Question

$A$ mixture in which the mole ratio of $H_2$ and $O_2$ is $2:1$ is used to prepare water by the reaction,
$2H_{2\left(g\right)}+O_{2\left(g\right)}\to 2H_2{O}_{\left(g\right)}$
The total pressure in the container is $0.8$ atm at $20{}^{\circ }C$ before the reaction. The final pressure at after $120{}^{\circ }C$ reaction is (assuming $80$ yield of water)

• A. $1.787$ atm
• B. $0.878$ atm
• C. $0.787$ atm
• D. $1.878$ atm

Answer 1

Answer: (C)

$2H_{2\left(g\right)}+O_{2\left(g\right)}$ $\to$ $2H_2{O}_{\left(g\right)}$
$\begin{array}{cccc}Initialmole& 2a& a& 0\\ Finalmole& \left(2a-2x\right)& \left(a-x\right)& 2x\end{array}$
Given $2x=2a\times \frac{80}{100}=1.6a$
$\therefore$ $x=0.8a$
Thus, after the reaction $H_2$ left =$2a-1.6a$ =$0.4a$ mole
$O_2$ left =$0.2a$ mole
$H_{2}O$ formed = $1.6a$ mole
$\therefore$ Total mole at $120C^{\circ }$ in gaseous phase
=$0.4a+0.2a+1.6a=2.2a$
Now, given at initial conditions $P=0.8$ atm, $T-293K$
$P\times T=nRT$
$0.8V=3a\times R\times 293$
$\therefore$ $V=\frac{3a\times R\times 293}{0.8}$
The volume of container remains constant.
Using, $P\times V=nRT$
$\therefore$ $P\times \frac{3a\times R\times 293}{0.8}=2.2a\times R\times 393$ $\left[T=393K\right]$
$P=\frac{393\times 0.8\times 2.2}{3\times 293}=0.787atm$