Consider the general reaction given below at 400 K: xA( g) yB( g)… — Solution | JEE Chemistry

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Consider the general reaction given below at $400 \, \text{K}$ : $xA(\text{g}) \rightleftharpoons yB(\text{g})$ . 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 $K_c = 2.586$ (ii) $K_p = 0.862$ and $K_c = 28.62$ . The values of $x$ and $y$ in (i) and (ii) respectively are :

A
$4,1$ $4,1$
B
$3,1$ $3,1$
C
$1,3$ $2,1$
D
$1,2$ $2,1$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The reaction is $xA(\text{g}) \rightleftharpoons yB(\text{g})$ at $400 \, \text{K}$ . For case (i), $K_p = 85.87$ and $K_c = 2.586$ . For case (ii), $K_p = 0.862$ and $K_c = 28.62$ .

Find: The values of $x$ and $y$ in cases (i) and (ii).

Use the relation

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

where

\Delta n_g = y - x

At $T = 400 \, \text{K}$ , taking $R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$ ,

RT = 0.0821 \times 400 = 32.84

For case (i),

\frac{K_p}{K_c} = \frac{85.87}{2.586} \approx 33.2

Since $RT \approx 32.84$ , we get

(RT)^{\Delta n_g} \approx 33.2 \implies \Delta n_g = 1

Therefore,

y - x = 1

Among the given options, $(1,2)$ satisfies this because $2 - 1 = 1$ .

For case (ii),

\frac{K_p}{K_c} = \frac{0.862}{28.62} \approx 0.030

Also,

\frac{1}{RT} = \frac{1}{32.84} \approx 0.030

Hence,

(RT)^{\Delta n_g} \approx 0.030 \implies \Delta n_g = -1

Therefore,

y - x = -1

Among the given options, $(2,1)$ satisfies this because $1 - 2 = -1$ .

Therefore, the values of $x$ and $y$ are $(1,2)$ for (i) and $(2,1)$ for (ii). The correct option is D.

Common mistakes

Using only the comparison $K_p > K_c$ or $K_p < K_c$ without applying the formula is incomplete. The sign of $\Delta n_g$ can be inferred qualitatively, but the exact pair $x, y$ must be checked using $K_p = K_c(RT)^{\Delta n_g}$ .
Taking $\Delta n_g = x - y$ instead of $\Delta n_g = y - x$ gives the opposite sign and leads to the wrong option. For gaseous reactions, always calculate change in moles as products minus reactants.
Forgetting to calculate $RT$ at the given temperature $400 \, \text{K}$ is a common error. The exponent is identified by comparing $\frac{K_p}{K_c}$ with powers of $RT$ , so $RT$ must be evaluated first.

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