Question

$N_2\left(g\right)+3H_2\left(g\right)\leftrightharpoons 2{NH}_3\left(g\right)$ 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 ${NH}_3$ 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 $\frac{p}{3}$

Given that initially, the mole ratio was $1:3$ of $N_2:H_2$.

$\underset{a}{N_2}+\underset{3a}{3H_2}⟶\underset{0}{2N{H}_3}$

Since, $50\%$ of the metal has reacted,

Total no. of moles reacted $=a-\frac{a}{2}+3a-\frac{3a}{2}+0+\frac{2a}{2}=5a-2a=3a$

No. of moles of $N{H}_3$ at equillibrium $=\frac{2a}{2}=a$

Equillibrium pressure $=p\left(Given\right)$

$\therefore$ Partial pressure of $N{H}_3$ at equillibrium $\left(P_{N{H}_3}\right)=\frac{\text{Moles of}N{H}_3\times \text{Total pressure}}{\text{Total no. of moles}}=\frac{a\times p}{3a}=\frac{p}{3}$