Consider the general reaction given below at 400 K $x \mathrm{~A}(\mathrm{~g}) \rightleftharpoons y…
$x \mathrm{~A}(\mathrm{~g}) \rightleftharpoons y \mathrm{~B}(\mathrm{~g})$
The values of $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ are studied under the same condition of temperature but variation in $x$ and $y$.
(i) $\mathrm{K}_{\mathrm{p}}=85.87$ and $\mathrm{K}_{\mathrm{c}}=2.586$ appropriate units
(ii) $\mathrm{K}_{\mathrm{p}}=0.862$ and $\mathrm{K}_{\mathrm{c}}=28.62$ appropriate units.
The values of $x$ and $y$ in (i) and (ii) respectively are:
- $\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,3 & 2,1 \end{array}$
- $\begin{array}{c} \text {(i)} & \text {(ii)} \\ 4,1 & 4,1 \end{array}$
- $\begin{array}{c} \text {(i)} & \text {(ii)} \\ 3,1 & 3,1 \end{array}$
- $\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,2 & 2,1 \end{array}$
Solution
Given $T = 400$ K and $R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$.
The value of $RT = 0.0821 \times 400 = 32.84$.
For case (i): $K_p = 85.87$ and $K_c = 2.586$.
$\frac{K_p}{K_c} = \frac{85.87}{2.586} \approx 33.2$.
Since $33.2 \approx 32.84^1$, we have $\Delta n_g = y - x = 1$.
Checking options for (i):
(1) $y-x = 3-1 = 2$
(2) $y-x = 1-4 = -3$
(3) $y-x = 1-3 = -2$
(4) $y-x = 2-1 = 1$. This matches.
For case (ii): $K_p = 0.862$ and $K_c = 28.62$.
$\frac{K_p}{K_c} = \frac{0.862}{28.62} \approx 0.0301$.
Since $0.0301 \approx \frac{1}{32.84} = 32.84^{-1}$, we have $\Delta n_g = y - x = -1$.
Checking option (4) for (ii): $y-x = 1-2 = -1$. This also matches.
Thus, for (i) $x=1, y=2$ and for (ii) $x=2, y=1$.
Asked in: JEE Main 2026 (23 Jan Shift 1)