Initially the molar ratio of N2 and H2 was 1: 3. At equilibrium, 50% of each has reacted. If the equilibrium pressure is P, the partial pressure of NH3 at equilibrium is

Question

Initially the molar ratio of N2 and H2 was 1: 3. At equilibrium, 50% of each has reacted. If the equilibrium pressure is P, the partial pressure of NH3 at equilibrium is (A) (P/3) (B) (P/4) (C) (P/6) (D) (P/8)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-fui8rmlbzejwb8cn.jpg" /> Sigma n=0.5+1.5+1=3 Mole fraction of ammonia = textx textNH text3 text= ( textmoles textof textammonia/ texttotal textmoles) . PN H3=xN H3× P =(1/3)× P =(P/3)

Q. 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 $N H_{3}$ at equilibrium is

$\frac{P}{3}$

$\frac{P}{4}$

$\frac{P}{6}$

$\frac{P}{8}$

$Σ n = 0.5 + 1.5 + 1 = 3$
Mole fraction of ammonia = $x_{NH_{3}} = \frac{moles of ammonia}{total moles}$ .
$P_{N H_{3}} = x_{N H_{3}} \times P$
$= \frac{1}{3} \times P$
$= \frac{P}{3}$

Q. Initially the molar ratio of N2​ and H2​ was 1 : 3. At equilibrium, 50% of each has reacted. If the equilibrium pressure is P, the partial pressure of NH3​ at equilibrium is

3P​

4P​

6P​

8P​