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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 8
Total: 39
In a pseudo first order hydrolysis of ester in water, the following results were obtained :
  t / s  0  30  60  90
  [ Ester] /mol L‌–1  0.55  0.31  0.17  0.085
(i) Calculate the average rate of reaction between the time interval 30 to 60seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis ofester.
Solution:  
(i) Average rate during the interval 30-60 sec
Rate =−
C2−C1
t2−t1
=−
0.17−0.31
60−30
=
0.14
30
 mol L−1 s−1
=4.67×10−3 mol L−1 s−1
(ii) Using formula, k=
2.303
t
‌log‌
[R0 ]
[R]
 in which [R0 ]=0.55 M
At t=30sec,[R]=0.31 mol L−1
∴‌k,=
2.303
30
‌log‌ ‌
0.55
0.31
=1.91×10−2 s−1
At t=60 sec,[R]=0.17 molL−1
∴‌k,=
2.303
60
‌log‌
0.55
0.17
=1.96×10−2 s−1
At t=90‌sec,[R]=0.085molL−1
∴‌k,=
2.303
90
‌log‌
0.55
0.085

=2.07×10−2 s−1
Hence, average K=
1.91+1.96+2.07
3
×10−2
=1.98×10−2s−1
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