Consider the following reversible chemical reactions: {A_2(g) + B_2(g)} {}{{k_1}{}} {2AB(g)} (1) {6AB(g)} {}{{k_2}{}} {3A_2(g) + 3B_2(g)} (2) The relation between K1 and K2 is:

JEE Main · 2019 · Shift-IIhardCK-084

Consider the following reversible chemical reactions: {A_2(g) + B_2(g)} {}{{k_1}{}} {2AB(g)} (1) {6AB(g)} {}{{k_2}{}}…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

Consider the following reversible chemical reactions: $\mathrm{A_2(g) + B_2(g)} \underset{}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{2AB(g)}\quad \cdots(1)$ $\mathrm{6AB(g)} \underset{}{\stackrel{k_2}{\rightleftharpoons}} \mathrm{3A_2(g) + 3B_2(g)}\quad \cdots(2)$

The relation between $K_1$ and $K_2$ is:

Options

a
$K_2 = K_1^{-3}$
✓
b
$K_1 K_2 = \frac{1}{3}$
c
$K_2 = K_1^3$
d
$K_1 K_2 = 3$

Correct Answera

$K_2 = K_1^{-3}$

Detailed Solution

Reaction (1): $\mathrm{A_2 + B_2 \rightleftharpoons 2AB}$ , equilibrium constant $K_1$

Reaction (2): $\mathrm{6AB \rightleftharpoons 3A_2 + 3B_2}$

Reaction (2) is the reverse of reaction (1) multiplied by 3:

Reverse of (1): $\mathrm{2AB \rightleftharpoons A_2 + B_2}$ , $K = K_1^{-1}$

Multiply by 3: $\mathrm{6AB \rightleftharpoons 3A_2 + 3B_2}$ , $K_2 = (K_1^{-1})^3 = K_1^{-3}$

Answer: Option (1) — $K_2 = K_1^{-3}$

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