Question

Determine the rate law and the value of k for the following reaction using the data provided. 2 N2O5(g) → 4 NO2(g) + O2(g) [ N2O5]i (M) Initial Rate (M-1s-1) 0.093 4.84 x 10-4 0.084 4.37 x 10-4 0.224 1.16 x 10-3 Determine the rate law and the value of k for the following reaction using the data provided. 2 N2O5(g) → 4 NO2(g) + O2(g) [ N2O5]i (M) Initial Rate (M-1s-1) 0.093 4.84 x 10-4 0.084 4.37 x 10-4 0.224 1.16 x 10-3 Rate = 5.2 × 10- 3 s-1[N2O5] Rate = 6.0 × 10-1 M-2s-1[N2O5]3 Rate = 1.6 × 10-3 M1/2s-1[N2O5]1/2 Rate = 1.7 × 10-2 M-1/2s-1[N2O5]3/2 Rate = 5.6 × 10-2 M-1s-1[N2O5]2

Answer 1

2N₂O₅(g) → 4NO₂(g) + O₂(g)

[ N2O5]i (M)	Initial Rate (M-1s-1)
0.093	4.84 x 10-4
0.084	4.37 x 10-4
0.224	1.16 x 10-3

For a reaction of the type

A + 2 B → 3 C

the rate law has the following form:

r = k [A]^a [B]^b

From the equation it is clear that the rate equation will be of the form

r = k [N₂O₅]^a

Now

r1 r2

=

k [N2O5]12 k [N2O5]22

4.84 x 10^-4/1.16 x 10^-3 = k x (0.093)^a/k x (0.224)^a

0.417 = 0.415^a

from this it is clear that a =1

so the rate law is r = k [N₂O₅]

now we can calculate k

4.84 x 10^-4 = k x 0.093

k = 4.84 x 10^-4 /0.093

k = 5.2 x 10^-3M^-2s^-1