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
