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 \(5.85 × 10 ^{- 2}\; M\).