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13 November, 12:38

The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibrium constant of 1.15 x 102 at a certain temperature. H2 (g) + F2 (g) ⇋ 2 HF (g) In a particular experiment, 3.00 mol H2 and 6.00 mol F2 are mixed in a 3.00 L flask at this temperature and allowed to come to equilibrium. Given this information, calculate the concentrations of H2, F2, and HF present in the flask at equilibrium.

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  1. 13 November, 12:57
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    [H2]eq = 0.0129 M

    [F2]eq = 1.0129 M

    [HF]eq = 0.9871 M

    Explanation:

    H2 (g) + F2 (g) ↔ 2HF (g)

    ∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

    experiment:

    ∴ n H2 = 3.00 mol

    ∴ n F2 = 6.00 mol

    ∴ V sln = 3.00 L

    ⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

    ⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

    [ ]i change [ ]eq

    H2 1 1 - x 1 - x

    F2 2 2 - x 2 - x

    HF - x x

    ⇒ K = (x) ² / (1 - x) * (2 - x) = 1.15 E2

    ⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

    ⇒ x² = (2 - 3x + x²) (115)

    ⇒ x² = 230 - 345x + 115x²

    ⇒ 0 = 230 - 345x + 114x²

    ⇒ x = 0.9871

    equilibrium:

    ⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

    ⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

    ⇒ [HF] = x = 0.9871 M
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