Question

A sample of HI(g) is placed in a 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? 2HI(g) ⇌ H2(g) + I2(g)

• A. 0.04
• B. 0.4
• C. 40
• D. 4

Answer 1

Answer: (D)

4

Answer 2

At equilibrium, 2HI(g) ⇌ H2(g) + I2(g)

Elements	HI	I2	H2
Initial	0.2	0	0
Change	-2x	x	x
Equilibrium	0.2-2x	x	x

It is given that the equilibrium****pressure of HI is 0.04 atm.
0.2 − 2x = 0.04
Therefore, 2x = 0.2−0.04 = 0.16
or x = 0.08 atm.
So, Kp = ${{PH_2} PI_2 \over P_{HI}^2}$= 0.08 x 0.08/0.042 $\Rightarrow$ Kp = 4