Question

Question 7.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

Class XI Equilibrium Page 225

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2HI(g) < = > H_2 (g) + I_2(g) \\ at \: \: t = 0 \: \: \: \: \: \: \: \: \: \: 0.2 \: atm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ at \: eqlm \: \: \: \: \: \: \: \: \: \: 0.04 \: atm \: \: \: \: \: \: \: \: \: \: \: \: 0.08atm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0.08atm$
at equilibrium ,
pressure of HI = 0.2 - 2x = 0.04
2x = 0.2 - 0.04 = 0.16
x = 0.08 atm

so, pressure of H2 and I2 at equilibrium is 0.08 atm .
e.g., at equilibrium ,
$P_{HI}=0.04 atm\\P_{I_2}=P_{H_2}=0.08atm$
$now, K_P=\frac{P_{H_2}.P_{I_2}}{P_{HI}^2}$
Kp = 0.08atm × 0.08atm/(0.04atm)²= 4

hence, Kp = 4

Answer 2

Answer: 4

Explanation:  $2HI\rightleftharpoons H_2+I_2$

initially pressure         p                        0         0

At eqm.                     p-2x                   x   x

The expression for dissociation constant is,

$K_p=\frac{x\times x}{(p-2x)^2}$

Given p= 0.2 atm

Also p -2x = 0.04

Thus 0.2- 2x= 0.04

x= 0.08

Thus $K_p=\frac{0.08\times 0.08}{(0.04)^2}$

$K_p=4$