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15 August, 07:20

A 2.320 mol quantity of NOCl was initially placed in a 1.750 L reaction chamber at 400°C. After equilibrium was established, it was found that 28.70 percent of the NOCl has dissociated: 2NOCl (g) ⇆ 2NO (g) + Cl2 (g) Calculate the equilibrium constant Kc for the reaction.

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  1. 15 August, 07:47
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    Kc = 0.0307

    Explanation:

    Step 1: Data given

    Number of moles NOCl = 2.320 moles

    Volume = 1.750 L

    Temperature = 400 °C

    NOCl was 28.70 % dissociated

    Step 2: The balanced equation

    2NOCl (g) ⇔ 2NO (g) + Cl2 (g)

    Step 3: Calculate molarity NOCl

    [NOCl] = moles / volume

    [NOCl] = 2.320 moles / 1.750 L

    [NOCl] = 1.326 M

    Step 4: Initial concentration

    [NOCl] = 1.326 M

    [NO] = 0M

    [Cl2] = 0M

    Step 5: Concentration at equilibrium

    For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

    [NOCl] = (1.326 - 2X) M

    [NO] = 2X M

    [Cl2] = XM

    Since NOCl is 28.70 % dissociated

    ([NOCl]0 - [NOCl]equi) = 0.2870[NOCl]0

    2x = 0.2870 (1.326) ⇒ x = 0.190

    Step 6: Calculate Kc

    Kc = [NO]²[Cl2] / [NOCl]²

    Kc = (2x) ²*x / (1.326 - 2x) ²

    Kc = [2 (0.190) ]² * (0.190) / [1.326 - 2 (0.190) ]²

    Kc = 0.0307
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