JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Equilibrium

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:

Answer: (D)

Solution

The pressure and concentration constants are linked by

K_{p} = K_{c} (R T)^{Δ n_{g}}

, where

Δ n_{g} = y - x

for

x A (g) ⇌ y B (g)

; at

400 K

R T = 0.0821 \times 400 \approx 32.8

. For case (i),

\frac{K _{p}}{K _{c}} = \frac{85.87}{2.586} \approx 33.2 \approx (R T)^{1}

, so

Δ n_{g} = + 1

and the smallest whole-number fit is

x = 1, y = 2

. For case (ii),

\frac{K _{p}}{K _{c}} = \frac{0.862}{28.62} \approx 0.0301 \approx (R T)^{- 1}

, so

Δ n_{g} = - 1

, giving

x = 2, y = 1

. The values are (i)

x = 1, y = 2

and (ii)

x = 2, y = 1

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option