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15 September, 20:08

Given:A (g) + 3B (g) ⇋ C (g) + 2D (g) 1.0 mole of A and 1.0 mole of B are placed in a 5.0-liter container. After equilibrium has been established, 0.50 mole of D is present in the container. Calculate the equilibrium constant, Kc, for the reaction.

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  1. 15 September, 23:00
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    Kc = 26.67

    Explanation:

    Step 1: Data given

    Number of moles of A = 1.0 mol

    Number of moles of B = 1.0 mol

    Number of moles of D = 0.50 moles

    Volume = 5.0 L

    Step 2: The balanced equation

    A (g) + 3B (g) ⇋ C (g) + 2D (g)

    Step 3: Number of initial moles

    Moles A = 1.0 mol

    Moles B = 1.0 mol

    Moles C = 0 moles

    Moles D = 0 moles

    [A] = 1.0 mol / 5.0 L = 0.2 M

    [B] = 1.0 mol / 5.0 L = 0.2 M

    [C] = 0 M

    [D] = 0 M

    Step 4: Moles at the equilibrium

    For 1 mol A we have 3 moles B to produce 1 mol C and 2 moles D

    There will react X of A, this give (1.0 - X) moles

    There will reacto 3X for B this gives (1.0 - 3X) moles

    There will be X produced for C this gives X moles

    There will be 2X produced for D this gives 2X moles

    After equilibrium has been established, 0.50 mole of D

    2X = 0.50 moles or X = 0.25 moles this gives 0.25 moles / 5.0 L = 0.05 M

    Step 5: Calculate Kc

    The concentration at the equilibrium

    [A] = 0.2 - 0.05 = 0.15M

    [B] = 0.2 - 0.15 = 0.05M

    [C] = 0.05M

    [D] = 0.1M

    Kc = (0.05) (0.1) ² / (0.15) (0.05) ³

    Kc = 26.67
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