Integrated rate law

Question

The following initial rate data are for the oxidation of arsenate ion by cerium(IV) ion in aqueous solution:

AsO₃^3- + 2 Ce⁴⁺ + H₂OAsO₄^3- + 2 Ce³⁺ + 2 H⁺

Experiment	[AsO33-]o, M	[Ce4+]o, M	Initial Rate, M s-1
1	1.72×10-2	0.184	2.89×10-4
2	3.44×10-2	0.184	5.79×10-4
3	1.72×10-2	0.367	1.15×10-3
4	3.44×10-2	0.367	2.30×10-3

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: AsO₃^3- doubles while Ce⁴⁺ remains constant and the rate doubles. This tells us the reaction is FIRST ORDER in AsO₃^3-.

Compare experiments 1 and 3: AsO₃^3- remains constant while Ce⁴⁺ doubles and the rate increases by 4 times. This tells us that the reaction is SECOND ORER in Ce⁴⁺.

Thus...

Rate = k[AsO₃^3-] [Ce⁴⁺]²

To find the rate constant (k), just use the rate equation above and use any rate from any experiment. I'll choose experiment 1...

2.89x10^-4 Ms^-1 = k [1.72x10^-2 M] [0.184 M]²

k = 2.89x10^-4 Ms^-1 / 5.82x10^-4 M³

k = 0.497 M^-2s^-1