Question

Please explain rules as well/ Calculations, and thank you for your patience

2. 2ClO2(aq)+2OH-(aq)-ClO3-(aq)+ClO2-(aq)+H2O(I)

Experiment number= 1,2 ,3
[ClO2](M) = 0.060,0.020,0.020
[OH-](M) =0.020,0.030,0.00276
Initial Rate(M/s)=0.0248,0.00276,0.00828

(A) What is the order of the overall reaction?

(B) What is the order of the reaction with respect to ClO2?

Answer 1

From experiment number 2 and 3
[ClO2] is kept constant at 0.020 M.
[OH-] is decreased from 0.030 M to 0.00276 M

The rate of the reaction increases from 0.00276 M/s to 0.00828 M/s.

Take log on both sides

Thus, the order with respect to hydroxide ion is -0.5.

From experiment number 1 and 2
[ClO2] is decreased from 0.060 M to 0.020 M.

[OH-] is increased from 0.020 M to 0.030 M

The rate of the reaction increases from 0.0248 M/s to 0.00276 M/s.

Take log on both sides

Thus, the order with respect to   ion is 2.

The overall order of the reaction is