QJEEM24Ch304f: JEE Main 2024 Chemistry - Class 11 JEEM24Ch304f Solution: For a certain reaction at , then for the same reaction is -…

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:

$$\mathrm{\Delta }{G}^{\circ }=-RT\ln K$$

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.

Puting the values we get,

$$\mathrm{\Delta }{G}^{\circ }=-8.314\times 300\times \ln (10)$$

We can convert the natural logarithm of 10 to base-10 logarithm, using the change of base formula, $\ln(10) = 2.3026$ . So,

$$\mathrm{\Delta }{G}^{\circ }=-8.314\times 300\times 2.3026$$

Calculating this yields:

$$\mathrm{\Delta }{G}^{\circ }=-8.314\times 300\times 2.3026=-5743.914Jmo{l}^{-1}$$

To convert this to kilojoules per mole, we divide by 1000:

$$\mathrm{\Delta }{G}^{\circ }=-5.743914\times {10}^{3}Jmo{l}^{-1}\times \frac{1kJ}{{10}^{3}J}=-5.743914kJmo{l}^{-1}$$

Now, when expressing this in terms of $\times 10^{-1} \ kJ mol^{-1}$ , we get:

$$\mathrm{\Delta }{G}^{\circ }=-57.43914\times {10}^{-1}kJmo{l}^{-1}$$

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.