Explanation:
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
