Question

Given the following sample data from experiments, determine the orders m and n.

Rate = k [Bleach]^m [Dye]^n

Run 1)
[Bleach] .08000 M
[Dye] .04000 M
Rate: 22.4 x 10^-3 (M/s)

Run 2)
[Bleach] .06500
[Dye] .02000
Rate: 3.00 x 10^-3 (M/s)

Run 3)
[Bleach] .02200
[Dye] .04000
Run: 0.466 x 10^-3

Run 4)
[Bleach] .08000
[Dye] .02000
Rate: 5.60 x 10^-3 (M/s)

Run 5)
[Bleach] .04000
[Dye] .04000
Rate: 2.80 x 10^-3 (M/s)

a) m = 0 and n =1
b) m = 1 and n = 2
c) m = 2 and n = 3
d) m = 3 and n = 4
e) m = 3 and n = 2

Answer 1

First take Run 1 and run 4:

[ Rate ]4/[ Rate ]1 = [0.0800]^m * [0.0200]^n/ [0.0800]^m * [0.04 00]^n

5.60 x 10^-3 (M/s) /22.4 x 10^-3 (M/s) = 0.0200]^n/ [0.04 00]^n

0.25 = [0.5)n

Thus n= 2

Order of reaction; with respect Dye is 2.

Now take Run 1 and run 3:

[ Rate ]3/[ Rate ]1 = [0.02200 ]^m * [0.0400]^n/ [0.0800]^m * [0.04 00]^n

0.466 x 10^-3 (M/s) /22.4 x 10^-3 (M/s) = 0.0220]^m/ [0.0800]^m

0.02080   = [0.275 )m

(0.275)^3 = [0.275 )m

Thus m= 3

Order of reaction ; with respect bleach is 3.

e) m = 3 and n = 2