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