Question

At a certain temperature and total pressure of 1.2 atm, the partial pressures of an equilibrium mixture for

2A(g)

→

B(g)

are P_A = 0.60 atm and P_B = 0.60 atm. After a disturbance, the system regains equilibrium with a total pressure of 1.6 atm. What is the partial pressure of A at the new equilibrium?

Answer 1

From the definition of Kp:
Kp = pB / (pA)^2

For the first set of equilibrium pressures:
Kp = (0.6 atm) / (0.6 atm)^2
Kp = 1.667 atm-1

All we know about the second set of equilibrium pressures is that Kp will remain constant at a given temperature. If the nature of the "disturbance" adds A or B to the mixture, we can find pA after a new equilibrium has been reached. I'll solve the case for "adding B." As it turns out, the case of "adding A" would result in the same final equilibrium pA (I won't show the work for this here, but this case could be solved similarly to what I'll do below).

Suppose x atm of B are added to the mixture. To reach equilibrium, some of B (y) would have to get converted to A (by Le Chatelier's principle):
Kp = (0.6 + x - y) / (0.6 + 2y)^2

Since the total pressure has to be 1.6 atm,
(0.6 + x - y) + (0.6 + 2y) = 1.6 atm
1.2 + x + y = 1.6 atm
x = 0.4 - y

Kp = (0.6 + (0.4 - y) - y) / (0.6 + 2y)^2
1.667 = (1.0- 2y) / (0.6 + 2y)^2
6.667y^2 + 6y - 0.4 = 0

By quadratic equation:
y = [-6 + sqrt(6^2 - 4*6.667*(-0.4))] / (2*6.667) (the "negative" answer would make no sense here)
y = 0.0623

pA = 0.6 + 2y
pA = 0.6 + 2(0.0623)
pA = 0.7246 atm