The data in the table below were obtained for the reaction: 2clo2 (aq) + 2 oh- (aq) --> clo3- (aq) + clo2- (aq) + h2o (l) 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 what is the order of the reaction with respect to clo2?

Lets organise the data given in the question
                [ClO₂] (m)       [OH⁻] (m)        initial rate (m/s)
                  0.060              0.030               0.0248
                  0.020              0.030              0.00276
                  0.020              0.090                0.00828
rate equation as follows 
rate = k [ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant 
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
order with respect to ClO₂ is 2

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