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12 October, 12:04

2 ClO2 (aq) + 2OH - (aq) → ClO3 - (aq) + ClO2 - + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

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  1. 12 October, 12:12
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    2 ClO2 (aq) + 2OH - (aq) → ClO3 - (aq) + ClO2 - + H2O (l)

    The data is given as;

    Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)

    1 0.060 0.030 0.0248

    2 0.020 0.030 0.00276

    3 0.020 0.090 0.00828

    a) Rate law is given as;

    Rate = k [ClO2]^x [OH-]^y

    From experiments 2 and 3, tripling the concentration of [OH-] also triples the rate of the reaction. This means the reaction is first order with respect to [OH-]

    From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]

    Rate = k [ClO2]² [OH-]

    b. Calculate the value of the rate constant with the proper units.

    Taking experiment 1,

    0.0248 = k (0.060) ² (0.030)

    k = 0.0248 / 0.000108

    k = 229.63 M-2 s-1

    c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

    Rate = 229.63 [ClO2]² [OH-]

    Rate = 229.63 (0.100) ² (0.050)

    Rate = 0.1148 M/s
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