Question

Exercise 12.2

The decomposition of N₂O₅ in the gas phase was studied at constant temperature.

                                    2 N₂O₅ (g) à 4 NO₂ (g)   + O₂ (g)

The following results were collected:

                                    [N₂O₅]                                 Time (s)

                                    0.1000                                        0

                                    0.0707                                     50

                                    0.0500                                     100

                                    0.0250                                     200

                                    0.0125                                     300

                                    0.00625                                   400

Determine the rate law and calculate the value of k.

Exercise 12.3

Using the data given in Ex. 12.2 above, calculate [N₂O₅] at 150 s after the start of the reaction.

Calculate the [N₂O₅] at the following times:

200 s

400 s

600 s

1,000 s

Answer 1

12.2) Prepare the following table.

Time (s)	[N2O5] (M)	ln [N2O5]	1/[N2O5] (M-1)
0	0.1000	-2.3026	10.0000
50	0.0707	-2.6493	14.1443
100	0.0500	-2.9957	20.0000
200	0.0250	-3.6889	40.0000
300	0.0125	-4.3820	80.0000
400	0.00625	-5.0752	160.0000

Plot [N₂O₅] vs time, ln [N₂O₅] vs time and 1/[N₂O₅] vs time as below.

Plot of [N₂O₅] vs time

Plot of ln [N₂O₅] vs time

Plot of 1/[N₂O₅] vs time

The plot of ln [N₂O₅] vs time is linear with a slope equal to -0.0069.

The integrated rate law for a first order reaction is of the form

ln [A] = -kt +ln [A]₀

The left side is dimensionless while t (time) has unit s; therefore, k must have unit s^-1. Comparing the regression equation with the integrated rate law, we can conclude that the first order rate constant is k = 0.0069 s^-1 (ans).