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	8.12×10-3	4.10×10-3	2.27×10-3
2	1.62×10-2	4.10×10-3	9.04×10-3
3	8.12×10-3	8.20×10-3	4.54×10-3
4	1.62×10-2	8.20×10-3	1.81×10-2

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

Compare experiment 1 and 2. The [NO] doubles and the [O₂] remains constant. The rate increases by 4 times. This tells us the reaction is 2nd order in NO, because the rate increases as the square of the concentration (rate = [NO]²).

Next, compare experiments 1 and 3. The [O₂] doubles and the [NO] remains constant. The rate increases by 2 times. This tell us the reaction is 1st order in [O₂] because the rate increases directly with the concentration (rate = [O₂]).

Overall rate equation:

Rate = k[NO]²[O₂]

To find the rate constant (k), just pick any experiment (I choose experiment #1) and plug in the values for concentration and rate, and solve for k.

Experiment 1:

Rate = k[NO]²[O₂]

2.27x10^-3 M/s = k(8.12x10^-3 M)2(4.10xx10^-3 M)

k = 8.40x10³ M^-2s^-1