JEE Chemistry 2023 Question with Solution

Given: $K_p = 2 \times 10^{12}$, $T = 300 \, \text{K}$, and $R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}$.

Find: $K_c$ for $\text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g)$.

Use the relation

$$K_p = K_c (RT)^{\Delta n_g}$$

For this reaction,

$$\Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 1 - \frac{1}{2} = -\frac{1}{2}$$

So,

$$2 \times 10^{12} = K_c \times (0.082 \times 300)^{-\frac{1}{2}}$$

Solving for $K_c$,

$$K_c = 9.92 \times 10^{12}$$

This is also written as

$$K_c = 0.992 \times 10^{13}$$

Hence, the nearest integer asked in the solution is $1$.

Therefore, the answer is $1$.

Given: $K_p = 2 \times 10^{12}$ at $300 \, \text{K}$.

Find: the value of $K_c$.

First compute the change in gaseous moles:

$$\Delta n = (1) - \left(1 + \frac{1}{2}\right) = -\frac{1}{2}$$

Now apply

$$K_p = K_c (RT)^{\Delta n}$$

Substitute the values shown in the solution:

$$2 \times 10^{12} = K_c (0.082 \times 300)^{-\frac{1}{2}}$$

After simplification, the solution gives

$$K_c = 9.92 \times 10^{12}$$

Then it is expressed as

$$0.992 \times 10^{13}$$

and the marked nearest integer is $1$.

Therefore, the correct numerical answer is $1$.