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9 March, 14:23

5. The rate constant for the reaction 3A  4B is 6.00 x 10-3 L•mol-1min-1. How long will it take the concentration of A to drop from 0.75 M to 0.25 M?

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  1. 9 March, 14:34
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    it will take 444.44 min to reach 0.25 M from 0.75 M

    Explanation:

    since the reaction constant is k = 6.00*10⁻³ L*mol⁻¹min⁻¹. For a reaction rate of the form

    -dCa/dt=k*Caⁿ

    doing a dimensional analysis

    [dCa/dt] = mol / (L*min) = [k]*[Caⁿ] = L / (mol*min) * (mol/L) ⁿ

    then only n=2 can comply with the dimensional analysis, therefore we get a the second order reaction. Thus

    -dCa/dt=k*Ca²

    -dCa/Ca² = k*dt

    -∫dCa/Ca² = k*∫dt

    (1/Ca₂ - 1/Ca₁) = k*Δt

    Δt = 1/k * (1/Ca₂ - 1/Ca₁)

    replacing values

    Δt = 1/k * (1/Ca₂ - 1/Ca₁) = 1 / (6.00*10⁻³ L*mol⁻¹min⁻¹) * (1/0.25 M - 1/0.75 M) = 444.44 min
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