Question

An equilibrium mixture for the following reaction: H2(g) + I2(g) <---> 2HI(g) is composed of the following: P(I2) = 0.08592 atm; P(H2) = 0.08592 atm; P(HI) = 0.5996 atm. If this equilibrium is disturbed by adding more HI so that the partial pressure of HI is suddenly increased to 1.0000 atm, what will the partial pressures of each of the gases be when the system returns to equilibrium?   

Answer 1

H₂(g) + I₂(g) <---> 2HI(g)

Given:

P(I₂) = 0.08592 atm.

P(H₂) = 0.08592 atm.

P(HI) = 0.5996 atm.

Calculate Kp of reaction

Kp = [P(HI)]^2/ P(I₂) P(H₂)

Kp = [0.5996]^2/([0.08592] [0.08592]) = 48.7

Now this equilibrium is disturbed by adding more HI so that the partial pressure of HI is suddenly increased to 1.0000 atm.

The partial pressures of each of the gases, when the system returns to equilibrium

Let the partial pressures of P(I₂) be x and P(H₂) = x

Then at equilibrium

Kp = [P(HI)]^2/ P(I₂) P(H₂)

48.7 = (1)^2/x*x

x^2 = (1)^2/ 48.7

x^2 = 0.020533

x = 0.14329 = P(I₂) = P(H₂)

The partial pressures of each of the gases, when the system returns to equilibrium

P(I₂) = 0.14329 atm.

P(H₂) = 0.14329 atm.

P(HI) = 1.0000 atm.