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20 August, 23:52

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.

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  1. 21 August, 00:10
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    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
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