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 : The order of reaction is 2 (second order) and the rate law is,

Explanation :

As the expression of rate is,

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

   .............(1)

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

   .............(2)

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

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

n = 2

Thus, the order of reaction = n = 2

The rate law will be,

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,

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