Question

33. At a certain temperature and total pressure of 1.2 atm, the partial pressures of an equilibrium mixture for 2A(g) → B(g) are PA = 0.60 atm and PB = 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? atm

Answer 1

The equilibrium expression for the given reaction is:

[B]/[A]^2 = Kp.

So, at equilibrium, Kp = 0.6 / (0.6^2) = 1/0.6 = 1.667 atm^-1. Note that because we are given the partial pressure of A directly, we don't need to worry about the "2" in front of it in the equation.

At the new equilibrium, we have a total pressure of 1.6 atm. So, if we have x atm of A, then we must have (1.6-x) atm of B.

So, [B] = (1.6-x)
[A] = x

Kp = (1.6-x)/x^2 = 1.667 atm

Now we just solve the equation:

1.667 x^2 = 1.6 - x

1.667 x^2 + x - 1.6 = 0

x^2 + 0.6 x - 0.96 = O.

This is a quadratic equation, so we use the quadratic formula to find the roots. We find the two roots are: -1.324 and 0.725.

The neative root is impossible, since pressures cannot be negative. So, let's test [A] = 0.725 atm:

[A] = 0.725 atm
[B] = 1.6-.725 = 0.875 atm

Kp = 0.875 / (.725)^2 = 1.664. The slight error is due to rounding of the roots.

Thus, the equilibrium pressure of A, to two significant figures, is 0.73 atm.