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7 December, 07:07

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 674°C. H2 (g) + I2 (g) ⇋ 2 HI (g) Kc = 49 at 674°C If the initial concentrations of hydrogen gas and iodine vapor are both 0.048 mol/L and the concentration of hydrogen iodine is 0.127 mol/L what is the equilibrium concentration of hydrogen gas? Enter a number to 4 decimal places.

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  1. 7 December, 07:10
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    [H₂] = 0.0248M

    Explanation:

    For the reaction:

    H₂ (g) + I₂ (g) ⇋ 2 HI (g)

    Kc is defined as:

    Kc = [HI]² / [I₂][H₂] = 49

    For the initial concentrations, Q of reaction is:

    Q = [0.127M]² / [0.048M][0.048M] = 7

    As Kc > Q, the reaction shift to the right, doing equilibrium concentrations as:

    [I₂] = 0.048M - x

    [H₂] = 0.048M - x

    [HI] = 0.127M + 2x

    Replacing in Kc:

    49 = (0.127+2X) ² / (0.048-X) ²

    49 = 0.0161 + 0.508X + 4X² / 0.002304 - 0.096X + X²

    0.1129 - 4.704X + 49X² = 0.0161 + 0.508X + 4X²

    0.0968 - 5.212X + 45X² = 0

    Solving for X:

    X = 0.0232M

    Thus, equilibrium concentration of hydrogen gas is:

    [H₂] = 0.048M - x = 0.048M - 0.0232M = 0.0248M
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