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.
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
N2 (g) + O2 (g) --> 2NO (g) The following data was recorded: Exper [N2] [O2]...
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