For a chemical reaction at 290K rate constant is 2x10-4 sec. What will be rate constant at 300K. If
Activation Energy of Reaction is zero
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Answer:
Given that at T
1
=300K, K
1
=1.6×10
−3
s
−1
and at T
2
=310K, K
2
=3.2×10
−3
s
−1
.
ln(
K
1
K
2
)=
R
E
a
[
T
1
1
−
T
2
1
]
Substituting the values in the above expression, we get
ln[
1.6
3.2
]=
2
E
a
[
300
1
−
310
1
]
E
a
=
10
2 (ln 2) 300×310
=12892.5 cal =12.892 Kcal.
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