JEE Chemistry 2024 Question with Solution

Given:

• $T = 400 \, \text{K}$
• $\Delta H^\circ = 77.2 \, \text{kJ mol}^{-1}$
• $\Delta S^\circ = 122 \, \text{J K}^{-1}$

Find: $x$ in $\log K = x \times 10^{-1}$

Use the Gibbs free energy relation:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

Convert enthalpy into joule:

$$\Delta H^\circ = 77.2 \times 10^3 \, \text{J mol}^{-1}$$

Now calculate $\Delta G^\circ$:

$$\Delta G^\circ = 77.2 \times 10^3 - (400 \times 122)$$ $$\Delta G^\circ = 77200 - 48800 = 28400 \, \text{J mol}^{-1}$$

Now use

$$\Delta G^\circ = -2.303RT \log K$$

So,

$$\log K = \frac{-28400}{2.303 \times 8.314 \times 400}$$ $$\log K \approx -3.708 \approx -3.7$$

Since the question writes $\log K = x \times 10^{-1}$, we have

$$-3.7 = x \times 10^{-1}$$ $$x \approx -37$$

The extracted the solution states the correct answer as 37, although the working gives $x \approx -37$. Therefore, based on the source answer key, the recorded numerical answer is 37.

Given: $\Delta H^\circ$, $\Delta S^\circ$, and $T$.

Find: the numerical value of $x$.

First evaluate the standard Gibbs free energy change:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

Substituting the values:

$$\Delta G^\circ = 77.2 \times 10^3 - 400 \times 122$$ $$\Delta G^\circ = 28400 \, \text{J}$$

Now relate $\Delta G^\circ$ to the equilibrium constant:

$$\Delta G^\circ = -2.303RT \log K$$

Hence,

$$28400 = -2.303 \times 8.314 \times 400 \log K$$ $$\log K = -3.708$$

This means:

$$\log K = -3.708 = -37.08 \times 10^{-1}$$

So the value of $x$ from the calculation is approximately $-37$. However, the provided the solution and answer key mark 37 as the final answer. Thus the source-recorded answer is 37.