The initial concentration of ethyl acetate is 0.8 M. During the acid catalysed hydrolysis the concentration of ester after 30 minutes is 0.05 M. Calculate the pseudo rate constant
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Answer:
(i) The average rate of reaction between the time interval 30 to 60 is
60−30
0.31−0.17
=4.67×10
−3
M/s.
(ii) k=
t
2.303
log
[Ester]
[Ester]
0
When t = 30 s
k=
30
2.303
×log(
0.31
0.55
)=1.91×10
−2
/s
When t = 60 s
k=
60
2.303
×log(
0.17
0.55
)=1.96×10
−2
/s
When t = 90 s
k=
90
2.303
×log(
0.085
0.55
)=2.07×10
−2
/s
Average rate constant k=
3
k1+k2+k3
=1.98×10
−2
/s
Explanation:
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