Answer:
The Arrhenius equation and temperature variation is given by the expression.
log(
k
′
)=
2.303R
E
a
[
TT
T
−T
]
Also, t
1/2
=
0.693
or t
∝
1
Hence, log(
t
20
40
2.303×8.314J/mol/K
300K×320K
320K−300K
0.3010=
19.147J/mol/K
[0.0002083/K]
=27664J/mol
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Verified answer
Answer:
The Arrhenius equation and temperature variation is given by the expression.
log(
k
k
′
)=
2.303R
E
a
[
TT
′
T
′
−T
]
Also, t
1/2
=
k
0.693
or t
1/2
∝
k
1
Hence, log(
t
1/2
′
t
1/2
)=
2.303R
E
a
[
TT
′
T
′
−T
]
log(
20
40
)=
2.303×8.314J/mol/K
E
a
[
300K×320K
320K−300K
]
0.3010=
19.147J/mol/K
E
a
[0.0002083/K]
E
a
=27664J/mol