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8 August, 17:39

At 350°c, keq = 1.67 * 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibrium if [ h2 ] is 2.44 * 10-3 m and [ i2 ] is 7.18 * 10-5 m?

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  1. 8 August, 17:56
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    According to the reversible reaction equation:

    2Hi (g) ↔ H2 (g) + i2 (g)

    and when Keq is the concentration of the products / the concentration of the reactants.

    Keq = [H2][i2]/[Hi]^2

    when we have Keq = 1.67 x 10^-2

    [H2] = 2.44 x 10^-3

    [i2] = 7.18 x 10^-5

    so, by substitution:

    1.67 x 10^-2 = (2.44 x 10^-3) * (7.18x10^-5) / [Hi]^2

    ∴[Hi] = 0.0033 M
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