Question

Given the initial rate data for the reaction, 4 A + 7 B

Answer 1

Let the rate law be r = k[A]^m[B]ⁿ                                              ---------------------- (1)

Where

r = rate of the reaction

k = rate constant

m = order of the reaction with respect to A

n = order of the reaction with respect to B

Now we can write the rate expression to the readings in the given tabular form one by one

For Reading (1)    rate law becomes 1.04 = k(0.050)^m (0.050)ⁿ              ------------------ (1)

For Reading (2)    rate law becomes   3.1 =k(0.050)^m (0.150)ⁿ              ------------------ (2)

For Reading (1)    rate law becomes 26 = k(0.250)^m (0.050)ⁿ              ------------------ (3)

Eqn(2) / Eqn(1) becomes ,

3.1/1.04 = (0.150 / 0.050 ) ⁿ

   3 = 3ⁿ

--> n = 1

Eqn(3) / Eqn(1) becomes ,

26/1.04 = (0.250 / 0.050 )^m

   25 = 5^m

            5²=5^m

--> m = 2

Therefore the rate law is r = k[A]²[B]¹          OR       r = k[A]²[B]