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5 July, 15:12

If the initial concentration of ab is 0.290 m, and the reaction mixture initially contains no products, what are the concentrations of a and b after 80 s?

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  1. 5 July, 15:22
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    According to the integrated second order rate law is:

    1/[AB] = 1/[AB]o + Kt

    when K is constant (given) = 5.5 x 10^-2

    and t is the time = 80 s

    and [AB]o is the initial concentration = 0.29 m

    so by substitution:

    1/[AB] = 1/0.29 + (5.5 x 10^-2) * 80

    ∴[AB] = 0.127 m

    from the reaction equation, we can see the molar ratio between AB& B & A is 1:1:1

    the [AB] lost = 0.29 m - 0.127m

    = 0.163 m

    so both A & B will gain the same number of moles

    ∴∴[A] = [B] = 0.163 m
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