2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: 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. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

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

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

answer seismograph;