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19 August, 16:51

An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.323 M HI, 4.34E-2 M H2 and 4.34E-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.228 mol of HI (g) is added to the flask?

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  1. 19 August, 17:51
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    [HI] = 0.5239M

    [H₂] = 7.05x10⁻²

    [I₂] = 7.05x10⁻²

    Explanation:

    The reaction of HI to produce H₂ and I₂ is:

    2HI → H₂ + I₂

    Where K of reaction is defined as:

    K = [H₂] [I₂] / [HI]²

    Replacing with the concentrations of the gases in equilibrium:

    K = [4.34x10⁻²] [4.34x10⁻²] / [0.323] ²

    K = 0.0181

    If you add 0.228 mol = 0.228M (Because volume of the flask is 1.0L), the concentration when the system reaches the equilibrium are:

    [HI] = 0.228M + 0.323M - X = 0.551M - X

    [H₂] = 4.34x10⁻² + X

    [I₂] = 4.34x10⁻² + X

    Where X is reaction coordinate.

    Replacing in K formula:

    K = 0.0181 = [4.34x10⁻² + X] [4.34x10⁻² + X] / [0.551 - X] ²

    0.0181 = 0.00188356 + 0.0868 X + X² / 0.303601 - 1.102 X + X²

    0.005495 - 0.01995 + 0.0181X² = 0.00188356 + 0.0868 X + X²

    0 = - 0.003611 + 0.10675X + 0.9819X²

    Solving for X:

    X = - 0.136 → False solution, there is no negative concentrations.

    X = 0.0271 → Right solution.

    Replacing for concentrations of each species:

    [HI] = 0.5239M[H₂] = 7.05x10⁻²[I₂] = 7.05x10⁻²
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