mol^{-1}$

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6-1.Equilibrium (Chemical Equilibrium)

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The $INCORRECT$ match in the following is

MediumJEE MAIN 2019

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At $320 \ K,$ a gas $A_2$ is $20 \%$ dissociated to $A_{(g)}.$ The standard free energy change at $320 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is approximately $(R = 8.314 \ J \ K^{-1} \ mol^{-1}; \ \ln \ 2 = 0.693; \ \ln \ 3 = 1.098).$

DifficultJEE MAIN 2018

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Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

DifficultMHT CET 2026

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At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

Difficult

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Calculate $\Delta G^\circ$ for the conversion of oxygen to ozone $\frac{3}{2} O_{2(g)} \to O_{3(g)}$ at $298 \ K$,if $K_p$ for this conversion is $2.47 \times 10^{-29}$.

Medium

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If the equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} C$, what is the standard free energy change of the process? $(R = 8.314 \ J \ mol^{-1} \ K^{-1}, \log 0.0038 = -2.42)$

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The $INCORRECT$ match in the following is

At $320 \ K,$ a gas $A_2$ is $20 \%$ dissociated to $A_{(g)}.$ The standard free energy change at $320 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is approximately $(R = 8.314 \ J \ K^{-1} \ mol^{-1}; \ \ln \ 2 = 0.693; \ \ln \ 3 = 1.098).$

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

At $227^{\circ} C$,dinitrogen tetraoxide is $60 \%$ dissociated. What is the standard free energy change at this temperature and at $1 \ atm$ pressure?

Calculate $\Delta G^\circ$ for the conversion of oxygen to ozone $\frac{3}{2} O_{2(g)} \to O_{3(g)}$ at $298 \ K$,if $K_p$ for this conversion is $2.47 \times 10^{-29}$.

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