Question

The value of $K_c$ 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 1. A mixture of 1 mole of water vapour and 3 moles of $CO$ is allowed to come to equilibrium at a total pressure of 2 $\text{atm}$. Calculate the partial pressure of hydrogen gas.

• A. $0.375\text{atm}$
• B. $0.75\text{atm}$
• C. $1.5\text{atm}$
• D. $2.5\text{atm}$

Answer 1

The correct option is A $0.375\text{atm}$

$\begin{array}{cccccccc}& CO\left(g\right)& +& H_{2}O\left(g\right)& \rightleftharpoons & C{O}_2\left(g\right)& +& H_2\left(g\right)\\ \text{(Initially)}& 3& & 1& & 0& & 0\\ \text{(At Equilibrium)}& 3-x& & 1-x& & x& & x\end{array}$
Applying law of mass action
$K_c=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[CO\right]\left[H_{2}O\right]}=\frac{x^2}{(3-x)(1-x)}$ or,
$1=\frac{x^2}{3-3x-x+x^2}$
or,
$x^2=x^2-4x+3x=0.75$
$\therefore$ Moles of hydrogen at equilibrium $=0.75$
Now, we know that partial pressure (p) $=\frac{\text{Moles of substance}}{\text{Total number of moles}}\times \text{Total Pressure}$
$\therefore p_{H_2}=\frac{0.75}{4}\times 2=0.375\text{atm}$