For the reaction $CO_{(g)} + 2H_{2_{(g)}} \rightleftharpoons CH_3OH_{(g)}$,the equilibrium constant $K_c$ is $0.5$. If the concentrations of $CO$ and $H_2$ at equilibrium are $0.18 \ M$ and $0.22 \ M$ respectively,what is the concentration of $CH_3OH$?

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(D) The equilibrium constant expression for the reaction is $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$.
Given $K_c = 0.5$,$[CO] = 0.18 \ M$,and $[H_2] = 0.22 \ M$.
Substituting the values: $0.5 = \frac{[CH_3OH]}{(0.18)(0.22)^2}$.
$[CH_3OH] = 0.5 \times 0.18 \times 0.0484$.
$[CH_3OH] = 0.004356 \ M$ or $4.356 \times 10^{-3} \ M$.

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