If temperature of a reaction increases from 27°c to 77°c, then rate of reaction increases by a factor of (consider temperature coefficient as 2.5)
And tha ans is (2.5)^5.
can any one explain how the ans comes pls.....
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2
Answer:
We are given that:
When T
1
=27+273=300K
Let k
1
=k
When T
2
=37+273=310K
k
2
=2k
Substituting these values the equation:
log(
k
1
k
2
)=
2.303
E
a
×(
T
1
T
2
T
2
–T
1
)
We will get:
log(
k
2k
)=
2.303×8.314
E
a
(
300×310
310−300
)
log(2)=
2.303×8.314
E
a
(
300×310
10
)
E
a
=53598.6 Jmol
−1
E
a
=53.6 kJmol
−1
Hence, the energy of activation of the reaction is 53.6 kJmol
−1
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