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10 June, 07:57

A student ran the following reaction in the laboratory at 1089 K: 2SO3 (g) 2SO2 (g) + O2 (g) When she introduced 8.39*10-2 moles of SO3 (g) into a 1.00 liter container, she found the equilibrium concentration of O2 (g) to be 1.78*10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

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  1. 10 June, 08:15
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    The equilibrium constant, Kc is 0.00967

    Explanation:

    Step 1: Data given

    Temperature = 1089 K

    Number of moles SO3 = 8.39 * 10^-2 moles = 0.0839 moles

    Volume = 1.0 L

    The equilibrium concentration of O2 (g) to be 1.78 * 10^-2 M

    Step 2: The balanced equation

    2SO3 (g) ⇆ 2SO2 (g) + O2 (g)

    Step 3: The initial concentration

    Concentration = moles / volume

    [SO3] = 0.0839 moles / 1 L = 0.0839 M

    [SO2] = 0M

    [O2] = 0M

    Step 4: The concentration at equilibrium

    For 2 moles SO3 we'll have 2 moles SO2 and 1 mol O2

    [SO3] = 0.0839 - 2X M

    [SO2] = 2X M

    [O2] = XM = 1.78 * 10^-2 M = 0.0178 M

    So X = 0.0178

    [SO3] = 0.0839 - 2*0.0178 = 0.0483 M

    [SO2] = 2*0.0178 = 0.0356 M

    [O2] = XM = 1.78 * 10^-2 M = 0.0178 M

    Step 5: Calculate the equilibrium constant, Kc

    Kc = [O2][SO2]² / [SO3]²

    Kc = (0.0178 * 0.0356²) / 0.0483²

    Kc = 0.00967

    The equilibrium constant, Kc is 0.00967
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