Question

Consider the following initial rate data for the decomposition of AB to yield A and B: [AB], mol/L Initial Rate, mol/(L. s) 0.200 0.00320 0.400 0.0128 0.600 0.0288 a. What is the order and rate law?

Answer 1

Answer : The order of reaction is 2 (second order) and the rate law is, $R=k[AB]^2$

Explanation :

As the expression of rate is,

$R=k[AB]^n$

where,

R = rate

k = rate constant

[AB] = concentration of AB

n = order of the reaction

First we have to calculate the rate for 0.200 mol/L 'AB' concentration with initial rate 0.00320 mol/L.s

$R_1=k[AB_1]^n$

$0.00320=k(0.200)^n$    .............(1)

Now we have to calculate the rate for 0.400 mol/L 'AB' concentration with initial rate 0.0128 mol/L.s

$R_2=k[AB_2]^n$

$0.0128=k(0.400)^n$    .............(2)

Now dividing the equation 1 by equation 2, we get the order of reaction.

$\frac{0.00320}{0.0128}=\frac{k(0.200)^n}{k(0.400)^n}$

By solving the term, we get the value of 'n'.

n = 2

Thus, the order of reaction = n = 2

The rate law will be,

$R=k[AB]^2$

So we conclude that,

For n = 0, the rate law will be zero order.

For n = 1, the rate law will be first order.

For n = 2, the rate law will be second order.

and so on.....

Hence, the order of reaction is 2 (second order) and the rate law is, $R=k[AB]^2$