Determine the rate law according to the following data. [H2] (M) [NO] (M) Rate (M/s) Trial 1 0.35 0.30 2.835 x 10-3 Trial 2 0.35 0.60 1.134 x 10-2 Trial 3 0.70 0.60 2.268 x 10-2

rate = k [H2O] [NO]²

Explanation:

[H2] (M) [NO] (M) Rate (M/s)

Trial 1 0.35 0.30 2.835 x 10-3

Trial 2 0.35 0.60 1.134 x 10-2

Trial 3 0.70 0.60 2.268 x 10-2

Comparing Trial 1 and 2 , the rate of the reaction increases by a factor of 4 when the concentration of NO was doubled. This means that the reaction is second order with respect to NO

Comparing Trial 2 and Trial 3, the rate of reaction increased by a factor of 2 when the concentration of H2O was doubled. This means that the reaction is first order with respect to H2O

The rate law fr the reaction is given as;

rate = k [H2O] [NO]²

the answer for sodium is 0.186

the answer for rubidium is 0.251

a:

by the combination of an unlimited number of elements