Consider the following reaction: AB (s) ⇌A (g) + B (g). At equilibrium, a 10.9 L container at 650 K contains A (g) at a pressure of 0.126 atm and B (g) at a pressure of 0.23 atm. The container is then compressed to half of its original volume. What is the pressure of B (g) after the reaction reattains equilibrium? Express your answer numerically in atm.

pA = 0.095 atm

pB = 0.303 atm

Explanation:

Step 1: the reaction

AB (s) ⇔ A (g) + B (g)

Kp = pA * pB

⇒ with Kp = equilibrium constant

Kp = 0.126 * 0.23 ⇒ Kp = 0.02898

Since the container will be compressed to half of its original volume, means that he pressure will be doubled.

⇒pA = 0.252

⇒pB = 0.46

To establish this equilibrium, each pressure has to be lowered by x

⇒pA = 0.252 - x

⇒pB = 0.46 - x

Kp = 0.02898 = (0.252 - x) (0.46-x)

0.02898 = 0.11592 - 0.252x - 0.46x + x²

-x² + 0.712x - 0.08694 = 0

D = b² - 4ac

⇒ D = 0.712² - 4 * (-1) * (-0.08694) = 0.506944‬ - 0.34776‬ = 0.159184

x = (-b ± √D) / 2a

x = (-0.712 ± √0.159184) / (2*-1) = (-0.712 ± 0.398978696) / -2

x = 0.156510652 or x = 0.555489348

x = 0.555489348 is impossble or the pressure would be negative

x=0.156510652

pA = 0.252 - 0.156510652 = 0.095489348 atm

pB = 0.46 - 0.156510652 = 0.303489348 atm