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JEE Main Chemistry Class 11 Chemical Kinetics Part 1 Questions
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© examsnet.com
Question : 8
Total: 100
The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation,
10
%
of the virus is inactivated. The rate constant for viral inactivation is __________
×
10
−
3
m
i
n
−
1
. (Nearest integer)
[Use :
ln
10
=
2.303
;
log
10
3
=
0.477
;property of logarithm
:
log
x
y
=
y
log
x
]
[20 Jul 2021 Shift 1]
Your Answer:
Validate
Solution:
As the unit of rate constant is
min
−
1
so it must be a first order reaction
K
×
t
=
2.303
log
A
0
∕
A
t
in 1 min
10
%
is in activated so tabing
A
0
=
100
A
t
=
90
in
1
m
i
n
So
K
×
1
=
2.303
×
log
100
90
=
2.303
×
(
log
10
−
2
log
3
)
=
2.303
×
(
1
−
2
×
0.477
)
=
0.10593
=
105.93
×
10
−
3
≈
106
© examsnet.com
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