Consider the following reversible chemical reactions: \[ \begin{aligned} & A_{2}(g)+B_{2}(g) \stackrel{K_{1}}{\rightleftharpoons} 2 A B(g) \\ & 6 A B(g) \stackrel{K_{2}}{\rightleftharpoons} 3 A_{2}(g)+3 B_{2}(g) \end{aligned} \] The relation between $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ is

(C) $\mathrm{K}_{2}=\mathrm{K}_{1}^{-3}$

(A) $\mathrm{K}_{1} \mathrm{~K}_{2}=\frac{1}{3}$

(B) $\mathrm{K}_{2}=\mathrm{K}_{1}^{3}$

(D) $\mathrm{K}_{1} \mathrm{~K}_{2}=3$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Equilibrium

Consider the following reversible chemical reactions:

A_{2} (g) + B_{2} (g) ⇌ K_{1} 2 A B (g) 6 A B (g) ⇌ K_{2} 3 A_{2} (g) + 3 B_{2} (g)

The relation between

K_{1}

and

K_{2}

Answer: (C)

The second equilibrium is just the first one reversed and tripled. Reversing a reaction inverts its constant, and multiplying a reaction by

3

raises the constant to the third power. Since the second reaction is

- 3

times the first (three times

2 AB <=> A X_{2} + B X_{2}

), its constant is

K_{2} = (\frac{1}{K _{1}})^{3} = K_{1}^{- 3}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option