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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 15
Total: 39
The experimental data for the decomposition of N2O5
‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌[2N2O5→4NO2+O2]
in gas phase at 318K are given below :
 t∕s  0  400  800  1200  1600  2000  2400  2800  3200
 102×[ N2O5 ]∕mol L−1  1.63  1.36  1.14  0.93  0.78  0.64  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 [N2O5] 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).
Solution:  
 t∕s  [ N2O5 ]×102∕ molL−1  log‌ [ N2O5 ]
 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   0.35  –2.46
Plot of [N2O5] vs time
(ii) Initial conc. ofN2O5=1.63×10–2M
Half of this concentration =0.815×10–2M
Time corresponding to this concentration = 1440 s. Hence, t1⁄2=1440s
(iii) Plot of log [N2O5] vs time
(iv) As plot of log [N2O5] vs time is a straight line. Hence it is a reaction offirst order, i.e. rate law is:
Rate =k [ N2O5 ]
(v) For first order reaction,
ogR=−
k
2.303
t+log‌R0
Therefore slope of the graph drawn between log R and t will be−
k
2.303
.
∴‌ Slope of the line =−
k
2.303
=
y2−y1
t2−t1
or, Slope =
−2.46−(−1.79)
3200−0
=−
0.67
3200
or,
k
2.303
=
0.67
3200
or, k=
0.67
3200
×2.303=4.82×10−4 s−1
(vi) t1∕2=
0.693
k
=
0.693
4.82×10−4
=1438s
The value of t1∕2 calculated from the value of k is very close to that obtainedfrom graph.
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