Question

Using the given data, determine the rate constant of this reaction.

A+2B ------> C+D

Trial [A](M) [B](M) Rate (M/s)

1. 0.270 0.380 0.0235

2 0.270 0.760 0.0235

3 0.540 0.380 0.0940

K=?

What units should be used?

Answer 1

In this problem we first write rate law and then find the order by using given data table.

Rate law :

Rate = k [ A]^m[B]ⁿ

We know k is rate constant , m and n are the orders with respect to A and B. In order to get the value of m and n we have to select two experimental trials in which one of the reactants concentration is constant.

Lets choose trial 1 and trial 2

Lets take ratio of tiral 2 : trial 1

Rate 2 / Rate 1 = k ([ A]^m [B]ⁿ)_{trial 2} /k ([ A]^m [B]ⁿ)_{trial 1}

Here k and concentration of A is constant so these are cancelled.

Lets plug the values

0.0235 / 0.0235 = ( 0.760 / 0.380 )ⁿ

n = 0

Now we calculate value of m

This time we use trial 3 : trial 1

Rate 3 / Rate 1 = k ([ A]^m [B]ⁿ)_{trial 3} /k ([ A]^m [B]ⁿ)_{trial 1}

0.0940 / 0.0235 = ( 0.540/0.270)^m

4 = 2^m

m = 2

so the rate law becomes

rate =k [B]²

to calculate the value of k we have to use trial 1

k = rate / [B]² = 0.0235 / ( 0.380)²= 0.1627

rate constant = 0.163