Question

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

Rate= k [Bleach]^m [Dye]ⁿ

Run	[Bleach] M	[Dye] M	Rate (M/s)
1	.08000	.04000	22.4 x 10-3
2	.06500	.02000	3.00 x 10-3
3	.02200	.04000	0.466 x 10-3
4	.08000	.02000	5.60 x 10-3
5	.04000	.04000	2.80 x 10-3

		m = 0 and n = 1
		m = 1 and n = 2
		m = 2 and n = 3
		m = 3 and n = 4
		m = 3 and n = 2

Answer 1

ans)

from above data that

from using above table

we have to take first Run 1 and run 4:

find out the rate of recation

[ Rate ]4/[ Rate ]1

using above values

= [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

we get n= 2

so from the above data

the Order of reaction; with respect Dye is 2.

Now we have to take Run 1 and run 3:

find the rate constant

[ 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

we can get m= 3

so the Order of reaction ; with respect bleach is 3.

so finally we get

m = 3 and n = 2

so option 5 is right