1. Tardigrade
  2. Question
  3. Chemistry
  4. A mixture in which the mole ratio of H2 and O2 is 2 :1 is used to prepare water by the reaction, 2 H2(g)+O2(g)arrow 2H2O(g) The total pressure in the container is 0.8 atm at 20 °C before the reaction. The final pressure at after 120 °C reaction is (assuming 80% yield of water)

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $A$ mixture in which the mole ratio of $H_{2}$ and $O_{2}$ is $2 :1$ is used to prepare water by the reaction,
$2 H_{2\left(g\right)}+O_{2\left(g\right)}\rightarrow 2H_{2}O_{\left(g\right)}$
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)

States of Matter

Solution:

$2H_{2\left(g\right)}+O_{2\left(g\right)}$ $\rightarrow$ $2H_{2}O_{\left(g\right)}$
$\begin{matrix}Initial mole&2a&a&0\\ Final mole&\left(2a-2x\right)&\left(a-x\right)&2x\end{matrix}$
Given $\quad$ $2x=2a\times\frac{80}{100}=1.6 a$
$\therefore $ $\quad$ $ x=0.8 a$
Thus, after the reaction $H_{2}$ left =$2a-1.6a$ =$0.4a$ mole
$O_{2}$ left =$0.2a$ mole
$H_{2}O$ formed = $1.6 a$ mole
$\therefore $ $\quad$ Total mole at $120 \,C^{\circ}$ in gaseous phase
=$0.4a+0.2a +1.6a=2.2a$
Now, given at initial conditions $P = 0.8$ atm, $T- 293 \,K$
$P\times T=nRT$
$0.8 V=3a\times R\times293$
$\therefore $ $\quad$ $V=\frac{3a\times R\times293}{0.8}$
The volume of container remains constant.
Using, $P\times V=nRT$
$\therefore $ $\quad$ $P\times\frac{3a\times R\times293}{0.8}=2.2a\times R\times393$ $\quad$ $\left[T=393 K\right]$
$P=\frac{393\times0.8\times2.2}{3\times293}=0.787 atm$