Question

# Partial pressures question? 2A ----> B...At a given temperature and total pressure of 1.2 atm the...

Partial pressures question?

2A ----> B...At a given temperature and total pressure of 1.2 atm the partial pressures of an equilibrium mixture for the rxn are Pa=.6 atm and Pb=.6 atm. After a disturbance the system regains equilibrium with a total pressure of 1.5 atm. What is the partial pressure of A?

I know the first answer is 1.7

The next answers are supposed to be PB= .81 and PA=.69

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 1.5 atm,
(0.6 + x - y) + (0.6 + 2y) = 1.5 atm
1.2 + x + y = 1.5 atm
x = 0.3 - y

Kp = (0.6 + (0.3 - y) - y) / (0.6 + 2y)^2
1.667 = (0.9 - 2y) / (0.6 + 2y)^2
6.667y^2 + 6y - 0.3 = 0

y = [-6 + sqrt(6^2 - 4*6.667*(-0.3))] / (2*6.667) (the "negative" answer would make no sense here)
y = 0.0475

pA = 0.6 + 2y
pA = 0.6 + 2(0.0475)
pA = 0.695 atm

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