Question

For the reaction:

3A + 2B -----> Products

Use the data below to determine the rate law

[A] (M)	[B] (M)	Rate (Ms-1)
0.300	0.200	3.067 x 10-3
0.600	0.400	8.680 x 10-3
0.900	0.200	5.131 x 10-3

Answer 1

            Let the rate (-r) be represented as

-r = K[A]^a [B]^b, where a and b are orders of reactions with respect to A and B respectively. K is the rate constant.

From 1^st data point

3.067*10^-3= K[ 0.3]^a[0.2]^b          (1)

From second data point

8.68*10^-3= K[ 0.6]^a[0.4]b          (2)

From 3^rd data point

5.131*10^-3= K[ 0.9]^a[0.2]^b      (3)      

Eq.3/Eq.1 gives

5.131/3.067= [3]^a,   

1.7 = [3]a,  

Taking ln, ln(1.7)= a ln3

And hence a=0.5

Eq.1 becomes

3.067*10^-3= K[ 0.3]^0.5[0.2]b          (1)

Eq.2 becomes

8.68*10^-3= K[ 0.6]^0.5[0.4]b          (2)

8.68/3.067= (0.6/0.3)^0.5*2^b    (4)

2^b=2, b=1

Eq.1 becomes

3.067*10^-3= K[0.3]^0.5 [0.2]

K= 0.0279/M^0.5sec

Hence the rate equation becomes –r= 0.0279*[A]⁰.⁵ [B]