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28 May, 14:17

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

A + B yields products

Trial [A] [B] Rate

1 0.30 M 0.25 M 1.2 * 10-2 M/min

2 0.30 M 0.50 M 4.8 * 10-2 M/min

3 0.60 M 0.50 M 9.6 * 10-2 M/min

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  1. 28 May, 14:26
    0
    Rate = (0.64 M²/min) [A]¹[B]²

    Explanation:

    1) Determination of the orders of A & B:

    The rate law of the reaction = k [A]ᵃ[B]ᵇ

    where, k is the rate law constant,

    a is the order of the reaction with respect to reactant A,

    b is the order of the reaction with respect to reactant B.

    This is initial rate method problem:

    From trial 1 & 2:

    Reactant [A] has the same concentration in both trials, but [B] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [B]. From trial 1, Rate₁ = k [A₁]ᵃ[B₁]ᵇ, [1.2 * 10⁻² M/min] = k [0.30 M]ᵃ[0.25 M]ᵇ From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 * 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ

    By dividing Rate₁ / Rate₂:

    Rate₁ / Rate₂ = k [A₁]ᵃ[B₁]ᵇ / k [A₂]ᵃ[B₂]ᵇ

    [1.2 * 10⁻² M/min] / [4.8 * 10⁻² M/min] = k [0.30 M]ᵃ[0.25]ᵇ / k [0.30 M]ᵃ[0.50 M]ᵇ

    0.25 = [0.50]ᵇ

    Taking log for both sides; log (0.25) = b log (0.5)

    b = log (0.25) / log (0.5) = 2.

    The reaction is second order with respect to reactant B.

    By the same way for reactant A:

    From trial 2 & 3:

    Reactant [B] has the same concentration in both trials, but [A] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [A]. From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 * 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ From trial 2, Rate₃ = k [A₃]ᵃ[B₃]ᵇ, [9.6 * 10⁻² M/min] = k [0.60 M]ᵃ[0.50 M]ᵇ

    By dividing Rate₂ / Rate₃:

    Rate₂ / Rate₃ = k [A₂]ᵃ[B₂]ᵇ / k [A₃]ᵃ[B₃]ᵇ

    [4.8 * 10⁻² M/min] / [9.6 * 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ / k [0.60 M]ᵃ[0.50 M]ᵇ

    0.50 = [0.50]ᵃ

    Taking log for both sides; log (0.50) = a log (0.5)

    a = log (0.50) / log (0.50) = 1.

    The reaction is first order with respect to reactant A.

    The rate law of the reaction will be: Rate = k [A]¹[B]² The overall order of the reaction is third order reaction.

    2) Determining the rate law constant:

    Taking data of trial 1:

    Rate = k [A]¹[B]²

    [1.2 * 10⁻² M/min] = k [0.30 M]¹[0.25 M]²

    k = [1.2 * 10⁻² M/min] / [0.30 M]¹[0.25 M]² = 0.64 M²/min.

    So, the rate law will be: Rate = (0.64 M²/min) [A]¹[B]²
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