Question

Initially the molar ratio of $N_{2}$ and $H_{2}$ was 1 : 3. 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

Answer: (A)

$\Sigma n=0.5+1.5+1=3$
Mole fraction of ammonia = $\text{x}_{\text{NH}_{\text{3}}} \text{=} \frac{\text{moles} \, \text{of} \, \text{ammonia}}{\text{total} \, \text{moles}}$ .
$P_{N H_{3}}=x_{N H_{3}}\times P$
$=\frac{1}{3}\times P$
$=\frac{P}{3}$