SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 7

####7. Consider the reaction,

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

The equilibrium constant of the above reaction is $K_{p}$ . If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $p_{N H_{3}} ≪< p_{total}$ at equilibrium)

(2019 Main, 11 Jan I)

(a) $\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{4}$

(b) $\frac{3^{3 / 2} K_{p}^{1 / 2} P^{2}}{16}$

(d) $\frac{K_{p}^{1 / 2} P^{2}}{4}$

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Answer:

Correct Answer: 7. (d)

Solution:

$\begin{aligned} N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) & At equilibrium: p_{N_{2}} = P, p_{H_{2}} = 3 P, p_{N H_{3}} = 2 P & \Rightarrow p_{(total)} = p_{N_{2}} + p_{H_{2}} + p_{N H_{3}} ≃ p_{N_{2}} + p_{H_{2}} & = p + 3 p = 4 p & [∵ P_{(total)} ≫ p_{N H_{3}}] & Now, K_{p} = \frac{p^{2} N H_{3}}{p_{N_{2}} \times p_{H_{2}}^{3}} = \frac{p_{N H_{3}}^{2}}{p \times (3 p)^{3}} & = \frac{p_{N H_{3}}^{2}}{27 \times p^{4}} = \frac{p_{N H_{3}}^{2}}{27 \times {(\frac{P}{4})}^{4}} & K_{p} = \frac{p_{N H_{3}}^{2} \times 4^{4}}{3^{2} \times 3 \times P^{4}} & \Rightarrow p_{N H_{3}}^{2} = \frac{3^{2} \times 3 \times P^{4} \times K_{p}}{4^{4}} & \Rightarrow p_{N H_{3}} = \frac{3 \times 3^{1 / 2} \times P^{2} \times K_{p}^{1 / 2}}{4^{2}} = \frac{3^{3 / 2} \times P^{2} \times K_{p}^{1 / 2}}{16} \end{aligned}$

Chemical and Ionic Equilibrium - Result Question 7

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Chemical and Ionic Equilibrium - Result Question 7

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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