Question

Total pressure of a mixture of $H_2$ and $O_2$ is 1.0 bar. The mixture is allowed to react to form water, which is completely removed to leave only pure $H_2$ at a pressure of 0.34 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same temperature and volume conditions, the composition of the original mixture is:

• A. $\begin{array}{cc}{X}_{H_2}& X_{O_2}\\ 0.50& 0.50\end{array}$
• B. $\begin{array}{cc}{X}_{H_2}& X_{O_2}\\ 0.22& 0.78\end{array}$
• C. $\begin{array}{cc}{X}_{H_2}& X_{O_2}\\ 0.78& 0.22\end{array}$
• D. $\begin{array}{cc}{X}_{H_2}& X_{O_2}\\ 0.35& 0.65\end{array}$

Answer 1

The correct option is C $\begin{array}{cc}{X}_{H_2}& X_{O_2}\\ 0.78& 0.22\end{array}$

Initially $H_2$ and $O_2$ both are present
So in $PV=nRT$,
$n_{H_2}+n_{O_2}=n_{\left(total\right)}=\frac{PV}{RT}=\frac{1\times V}{RT}=\frac{V}{RT}$
After reaction is complete, only $H_2$ (unreacted ) is left
Moles of unreacted $H_2=\frac{PV}{RT}=\frac{0.34V}{RT}$
Thus, $n_{H_2}+n_{O_2}+\text{unreacted}{n}_{H_2}=\frac{V}{RT}$
$n_{H_2}+n_{O_2}=\frac{V}{RT}-\frac{0.34V}{RT}$
$\Rightarrow n_{H_2}+n_{O_2}=\frac{0.66V}{RT}$ --- $\left(i\right)$
Also, $2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(l\right)$
$\text{Moles of}{H}_2=2\times \text{moles of}{O}_2$
so, $n_{H_2}=2n_{O_2}$
On putting this value in $\left(i\right)$, we get:
$3n_{O_2}=\frac{0.66V}{RT}$
$\Rightarrow n_{O_2}=\frac{0.66V}{3RT}$
$\therefore X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.66V/3RT}{V/RT}=0.22$
$X_{H_2}=1-0.22=0.78$