(A) The relationship between standard Gibbs energy change $(\Delta G^\circ)$ and equilibrium constant $(K)$ is given by: $\Delta G^\circ = -RT \ln K$.

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(A) The relationship between standard Gibbs energy change $(\Delta G^\circ)$ and equilibrium constant $(K)$ is given by: $\Delta G^\circ = -RT \ln K$.Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K = 100$.$\Delta G^\circ = -8.314 \times 298 \times \ln(100)$.Since $\ln(100) = 2.303 \times \log_{10}(100) = 2.303 \times 2 = 4.606$.$\Delta G^\circ = -8.314 \times 298 \times 4.606 \text{ J/mol}$.$\Delta G^\circ = -11411.4 \text{ J/mol} = -11.411 \text{ kJ/mol}$.

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

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The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$

MHT CET 2026 All MHT CET

A
-$11.411$ \text{ kJ/mol}
B
-$5.744$ \text{ kJ/mol}
C
-$570.584$ \text{ kJ/mol}
D
-$57.058$ \text{ kJ/mol}

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6-1.Equilibrium (Chemical Equilibrium)

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At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

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The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is $13.8 \, kJ \, mol^{-1}$. Find the value of $K_{c}$ at $298 \, K$.

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The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$

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At $298 \ K$, if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$, the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

At $60^{\circ} C$,dinitrogen tetroxide is $50$ per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

In the glycolysis process,during the phosphorylation of glucose,the equilibrium constant at $298 \ K$ is $3.6 \times 10^{-3}$. Find the value of $\Delta G^{\Theta}$. What does this indicate? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is $13.8 \, kJ \, mol^{-1}$. Find the value of $K_{c}$ at $298 \, K$.

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