The reaction: 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.

The rate law is defined as the rate of a chemical reaction, which is dependent on the concentration of the reactants.

It can be calculated by the formula:

where,

In the given reaction:

Using the rate law formula, we get:

Rate = K [Cl]ⁿ  [OH⁻]ⁿ

0.0248 =  K [0.060]ⁿ  [0.030 ]ⁿ       ........(1)

0.00276 = K  [0.020]ⁿ  [0.030 ]ⁿ   .........(2)

0.00828 = K  [0.020]ⁿ  [0.090]ⁿ     ..........(3)

Dividing (1) equation by (2), we get:

Dividing the equation (2) by (3), we get:

Based on the rate law:

a. Rate = K [ClO₂]²⁻ [OH⁻]¹

b. 0.0248 = K [0.06]² [0.03]¹

   K =  = 229.63 m⁻²s⁻¹

c. Rate = K [ClO]₂ [OH⁻]ⁿ

  Rate = 229.62 x (0.2)² (0.05)¹

Therefore, the rate of the reaction is 0.45926 m/sec.

To know more about the rate law of the reaction, refer to the following link:

link

Explanation:

question solved.

i know insulin is primary but i dont know the others im sorry

explanation: