Question

The reaction, $CO(g) + 3H_2(g) \leftrightarrow CH_4(g) + H_2O(g)$ is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of $H_2$ and 0.02 mol of $H_2O$ 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.

Answer 1

Let the concentration of methane at equilibrium be $x$.
$CO(g) + 3H_2(g) \leftrightarrow CH_4(g) + H_2O(g)$
At equilibrium $\frac{0.3 }{ 1} = 0.3\,M\; \frac{0.1 }{ 1} = 0.1\,M$ x $\frac{0.02 }{ 1} = 0.02\,M$
It is given that $K_c$= $3.90$.

Therefore, $\frac{\bigg[CH_4(g)\bigg]\bigg[H_2O(g)\bigg]}{ \bigg[CO(g)\bigg]\bigg[H_2(g)\bigg]^3}$ = $K_c$

$\Rightarrow$ $x × \frac{0.02 }{ 0.3 × (0.1)^3}$ = $3.90$

$\Rightarrow$ $x$ = $\frac{3.90 × 0.3 × (0.1)^3 }{ 0.02}$

= $\frac{0.00117 }{ 0.02}$

=$0.0585 M$
= $5.85 × 10 ^{- 2}\; M$

Hence, the concentration of CH4 at equilibrium is $5.85 × 10 ^{- 2}\; M$.