For a certain reaction at $300 \mathrm{~K}, \mathrm{~K}=10$, then $\Delta \mathrm{G}^{\circ}$ for the same reaction is - ____________ $\times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}$.

(Given $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

<p>To determine the change in standard Gibbs free energy ($\Delta G^{\circ}$) for the reaction at a given temperature when the equilibrium constant ($K$) is known, we can use the following relationship:</p> <p>$\Delta G^{\circ} = -RT \ln K$ </p> <p>Here, $R$ is the universal gas constant ($8.314 \ J K^{-1} mol^{-1}$), $T$ is the temperature in Kelvin ($300 \ K$), and $K$ is the equilibrium constant.</p> <p>Puting the values we get,</p> <p>$\Delta G^{\circ} = -8.314 \times 300 \times \ln(10)$</p> <p>We can convert the natural logarithm of 10 to base-10 logarithm, using the change of base formula, $\ln(10) = 2.3026$. So,</p> <p>$\Delta G^{\circ} = -8.314 \times 300 \times 2.3026$</p> <p>Calculating this yields:</p> <p>$\Delta G^{\circ} = -8.314 \times 300 \times 2.3026 = -5743.914 \ J mol^{-1}$</p> <p>To convert this to kilojoules per mole, we divide by 1000:</p> <p>$$ \Delta G^{\circ} = -5.743914 \times 10^{3} \ J mol^{-1} \times \frac{1 \ kJ}{10^{3} J} = -5.743914 \ kJ mol^{-1} $$</p> <p>Now, when expressing this in terms of $\times 10^{-1} \ kJ mol^{-1}$, we get:</p> <p>$\Delta G^{\circ} = -57.43914 \times 10^{-1} \ kJ mol^{-1}$</p> <p>Therefore, $\Delta G^{\circ}$ for the reaction is approximately $-57.43914 \times 10^{-1} \ kJ mol^{-1}$ or $-57.4 \times 10^{-1} \ kJ mol^{-1}$ when rounded to three significant figures.</p>