Determine Rate Law: Initial Rates

Question

The following initial rate data are for the oxidation of nitrogen monoxide by oxygen at 25 ^oC:

2 NO + O₂2 NO₂

Experiment	[NO]o, M	[O2]o, M	Initial Rate, M s-1
1	2.02×10-3	1.67×10-3	5.34×10-5
2	4.04×10-3	1.67×10-3	2.14×10-4
3	2.02×10-3	3.34×10-3	1.07×10-4
4	4.04×10-3	3.34×10-3	4.27×10-4

Complete the rate law for this reaction in the box below.

Use the form k[A]^m[B]ⁿ , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n.

Rate =

From these data, the rate constant is ______M^-2s^-1.

Answer 1

Comparing Exp 1 and 2 we see that [NO] doubles and [O₂] remains constant. The rate increases by 4 times (2.14x10^-4 / 5/34x10^-5). This tells us the reaction is SECOND order in [NO]

Comparing Exp 1 and 3 we see that [O₂] doubles and [NO] remains constant. The rate increases by 2 times (1.07x10^-4 / 5.34x10^-5). This tells us the reaction is FIRST order in [O₂]

Rate law:

Rate = k [NO]²[O₂]

Calculation of rate constant (k); Choose any experiment and use the rate and concentrations to find k...

From Exp. 1: 5.34x10^-5 Ms^-1 = k [2.02x10^-3 M]²[1.67x10^-3 M]

k = 5.34x10^-5 Ms^-1 / [2.02x10^-3 M]²[1.67x10^-3 M] = 5.34x10^-5 Ms^-1 / 6.8x10-9 M³

k = 7.85x10³ M^-2s^-1

(as always, please check the math)