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Today, 08:06

A (g) + 2B (g) → C (g) + D (g)

If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?

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  1. Today, 08:20
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    0.169

    Explanation:

    Let's consider the following reaction.

    A (g) + 2B (g) ⇄ C (g) + D (g)

    We can find the pressures at equilibrium using an ICE chart.

    A (g) + 2 B (g) ⇄ C (g) + D (g)

    I 1.00 1.00 0 0

    C - x - 2x + x + x

    E 1.00-x 1.00-2x x x

    The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

    The pressures at equilibrium are:

    pA = 1.00-x = 1.00-0.211 = 0.789 atm

    pB = 1.00-2x = 1.00-2 (0.211) = 0.578 atm

    pC = x = 0.211 atm

    pD = x = 0.211 atm

    The pressure equilibrium constant (Kp) is:

    Kp = pC * pD / pA * pB²

    Kp = 0.211 * 0.211 / 0.789 * 0.578²

    Kp = 0.169
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