JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

JEE Main2026Jan 23, Shift 1Chemistry

Consider the general reaction given below at 400 K

$x A (g) ⇌ y B (g)$

The values of $K_{p}$ and $K_{c}$ are studied under the same condition of temperature but variation in x and y.

(i) $K_{p} = 85.87$ and $K_{c} = 2.586$ appropriate units

(ii) $K_{p} = 0.862$ and $K_{c} = 28.62$ appropriate units

The value of x and y in (i) and (ii) respectively are:

A
(i) 3, 1 ; (ii) 3, 1
B
(i) 4, 1 ; (ii) 4, 1
C
(i) 1, 3 ; (ii) 2, 1
D
(i) 1, 2 ; (ii) 2, 1

Solution

The standard link is $K_{p} = K_{c} (R T)^{Δ n_{g}}$ , where $Δ n_{g} = y - x$ .

At 400 K, $R T \approx 0.0821 \times 400 = 32.84$ .

Case (i): $\frac{K _{p}}{K _{c}} = \frac{85.87}{2.586} \approx 33.2 \approx R T$ , so $Δ n_{g} = + 1$ ⇒ $y - x = 1$ . Among the given options, $(x, y) = (1, 2)$ fits.

Case (ii): $\frac{K _{p}}{K _{c}} = \frac{0.862}{28.62} \approx 0.0301 \approx 1/ R T$ , so $Δ n_{g} = - 1$ ⇒ $y - x = - 1$ . Among the options, $(x, y) = (2, 1)$ fits.

$∴$ Option (4): (i) 1, 2; (ii) 2, 1.

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