# Chemistry

i need help with a rate law question:

The reaction is

2ClO2(aq) + 2OH -(aq)---> ClO3-(aq)+ ClO2-(aq) + H2O(l)

Experiment 1:

[ClO2]0= 0.0500
[OH-]0= 0.100
INITIAL RATE= 5.75*10^-2

Experiment 2:

[ClO2]0= 0.100
[OH-]0= 0.100
INITIAL RATE= 2.30*10^-1

Experiment 3:

[ClO2]0= 0.100
[OH-]0= 0.050
INITIAL RATE= 1.15*10^-1

a) determine the rate law and the value of the rate constant.

b) what would be the initial rate for an experiment with [ClO2]0= 0.175 mol/L and [OH-]0= 0.0844 mol/L?

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1. WHen i did it the answers i got was order of reaction m= 2 and k = 4.6

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I don't get that.
I have an order of 3 and k = 230
rate = k(ClO2^-)x(OH^-)y.
For x I have 2. For y I have 1. The order overall is 3. Check my work.

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