Easy
The equilibrium constant of the following reaction is $0.5$. $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$. At equilibrium,$[CO] = 0.18 \ mol \ L^{-1}$ and $[H_2] = 0.22 \ mol \ L^{-1}$. Calculate the concentration of $CH_3OH$.
The equilibrium constant expression for the reaction is $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ mol \ L^{-1}$,and $[H_2] = 0.22 \ mol \ L^{-1}$.
Substituting the values into the expression: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times (0.22)^2$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ mol \ L^{-1}$ or $4.356 \times 10^{-3} \ mol \ L^{-1}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ mol \ L^{-1}$,and $[H_2] = 0.22 \ mol \ L^{-1}$.
Substituting the values into the expression: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times (0.22)^2$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ mol \ L^{-1}$ or $4.356 \times 10^{-3} \ mol \ L^{-1}$.
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