The reaction,$CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is at equilibrium at $1300 \, K$ in a $1 \, L$ flask. It also contains $0.30 \, mol$ of $CO$,$0.10 \, mol$ of $H_{2}$,and $0.02 \, mol$ of $H_{2}O$ and an unknown amount of $CH_{4}$ in the flask. Determine the concentration of $CH_{4}$ in the mixture. The equilibrium constant,$K_{c}$ for the reaction at the given temperature is $3.90$.

(D) Let the concentration of methane at equilibrium be $x$.
The equilibrium expression for the reaction $CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is:
$K_{c} = \frac{[CH_{4}][H_{2}O]}{[CO][H_{2}]^3}$
Given concentrations at equilibrium in a $1 \, L$ flask:
$[CO] = 0.30 \, M$,$[H_{2}] = 0.10 \, M$,$[H_{2}O] = 0.02 \, M$,$[CH_{4}] = x \, M$
Substituting the values into the $K_{c}$ expression:
$3.90 = \frac{x \times 0.02}{0.30 \times (0.10)^3}$
$3.90 = \frac{0.02x}{0.30 \times 0.001}$
$3.90 = \frac{0.02x}{0.0003}$
$x = \frac{3.90 \times 0.0003}{0.02}$
$x = \frac{0.00117}{0.02} = 0.0585 \, M$
Thus,the concentration of $CH_{4}$ is $5.85 \times 10^{-2} \, M$.