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Determine the rate law and the value of k for the following reaction using the data provided
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
Expert Solution
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 = 0.415a
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
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