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JEE Advanced levelSingle correct optionDifficulty 3 of 3ce095

Pure $A (g)$ at 2.0 atm is introduced into a rigid vessel at constant temperature and dissociates by $A (g) B (g) + C (g)$ . At equilibrium the total pressure is 2.5 atm. $K_{p}$ (in atm) is

A $0.208$
B $0.167$
C $0.125$
D $0.133$

Answer: B

Let

x

be the pressure of

A

that reacts: total pressure

= 2.0 + x = 2.5

, so

x = 0.5

atm. Then

p_{A} = 1.5

,

p_{B} = p_{C} = 0.5

atm and

K_{p} = (0.5) (0.5) /1.5 = 0.167

atm.

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