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21 October, 14:45

A mixture of 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00-L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was 0.233 M. What is the value of Kc for this reaction? CO (g) + Br2 (g) COBr2 (g)

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  1. 21 October, 15:11
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    The value for the equilibrium constant Kc is 5.23

    Explanation:

    Step 1: Data given

    Number of moles CO = 0.500 moles

    Number of moles Br2 = 0.400 moles

    Volume = 1.0 L

    The equilibrium concentration of COBr2 was 0.233 M.

    Step 2: The balanced equation

    CO (g) + Br2 (g) ⇆ COBr2 (g)

    Step 3: The initial concentrations

    Concentration = mol / volume

    [CO] = 0.500 moles / 1L = 0.500 M

    [Br2] = 0.400 moles / 1L = 0.400 M

    [COBr2] = 0M

    Step 4: The concentration at the equilibrium

    [CO] = 0.500 - X M

    [Br2] = 0.400 - X M

    [COBr2] = XM = 0.233 M

    [CO] = 0.500 - 0.233 = 0.267 M

    [Br2] = 0.400 - 0.233 = 0.167 M

    [COBr2] = XM = 0.233 M

    Step 5: Calculate Kc

    Kc = [COBr2]/[CO][Br2]

    Kc = 0.233 / (0.267*0.167)

    Kc = 5.23

    The value for the equilibrium constant Kc is 5.23
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