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