Dihydrogen gas is obtained from water gas by the following equation:
$\underbrace{CO_{(g)} + H_2O_{(g)}}_{water\,gas} \rightleftharpoons_{500^{\circ}C} CO_{2_{(g)}} + H_{2_{(g)}}$
At $733 \ K$,the concentrations of $CO$,$H_2O$,$CO_2$,and $H_2$ in water gas are $0.18$,$0.0412$,$0.15$,and $0.2 \ mol \ L^{-1}$ respectively. Find $K_c$.

(N/A) The equilibrium constant $K_c$ for the reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + H_{2_{(g)}}$ is given by the expression:
$K_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$
Given concentrations are:
$[CO] = 0.18 \ mol \ L^{-1}$
$[H_2O] = 0.0412 \ mol \ L^{-1}$
$[CO_2] = 0.15 \ mol \ L^{-1}$
$[H_2] = 0.2 \ mol \ L^{-1}$
Substituting these values into the expression:
$K_c = \frac{0.15 \times 0.2}{0.18 \times 0.0412}$
$K_c = \frac{0.03}{0.007416}$
$K_c \approx 4.045$