Question

Consider the base­catalyzed​reaction OCl­ (aq) + I­ (aq) → OH­ (aq) + Cl­ (aq) Determine the rate law and the rate constant, using the following initial rate data. (Hint: base­catalyzed​indicates that OH ­ affects the rate despite appearing as a product.)

[OCl-]	1.62e-3	1.62e-3	.00271	.00162
[I-]	1.62e-3	2.88e-3	.00162	.0028
[OH-]	.52	.52	.84	.91
initil rate m/s	.000306	.00054	.000316	.000311

Answer 1

Solution:

Rate law:

Since given given reaction is base catalyzed and concentration of OH- also affects the overall rate.

so we write rate equation :

Rate = k [OCl-]^l [I- ]^m [ OH-]ⁿ

Here k is rate constant , l , m and n are the orders with respect to OCl- , I- , and OH- respectively.

We use given experimental data to find the orders.

Lets number of the experiment.

Lets compare experiment 1 and 2 and use ratio of rate law.

Lets take ln of both side

Now ration of rate 4 and rate 2

Lets find value of n

n = -1

Lets find value of l

We use rate 3 and rate 1

l = 1

Hence the rate law

Rate = k [OCl-] [I- ] [ OH-]^-1

Calculation of rate constant:

Lets use rate 1 to get rate constant

k = 60.631