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30 August, 09:19

Consider the following for the reaction at 300 K:3 ClO - (aq)   ClO3 - (aq) + 2 Cl - (aq) ExperimentInitial [ClO-] (M) Initial Rate of Formation of ClO3 - (aq) (M/min) 10.4521.048 * 10-420.9034.183 * 10-4 (7) (4 pts) What is the order of the reaction with respect to ClO - (aq) ? A) 0B) 1C) 2 (8) (4 pts) For experiment #2, what is the initial rate of consumption of ClO-

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  1. 30 August, 10:23
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    Second order

    Δ[ClO⁻]/Δt = - 4.183 x 10⁻⁴ M/min

    Explanation:

    Given the dа ta:

    Experiment # [ClO-] (M) Initial Rate of Formation of ClO3 - (M/min)

    1 10.452 1.048 x 10⁻⁴

    2 20.903 4.183 x 10⁻⁴

    we need to determine the order of the reaction with respect to ClO⁻.

    We know the rate law for this reaction will have the form:

    Rate = k [ClO⁻]^n

    where n is the order of the reaction. Thus, what we need to do is to study the dependence of the initial rate on n for the experiment.

    If the reaction were zeroth order the rate would not change, so we can eliminate n = 0

    If the reaction were first order, doubling the concentration of [ClO-], as it was done exactly in experiment # 2, the initial rate should have doubled, which is not the case.

    If the reaction were second order n: 2, doubling the concentration of [ClO-], should quadruple the initial rate of formation of ClO3-, which is what it is observed experimentally. Therefore the reaction is second order respect to ClO-.

    The initial rate of consumption of ClO⁻ is the same as the rate of formation of ClO₃⁻ since:

    Δ = - Δ[ClO⁻]/Δt = + Δ[ClO₃⁻]/Δt = + 1/2 [Cl⁻] / Δt

    where t is the time.

    from the coefficients of the balanced chemical equation.

    - Δ[ClO⁻]/Δt = + Δ[ClO₃⁻]/Δt = + 1/2 [Cl⁻ ] = rate

    Δ[ClO⁻]/Δt = - 4.183 x 10⁻⁴ M/min
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