Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given the data in the accompanying table, what is the rate law for the reaction?

The rate equation is as follows;

Rate = k [A]ˣ[B]ⁿ

where k - rate constant

x and n are order of the reaction with respect to [A] and [B] respectively

the data is tabulated as follows

experiment [A] [B] rate mol/Ls

1 0.20 0.10 20

2 0.20 0.20 40

3 0.40 0.20 160

to find x lets take the data for experiment 2 and 3 where [B] is constant for both

40 mol/Ls = k [0.20 mol/L] ˣ[0.20 mol/L]ⁿ - - - 1)

160 mol/Ls = k [0.40 mol/L]ˣ[0.20 mol/L]ⁿ - - 2)

divide 2) by 1)

160/40 = (0.4/0.2) ˣ

4 = 2ˣ

x = 2

to find n we have to take data from experiment 1 and 2 where [A] is constant

40 mol/Ls = k [0.20 mol/L]ˣ[0.20 mol/L]ⁿ - - - 1)

20 mol/Ls = k [0.20 mol/L]ˣ[0.10 mol/L]ⁿ - - - 2)

divide 1) by 2)

40/20 = (0.2/0.1) ⁿ

2 = 2ⁿ

n = 1

order of the reaction with respect to [A] is 2

and order of the reaction with respect to [B] is 1

rate law for the equation is

Rate = k [A]²[B]¹

This a third order reaction as the sum of the orders of the reactants is 3