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Applications of equilibrium constant

At T~K, a compound AB_2(g) dissociates according to the reacti

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Solving time: 3 mins

At $T K$ , a compound $A B_{2} (g)$ dissociates according to the reaction $2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$ , with a degree of dissociation $‘ x ’$ which is small compared with unity. The expression for $‘ x ’$ in terms of the equilibrium constant, $K_{P}$ and the total pressure $P$ is

\frac{K _{P}}{P}

(K_{P})^{1/3}

(\frac{2 K _{P}}{P})^{1/3}

(\frac{K _{P}}{P})^{1/3}

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Updated on: Apr 20, 2023

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Theodore Discussed

T K

, a compound

A B_{2} (g)

dissociates according to the reaction

2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)

, with a degree of dissociation

‘ x ’

which is small compared with unity. The expression for

‘ x ’

in terms of the equilibrium constant,

K_{P}

and the total pressure

P

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

Eq u i l ib r i u m 2 A B_{2} 1 1 - x ⇌ 2 A B 0 x + B_{2} 0 \frac{x}{2}

K_{p} = \frac{( \frac{x}{1 + \frac{x}{2}} \times P ) ^{2} \cdot ( \frac{x /2}{1 + x /2} \cdot P )}{( \frac{1 - x}{1 + x /2} \cdot P ) ^{2}}

= \frac{x ^{3} \cdot P}{2 ( 1 - x ) ^{2} ( 1 + \frac{x}{2} )}

x << 1, K_{p} = \frac{x ^{3} \cdot P}{2} \Rightarrow x = (\frac{2 K p}{P})^{1/3}

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Question Text	At $T K$ , a compound $A B_{2} (g)$ dissociates according to the reaction $2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$ , with a degree of dissociation $‘ x ’$ which is small compared with unity. The expression for $‘ x ’$ in terms of the equilibrium constant, $K_{P}$ and the total pressure $P$ is
Updated On	Apr 20, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 13
Upvotes	1530
Avg. Video Duration	11 min

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