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11 September, 20:39

In an analysis of interhalogen reactivity, 0.440 mol of ICl was placed in a 5.00-L flask, where it decomposed at a high T: 2 ICl (g) ⇌ I2 (g) + Cl2 (g) Calculate the equilibrium concentrations of I2, Cl2, and ICl (Kc = 0.110 at this temperature). [I2]eq: M [Cl2]eq: M [ICl]eq: M

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  1. 11 September, 21:05
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    [I2] = [Cl2] = 0.0176 M

    [ICl] = 0.0528 M

    Explanation:

    Step 1: Data given

    Number of moles ICI = 0.440 moles

    Volume = 5.00 L

    Kc = 0.110

    Step 2: The balanced equation

    2 ICl (g) ⇌ I2 (g) + Cl2 (g)

    Step 3: The initial concentrations

    [[ICl] = 0.440 moles/5.00 L = 0.088 M

    [I2[ = 0M

    [Cl2] = 0M

    Step 4: The concentrations at the equilibrium

    [[ICl] = (0.088 - 2x) M

    [I2[ = xM

    [Cl2] = xM

    Step 5: Define Kc

    Kc = 0.110 = [I2][Cl2] / [ICl]²

    0.110 = x² / (0.088 - 2x) ²

    0.332 = x (0.088 - 2x)

    x = 0.332 * (0.088 - 2x)

    x = 0.029216 - 0.664x

    1.664x = 0.029216

    x = 0.0176

    [I2] = [Cl2] = 0.0176 M

    [ICl] = 0.088 - 2 (0.0176) = 0.0528 M
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