CBSE Class 12 Chemistry 2017 Delhi Set 1 Solved Paper

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Question : 15
Total: 26
Following data are obtained for the reaction:
N2O52NO2+12 O2
 ts  0  300  600
 [N2O5]molL1  1.6×102  0.8×102  0.4×102

(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log2=0.3010,log4=0.6021)
Solution:  
(a) For first order reaction the integral rate law is:
kt=ln(
a0
a1
)

Given, a0=1.6×102molL1
For t=300 s,at=0.8×102molL1
For t=600 s,at=0.4×102molL1
Using first set of data in the rate law,
k×300=ln
1.6×102
0.8×102

k=0.00231 s1
Using second set of data in the rate law,
k×600=ln
1.6×102
0.4×102

k=0.00231 s1
The value of k is consistent,therefore it follows first order reaction.
(b) The half-life of first order reaction is given by the following equation :
t12=
ln2
k
=2.303×
log2
k

t12=2.303×
log2
0.00231
=300.08 s.
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