Question

Consider the following reaction: H2(g) + I2(g) ⇌ 2HI(g) At a particular temperature, a reaction mixture at equilibrium contains pressures of H2 = 0.958 atm, I2 = 0.877 atm, and HI = 0.020 atm. At the same temperature, a second reaction mixture (not at equilibrium) contains pressures of H2 = 0.621 atm, I2 = 0.621 atm, and HI = 0.101 atm. What will be the partial pressure of HI when the second reaction mixture reaches equilibrium? a) 0.0144 atm b) 0.0433 atm c) 0.0577 atm d) 0.101 atm

Answer 1

from 1st reaction pressures:

Kp = p(HI)^2 / p(H2)*p(I2)

Kp = 0.020^2 / (0.958*0.877)

Kp = 4.761*10^-4

For the one that is not at equilibrium:

Qp = p(HI)^2 / p(H2)*p(I2)

Qp = 0.101^2 / (0.621*0.621)

Qp = 2.64*10^-2

Since Qp is greater than Kp, equilibrium will move to left

H2 (g) + I2 (g)   <—> 2HI (g)

0.621   0.621       0.101   (initial)

0.621+x   0.621+x       0.101-2x (at equilibrium)

Kp = p(HI)^2 / p(H2)*p(I2)

4.761*10^-4 = (0.101-2x)^2 / (0.621+x)^2

sqrt(4.761*10^-4) = (0.101-2x) / (0.621+x)

0.0218 = (0.101-2x) / (0.621+x)

0.0135 + 0.0218*x = 0.101 - 2x

2.0218*x = 0.0875

x = 4.33*10^-2 atm

At equilibrium

p(HI) = 0.101 - 2x = 0.101 - 2*(4.33*10^-2) = 0.0144 atm

Answer: a