JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Consider the general reaction given below at 400 K

$x \mathrm{~A}(\mathrm{~g}) \rightleftharpoons y \mathrm{~B}(\mathrm{~g})$

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:

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,3 & 2,1 \end{array}$
B
$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 4,1 & 4,1 \end{array}$
C
$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 3,1 & 3,1 \end{array}$
D
$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,2 & 2,1 \end{array}$

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Solution

The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$ , where $\Delta n_g = y - x$ .

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) $x=2, y=1$ .

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Options

$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,3 & 2,1 \end{array}$

$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 4,1 & 4,1 \end{array}$

$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 3,1 & 3,1 \end{array}$

$\begin{array}{c} \text {(i)} & \text {(ii)} \\ 1,2 & 2,1 \end{array}$

Solution

The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$ , where $\Delta n_g = y - x$ .

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) $x=2, y=1$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry