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The value of equilibrium constant of the reaction,
$HI(g) \rightleftharpoons \dfrac{1}{2}H_2(g) + \dfrac{1}{2}I_2(g)$ is 8.0.
The equilibrium constant of the reaction,
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be -
$HI(g) \rightleftharpoons \dfrac{1}{2}H_2(g) + \dfrac{1}{2}I_2(g)$ is 8.0.
The equilibrium constant of the reaction,
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be -
A
16
B
$\dfrac{1}{8}$
C
$\dfrac{1}{16}$
D
$\dfrac{1}{64}$
Detailed Solution
$HI(g) \rightleftharpoons \dfrac{1}{2}H_2(g) + \dfrac{1}{2}I_2(g)$
$K = \dfrac{[H_2]^{1/2}[I_2]^{1/2}}{[HI]} = 8$
The required reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, with $K' = \dfrac{[HI]^2}{[H_2][I_2]}$
This is the first reaction reversed (so the constant becomes $\dfrac{1}{K}$) and multiplied by 2 (so the constant is squared).
$K' = \left(\dfrac{1}{K}\right)^2 = \left(\dfrac{1}{8}\right)^2$
$K' = \dfrac{1}{64}$
$K = \dfrac{[H_2]^{1/2}[I_2]^{1/2}}{[HI]} = 8$
The required reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, with $K' = \dfrac{[HI]^2}{[H_2][I_2]}$
This is the first reaction reversed (so the constant becomes $\dfrac{1}{K}$) and multiplied by 2 (so the constant is squared).
$K' = \left(\dfrac{1}{K}\right)^2 = \left(\dfrac{1}{8}\right)^2$
$K' = \dfrac{1}{64}$