Question

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?

Answer 1

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