Equilibrium Question 660
Question: The dissociation equilibrium of a gas $ AB_2 $ can be represented as $ 2AB_2(g) \leftrightharpoons 2AB(g)+B_2(g) $ The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant $ K_{P} $ and total pressure p is
Options:
A) $ {{(2K_{P}/P)}^{1/3}} $
B) $ {{(2K_{P}/P)}^{1/2}} $
C) $ (K_{P}/P) $
D) $ (2K_{P}/P) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{\underset{1-x}{\mathop{1}},}{\mathop{2AB_2}},(g)\rightarrow \underset{\underset{x}{\mathop{2}},}{\mathop{2AB}},(g)+\underset{\underset{x/2}{\mathop{1}},}{\mathop{B_2}},(g) $
$ \therefore Kp=\frac{x^{2}}{2{{(1-x)}^{2}}}.[ \frac{P}{1+\frac{x}{2}} ]=\frac{x^{2}.P}{2} $
$ (1-x\approx 1,and,1+\frac{x}{2}\approx 1,since,x«1) $ Or $ P=\sqrt[3]{\frac{2Kp}{P}} $
BETA
