Question

$N_{2}+3H_{2} \rightarrow 2NH_{3}$

$1$ mole $N_{2}$ and $4$ moles $H_{2}$ are taken in $15 \text{L}$ flask at $27^{o}C$ . After complete conversion of $N_{2}$ into $NH_{3}$ , $5 \text{L}$ of $H_{2}O$ is added. Pressure set up in the flask is (assume water dissolves $\text{N}\text{H}_{3}$ )

• A. $\frac{3 \times 0.0821 \times 300}{15} \, atm$
• B. $\frac{2 \times 0.0821 \times 300}{10} \, atm$
• C. $\frac{1 \times 0.0821 \times 300}{15} \, atm$
• D. $\frac{1 \times 0.0821 \times 300}{10} \, atm$

Answer 1

Answer: (D)

$\begin{array}{ccccc} & N _{2}+ & 3 H _{2} & \longrightarrow & 2 NH _{3} \\ \text { Initial mole } & 1 & 4 & & 0 \\ \text { Reacted mole } & 1 & 3 & & \\ \text { Left mole } & 0 & 4-3=1 & & 2\end{array}$
$H_{2}$ left = 1 mol
$NH_{3}$ formed = 2 mol
But $NH_{3} \, \left(g\right)$ dissolves in water forming $NH_{4}OH$ . Residual gas is only $H_{2}$ which exerts pressure.
Volume $=15-5=10 \, L$ (5L occupied by $H_{2}O$ added)
$\therefore p=\frac{n}{V}RT=\frac{1 \times 0.0821 \times 300}{10} \, atm$