JEE MAIN - Chemistry (2022 - 30th June Morning Shift - No. 3)
The equilibrium constant for the reversible reaction
2A(g) $$\rightleftharpoons$$ 2B(g) + C(g) is K1
$${3 \over 2}$$A(g) $$\rightleftharpoons$$ $${3 \over 2}$$B(g) + $${3 \over 4}$$C(g) is K2.
K1 and K2 are related as :
$${K_1} = \sqrt {{K_2}} $$
$${K_2} = \sqrt {{K_1}} $$
$${K_2} = K_1^{3/4}$$
$${K_1} = K_2^{3/4}$$
Explanation
$2 \mathrm{A}(\mathrm{g})=2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \quad K_{1} \quad\quad ...(i)$
$$ \frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii) $$
eq. (ii) is $\frac{3}{4}$ times of eq. (i), hence,
$K_{2}=\left(K_{1}\right)^{\frac{3}{4}}$
$$ \frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii) $$
eq. (ii) is $\frac{3}{4}$ times of eq. (i), hence,
$K_{2}=\left(K_{1}\right)^{\frac{3}{4}}$
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