question_answer 24)

The experimental data for decomposition of N2O5 [2N2O5  4NO2 + O2]

t/s
0 400 800 1200 1600 2000 2400 2800 3200	1.63 1.36 1.14 0.93 0.78 0.65 0.53 0.43 0.35

  (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log  and t. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

Answer:

The available data is

Time (s)
0	1.63	- 1.79
400	1.36	- 1.87
800	1.14	- 1.94
1200	0.93	- 2.03
1600	0.78	- 2.11
2000	0.64	- 2.19
2400	0.53	-2.28
2800	0.43	- 2.37
3200	0.35x 10-2	- 2.46

(i) The plot of [N2O5] vessus time is as follows:   (ii) Initial conc. of N2O5 = 1.63 x 10?2M. Half of initial conc. = 1/2 x (1.63 x 10?2 M) = 0.815 x 10?2 M Time corresponding to half of initial concentraction (t/2) from the plot (a) = 1400 (s) approximately (iii) The graph of log [N2O5] Vs. time has been plotted. (iv) Since the graph between log [N2O5] and time is a straight line the reaction is of first order. The rate equation: rate (r) = k[N2O5] (v) Slope of the line But slope (vi) Half life period (t1/2) The half life period (t1/2) as calculated from the slope is nearly the same as already predicted.