Question

The equilibrium constant for the reaction $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$ is 3 at 500K. In a 2 litre vessel 60gm of water gas [equimolar mixture of CO(g) and $H_2\left(g\right)$] and 90 gm of steam is initially taken.

What is the equilibrium concentration of $H_2\left(g\right)$ at equilibrium (mole/L)?

• A. 1.75
• B. 3.5
• C. 1.5
• D. 0.75

Answer 1

The correct option is B 3.5

$60g$ of water gas correspond to $\frac{60}{30}=2$ moles of $CO$ and 2 moles of $H_2$

$90g$ of steam corresponds to $\frac{90}{18}=5$ moles of $H_2$

The initial moles of $CO,H_{2}O,C{O}_{2}and{H}_2$ are $2,5,0$ and $2$ moles respectively.

The moles of $CO,H_{2}O,C{O}_{2}and{H}_2$ present at equilibrium are $2-x,5-x,x$ and $2+x$ moles respectively.

The expression for the equilibrium constant is $K_c=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[CO\right]\left[H_{2}O\right]}$

Substitute values in the above expression.

$3=\frac{\left(\frac{x}{2}\right)\left(\frac{2+x}{2}\right)}{\left(\frac{2-x}{2}\right)\left(\frac{5-x}{2}\right)}$

Thus $x=1.5$ and $\left[H_2\right]=2+1.5=3.5M$