Pure $A (g)$ at 2.0 atm is introduced into a rigid vessel at constant temperature, where $A (g) 2 B (g)$ is established. The total pressure at equilibrium is 3.0 atm. The value of $K_{p}$ (in atm) is

Answer: B

Let

A

lose

x

atm:

p_{A} = 2.0 - x

p_{B} = 2 x

, total

= 2.0 + x = 3.0

, so

x = 1.0

atm. Then

p_{A} = 1.0

and

p_{B} = 2.0

atm, and

K_{p} = p_{B}^{2} / p_{A} = (2.0)^{2} /1.0 = 4.00

atm.