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30 November, 23:31

At equilibrium, 0.170 mol of o2 is present. calculate kc.

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  1. 1 December, 00:42
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    (missing in your question):

    0.66 mol of SO3 is placed in 4.5 L container according to this reaction

    2SO3 (g) ↔ 2SO2 (g) + O2 (g)

    So the answer:

    for the equilibrium reaction so:

    Kc = [SO2]^2 * [O2] / [ SO3]^2

    when concentration (c) = n / V when n = no. of moles / v (volume per L)

    ∴ [ SO3]° = 0.66 / 4.5 = 0.147 M

    by using Ice table we can determine the concentrations:

    2SO3 (g) ↔ 2SO2 (g) + O2 (g)

    0.147-2X 2x X

    ∴ [ O2] = 0.17 M / 4.5 L = 0.0378 M

    So X = 0.0378

    ∴ [SO3] = 0.147 - 2 * 0.0378 = 0.0714 M

    and [ SO2] = 2 * 0.0378 = 0.0756 M

    ∴ by substitution in Kc formula

    ∴Kc = [0.0756]^2 * [0.0378] / [0.0714] = 0.003
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