since they both = k x [NO]^m.. they = each other
rate 1 / [H2 #1]^n = rate 2 / [H2 #2]^n
rearranging...
[H2 #1]^n / [H2 #2]^n = rate 1 / rate 2
([H2 #1] / [H2 #2]) ^n = rate 1 / rate 2
solving...by plugging in [H2] and rate ..and I canceled out the x10^-3
(2/4)^n = (4/8)
(1/2)^n = (1/2)
so n must = 1
and now we know..
rate = k x [H2] x [NO]^m
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likewise...
from experiment 1 to 4.. [H2] was constant.. and [NO] varied..
([NO #1] / [NO #2]) ^m = rate 1 / rate 2
solving..
(4/2)^m = (4/1)
2^m = 4
m must = 2
now we have this equation..
rate = k x [H2] x [NO]²
*********
what's next?
a).. find k
b).. use that k and [H2] and [NO] to fill out the table...
k = rate / ([H2] x [NO]²)
you can pick any of the 5 experiments that you know rate, [H2] and [NO] to calc k.. I'll pick the first.
k = 4.0x10^-3 M/s / ((2.0x10^-3 M) x (4.0x10^-3 M)²)
k = 1.25x10^5 /(M² sec)
notice the units? moles / dm3 = M.. molarity.. moles / L.. you can write that as 1.25x10^5 L² / mole² sec.. if you wish.
********
then...
rate 3 = 1.25x10^5 /(M² sec) x (6.0x10^-3M) x (4.0x10^-3M)² = 12.0 x10^-3
rate 5 = 1.25x10^5 /(M² sec) x (6.0x10^-3M) x (4.0x10^-3M)² = 0.25 x10^-3
ok?
****************
questions?
can you deduce the mechanism? n=1.. m=2?