Question

For the following combination reaction,

2 NO2(g) + F2(g) = 2 NO2F(g)

The initial rates of this reaction were determined for several concentrations of the reactants. All of the data is summarized via the following table:

Experiment	[NO2] (M)	[F2] (M)	Rate (M/s)
1	0.100	0.100	0.026
2	0.200	0.100	0.051
3	0.200	0.200	0.103
4	0.400	0.400	0.411

a) Determine the rate law for this reaction based on the data above.

b) Calculate the rate constant, with appropriate units.

Answer 1

Answer – a) We are given, reaction - 2 NO₂(g) + F₂(g) ----> 2 NO₂F(g)

First we need to calculate the order for each reactant, we assume rate law is

Rate = k [NO₂]^m [F₂]ⁿ

So, m and n order with respect to NO₂ and F₂, so rate law are as follow -

Rate1 = k [NO₂]₁^m [F₂]₁ⁿ

Rate2 = k [NO₂]₂^m [F₂]₂ⁿ

Rate 3 = k [NO₂]₃^m [F₂]₃ⁿ

Rate 4 = k [NO₂]₄^m [F₂]₄ⁿ

Now first we need to calculate order with respect to NO₂

So, we need to take the same concentration of F₂

Rate2/ Rate1 = k [NO₂]₂^m [F₂]₂ⁿ / k [NO₂]₁^m [F₂]₁ⁿ

0.051 / 0.026 = (0.200)^m /(0.100)^m * (0.100)ⁿ / (0.100)ⁿ

2 = (2)^m

So, m = 1

So order with respect to NO₂ is first order

Now order with respect to F₂

Rate3/Rate2 = k [NO₂]₃^m [F₂]₃ⁿ / k [NO₂]₂^m [F₂]₂ⁿ

0.103 / 0.051 = (0.200)^m /(0.200)^m *(0.200)ⁿ /(0.100)ⁿ

2 = (2)ⁿ

So, n = 1

So order with respect to F₂ is first

So overall order = 1 + 1 = 2

So rate law, Rate = k [NO₂] [F₂]

b) Now we need to calculate the rate constant

We know rate law

Rate = k [NO₂] [F₂]

0.026 M/s = k *(0.100 M) (0.100 M)

k = 0.026 M.s^-1/ 0.100 M *0.100 M

= 2.6 M^-1 s^-1