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15 November, 06:36

Hydrogen gas and gaseous iodine will react to form hydrogen iodide, as described by the following chemical equation. H2 (g) + I2 (g) ↽--⇀HI (g) Kc (400∘C) = 50.0Assume that all of the HI (g) is removed from a vessel containing this reaction, and equilibrium is re-established. What will be the new equilibrium concentration of HI if the equilibrium concentrations of H2 and I2 are both equal to 0.450M?

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  1. 15 November, 07:07
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    3.18 M

    Explanation:

    Given data

    [H₂]eq = [I₂]eq = 0.450 M

    [HI]eq = ?

    Let's consider the following reaction at equilibrium.

    H₂ (g) + I₂ (g) ⇄ 2 HI (g) Kc (400°C) = 50.0

    The concentration equilibrium constant (Kc) is:

    Kc = [HI]²eq / [H₂]eq * [I₂]eq

    [HI]²eq = Kc * [H₂]eq * [I₂]eq

    [HI]eq = √ (Kc * [H₂]eq * [I₂]eq)

    [HI]eq = √ (50.0 * 0.450 * 0.450)

    [HI]eq = 3.18 M
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