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18 July, 09:05

A reversible reaction has a forward rate constant of 0.412 mol/L/s and a reverse reaction rate constant of 0.827 mol/L/s. What's the equilibrium constant for this reaction?

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  1. 18 July, 11:00
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    Answer: 0.498

    Explanation:

    Given the following:

    Forward rate constant (Kf) = 0.412 mol/L/s

    Reverse rate constant (Kr) = 0.827 mol/L/s

    The ratio of the rate constant for the forward reaction and the rate constant for the reverse reaction produces the equilibrium constant.

    Therefore,

    Keq = (forward rate constant (Kf) / Reverse rate constant)

    Keq = 0.412mol/L/s / 0.827mol/L/s

    Keq = 0.498
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