JEE Chemistry 2024 Question with Solution

Given: $\Delta H^\circ = 77.2 \times 10^3 \, \text{J mol}^{-1}$, $\Delta S^\circ = 122 \, \text{J K}^{-1}$, and $T = 400 \, \text{K}$.

Find: the value of $x$ when $\log K = x \times 10^{-1}$.

Use the relation

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

So,

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

Now use

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

Therefore,

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

Given that

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

we write

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

Hence,

$$x = -37$$

Therefore, the computed value from the working is $-37$. The solution lists the final answer as 37, so there is a sign discrepancy in the source. Following the provided final answer on the page, the accepted answer is $37$.

Given: $\Delta H^\circ = 77.2 \times 10^3 \, \text{J mol}^{-1}$, $\Delta S^\circ = 122 \, \text{J K}^{-1}$, and $T = 400 \, \text{K}$.

Find: the value of $x$.

First calculate the standard Gibbs free energy change:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 77200 - 400 \times 122 = 28400 \, \text{J mol}^{-1}$$

Now use

$$\Delta G^\circ = -RT \ln K$$

So,

$$\ln K = -\frac{28400}{8.314 \times 400}$$ $$\ln K \approx -8.544$$

Convert to common logarithm:

$$\log K = \frac{\ln K}{2.303} \approx \frac{-8.544}{2.303} \approx -3.71$$

Comparing with

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

we get

$$x \approx -37.1$$

So the numerical value from the thermodynamic calculation is $-37$ approximately, while the solution's marks 37 as the accepted answer.