For the reaction 2CIO_2 + 2OH^- rightarrow CIO_3^- + CIO_2^- + H_2CO(I) A. Use t

Question

For the reaction 2CIO_2 + 2OH^- rightarrow CIO_3^- + CIO_2^- + H_2CO(I) A. Use the table below to determine the rate law, the overall order, value of the rate constant k, and the rate of the fourth experiment and the energy of activation. B. The following mechanism proposed: CIO_2 + CIO_2 rightarrow CI_2O_4 CI_2O_4 + OH^- rightarrow CIO_3^- + HCIO_2 HCIO_2 + OH^- rightarrow CIO_2^- + H_2O Does this mechanism agree with the overall reaction and the rate law determined from the data above? Show reasoning.

Explanation / Answer

Let the rate law be –r = K[ClO2]a [OH-]b,

From experiment1, 1.4*10-3 = K[0.020]a [0.050]b                    (1)

From experiment 2, 1.4*10-3 = K[0.020]a [0.1000]b                  (2)

Eq.2/Eq.1 gives 2b=1, b=0 hence zero order with respect to [OH-]

From experiment3, 7*10-4 = K[0.010]a [0.050]b                                 (3)

Eq.3/ Eq.1 gives 2= 2a, a= 1

Hence the rate law becomes –r = K[ClO2]1

From Eq.1, 1.4*10-3 = K[0.020]

K= 0.07/sec the rate law becomes – r= 0.07[ClO2]

For 4th experiment, r =0.07*0.035=0.00245 M/s

in a series of reaction, the slowest step will be the rate limiting step.

Accordingly, the rate will be r = K[ClO2]2 which is different from given mechanism.

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