Question

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ for the reaction initially the mole ratio was $1:3$ of $N_2:H_2$. At equilibrium $50$% of each has reacted. If the equilibrium pressure is $p$, the partial pressure of $N{H}_3$ at equilibrium is__________.

• A. $\frac{P}{3}$
• B. $\frac{P}{4}$
• C. $\frac{P}{6}$
• D. $\frac{P}{8}$

Answer 1

The correct option is A $\frac{P}{3}$

$N_2+{3H}_2\rightleftharpoons {2NH}_3$

$t=0$ $a$ $3a$ $0$

$t=t_{eq}$ $\left(a\right)(1-\alpha )$ $3a(1-\alpha )$ $2a\alpha$

Given $50$% reacted $\alpha =1/2$

${pNH}_3=\left(molefraction\right)\times \left(totalpressure\right)$

$=2\times \alpha \times \frac{1}{2}\times P=\frac{P}{3}$

$=\frac{2\times \alpha \times \frac{1}{2}\times P}{\alpha \times (4-2\times 1/2)}=\frac{P}{3}$