Question
The experimental data for decomposition of N2O5 [2N2Os → 4NO2 + O2] in gas phase at 318 k are given below:
(a) Plot [N2O5] again t.
(b) Find the half life period for the reaction.
(c) Draw a graph between log [N2O5] and t.
(d) What is rate law?
(e) Calculate the rate constant.
(f) Calculate the half life period from K and compare it with (ii).
Solution
(a) Plot of [N2O5] vs. time.
(c)
d) The given reaction is of the first order as the plot log[N2O5] v/s time, is a straight line.
Therefore the rate of reaction is
Rate = k[N2O5]
e) Form the plot, log [N2O5] v/s T, we obtained
slope=
f) Half-life is given by
This value, 1438 second is very closed to the value that was obtained form the graph.
(b) Time taken for the concentration of N2O5 to change from 1.63 x 10–2 mol L–1 to half the value.
t0.5 = 1420 s [from the graph] (c) Plot of log (N2O5) vs. time.
|
t |
log10[N2O5] |
|
0 400 800 1200 1600 2000 2400 2800 3200 3600 |
- 1.7918 - 1.8665 - 1.9431 - 2.0315 - 2.1079 - 2.1938 - 2.2757 - 2.3752 - 2.4559 - 2.5376 |
(c)
d) The given reaction is of the first order as the plot log[N2O5] v/s time, is a straight line.
Therefore the rate of reaction is
Rate = k[N2O5]
e) Form the plot, log [N2O5] v/s T, we obtained
slope=
f) Half-life is given by
This value, 1438 second is very closed to the value that was obtained form the graph.
