Medium
At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.
(N/A) The equilibrium constant expression for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is given by:
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(0.14)^2}{(0.6)(0.8)}$
$K_c = \frac{0.0196}{0.48}$
$K_c \approx 0.04083$ or $4.08 \times 10^{-2}$
$K_c = \frac{[HI]^2}{[H_2][I_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(0.14)^2}{(0.6)(0.8)}$
$K_c = \frac{0.0196}{0.48}$
$K_c \approx 0.04083$ or $4.08 \times 10^{-2}$
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