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Class 12 NEET Chemistry Chemical Kinetics Questions
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© examsnet.com
Question : 3
Total: 73
Rate constants of a reaction at 500 K and 700 K are
0.04
s
−
1
and
0.14
s
−
1
, respectively; then, activation energy of the reaction is :
(Given:
log
3.5
=
0.5441
,
R
=
8.31
J
K
−
1
mol
−
1
)
[NEET 2024 Re]
182310 J
18500 J
18219 J
18030 J
Validate
Solution:
K
=
Ae
−
E
a
∕
R
T
After taking In both side
ln
K
=
ln
A
−
E
a
R
T
ln
K
1
=
ln
A
−
E
a
R
T
1
at temp.
T
1
.......(i)
In
K
2
=
ln
A
−
E
a
R
T
2
at temp.
T
2
.......(ii)
(ii) - (i)
In
K
2
−
ln
K
1
=
E
a
R
[
1
T
1
−
1
T
2
]
ln
K
2
K
1
=
E
a
R
[
1
500
−
1
700
]
ln
0.14
0.04
=
E
a
R
[
700
−
500
500
×
700
]
ln
14
4
=
E
a
R
[
200
500
×
700
]
log
3.5
=
E
a
2.303
×
R
[
1
250
×
7
]
0.5441
=
E
a
2.303
×
8.31
[
1
250
×
7
]
E
a
=
0.5441
×
8.31
×
250
×
7
×
2.303
=
0.5441
×
83.1
×
25
×
7
×
2.303
=
18222.65
≈
18219
J
© examsnet.com
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