Question

$N_2+3H_2\to 2N{H}_3$
1 mol of $N_2$ and 4 mol of $H_2$ are taken in a $15\text{L}$ flask at ${27}^{\circ }\text{C}$. After the complete conversion of $N_2$ into $N{H}_3$, $5\text{L}$ of $H_{2}O$ is added. The pressure set up in the flask is:

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

Answer 1

The correct option is D $\frac{1\times 0.0821\times 300}{10}\text{atm}$

$\begin{array}{cccccc}& N_2& +& 3H_2& \to & 2N{H}_3\\ t=0& 1\text{mole}& & 4\text{mole}& & 0\\ t=t_{final}& 0& & 1\text{mole}& & 2\text{mole}\end{array}$

$N{H}_3$ will be absorbed by water and the volume will be $15-5=10\text{L}$.

$P=\frac{nRT}{V}=\frac{1\times 0.0821\times 300}{10}\text{atm}$