Question

Given the following data collected from experiments, determine the rate law, overall order, and the numerical value for the rate constant.

Experiment	[A]	[B]	Initial Rate (M sec^-1)
1	0.250	0.400	0.400
2	0.0500	0.800	0.0320
3	0.0500	0.100	0.00400

Answer 1

Let the rate law be:

rate = k [A]^x [B]^y

put values from experiment 2:

0.0320 = k [0.0500]^x [0.800]^y …eqn 1

put values from experiment 3:

0.00400 = k [0.0500]^x [0.100]^y ..eqn 2

divide eqn 1 by eqn 2

0.0320/0.00400 =[0.800/0.100]^y

8 =[8]^y

y=1

rate law now is:

rate = k [A]^x [B]

put values from experiment 1

0.400 = k [0.250]^x 0.400 ..eqn 3

eqn 1 becomes

0.0320 = k [0.0500]^x 0.800 ..eqn 4

divide eqn 3 by eqn 4

0.400/0.0320 = (0.250/0.0500)^x * (0.400/0.800)

12.5 = (5)^x * (1/2)

25 = (5)^x

x = 2

so, rate law is:

rate = k [A]^2 [B]

overall order = 2+1 = 3

to find K put values from experiment 1:

0.400 M/s = k (0.250 M)^2 (0.400 M)

K = 16.0 M-2.s-1

Answer:

1)rate = k [A]^2 [B]

2) order = 3

3) 16.0 M-2.s-1