Question

N2 (g) + O2 (g) --> 2NO (g)

The following data was recorded:

Exper	[N2]	[O2]	Rate
1	0.04	0.02	707 mol/Lsec
2	0.04	0.01	500 mol/Lsec
3	0.01	0.01	125 mol/Lsec

Determine the rate law.

Answer 1

Let:
rate = K [N2]^x * [O2]^y

using experiment 1:
rate = K [N2]^x * [O2]^y
707 = k * 0.04^x * 0.02^y ....eqn 1

using experiment 2:
rate = K [N2]^x * [O2]^y
500 = k * 0.04^x * 0.01^y ....eqn 2

using experiment 3:
rate = K [N2]^x * [O2]^y
125 = k * 0.01^x * 0.01^y ....eqn 3

divide eqn2 by eqn 1
500/125 = (0.04/0.01)^x
4 = 4^x
a0,
x = 1

divide eqn 1 by eqn 2
707/500 = (0.02/0.01)^y
1.414 = 2^y
(2)^1/2 = 2^y
y=1/2

so,
rate = k [N2]^1 [O2]^1/2
put values from experiment 1
707 = k * (0.04)^1 * (0.02)^(1/2)
k = 124981

So rate law is:
rate = 124981 [N2] [O2]^1/2