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 Pa = 0.60 atm and Pb = 0.60 atm. After a disturbance, the system regains equilibrium with a total pressure of 1.9 atm. What is the partial pressure of A at the new equilibrium?

Answer in atm

Please show all work I have tried this problem twice and have gotten wrong answers would like to see the differences in steps.

Answer 1

Kp = P(B) / P(A)^2
Kp = (0.60) / (0.60)^2
Kp = 1 / 0.60
Kp = 1.667 atm^-1

The increase in pressure is due to the decrease in the volume of the vessel. According to LeChatelier's Principle the system will try to decrease this pressure and the equilibrium will shift to write. Why? Because from left to right the number of moles decreases.

When the pressure is increased from 1.2 to 1.5, the rise in pressure is 0.3 atm. Since the mole ratio of A to B is 2/1, the increase in pressure in A will be 0.2 atm, in B will be 0.1 atm. Therefore , the new pressures will be ;
A = 0.6 + 0.2 = 0.8 atm
B = 0.6 + 0.1 = 0.7 atm.
Since the equilibrium will shift to right;
Equilibrium reaction : ..........2A(g) <---> B(g)
Final equilib. pressures : .... 0.6 ........... 0.6
Increase in pressure : ......... 0.2 ........... 0.1
Total pressure : .................... 0.8 ........... 0.7
Change due to shift : ........... -2x ............ +x
New equilib. pressures : .... 0.8 - 2x ...... 0.7 + x

Kp = P(B) / P(A)^2
Kp = (0.7 + x) / (0.8 - 2x)^2 = 1/0.6

4x^2 - 3.8x + 0.22 = 0

Solving the equation , one of the roots will satisfy the conditions ( x = 0.0625 atm)
P(B) = 0.7 + x = 0.7 + 0.0625 = 0.7625 atm
P(A) = 0.8 - 2x = 0.8 - 0.125 = 0.675 atm

If you put these values in Kp expression, they will satisfy Kp value.