Question

1. Dinitrogen pentioxide decomposes by the following reaction: 2N2O5 (g) --> 4NO2(g) + O2 (g)

time (s)   [N2O5]
4.00E-04   3.00E-03
1.60E-03   2.00E-03
2.20E-03   1.50E-03
2.80E-03   1.00E-03
3.09E-03   7.60E-04

Make three graphs, one for zero order, one for first order and one for second order kinetics. Print them out with a label showing which is zero, which first, which second. The equations for the straight line and the correlation coefficient should be shown on each graph.

What order is the reaction with respect to [N2O5]?

What is the rate constant k?

Answer 1

As per the data given

S.No	Time (s)	[A]	1/[A]	ln[A]
1.	0.0004	0.003	333.33	-5.8091
2	0.0016	0.002	500	-6.215
3	0.0022	0.0015	666.67	-6.5023
4	0.0028	0.001	1000	-6.9078
5	0.00309	0.00076	1315.79	-7.1822

In the graph of zero order, R2 = 1

rate constant, k = -slope

                      = -(-0.832)

                    k = 0.832 mol/lit/sec

from the graph for I order, R2 = 0.96

Rate constant = -slope

                     = -(-498.7)

                  k = 498.7 sec-1

from the graph for I order, R2 = 0.855

Rate constant = slope

                k = 34133 mol-1sec-1

The order of decomposition of N2O5 is first order. i.e., the rate of decomposition reaction is doubled, when the concentration N2O5 is doubled.