Question

$N_2+3H_2\to 2N{H}_3$

$1$ 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 $N{H}_3$ , $5\text{L}$ 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)

$\begin{array}{ccccc}& N_2+& 3H_2& ⟶& 2N{H}_3\\ \text{ Initial mole }& 1& 4& & 0\\ \text{ Reacted mole }& 1& 3& & \\ \text{ Left mole }& 0& 4-3=1& & 2\end{array}$
$H_2$ left = 1 mol
$N{H}_3$ formed = 2 mol
But $N{H}_3\left(g\right)$ dissolves in water forming $N{H}_{4}OH$ . Residual gas is only $H_2$ which exerts pressure.
Volume $=15-5=10L$ (5L occupied by $H_{2}O$ added)
$\therefore p=\frac{n}{V}RT=\frac{1\times 0.0821\times 300}{10}atm$