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BITSAT 2014 Solved Question Paper
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© examsnet.com
Question : 63
Total: 150
Two substances R and S decompose in solution independently, both following first order kinetics. The rate constant of R is twice that of S. In an experiment, the solution initially contained 0.5 millimoles of R and 0.25 of S. The molarities of R and S will be equal just at the end of time equal to
twice the half life of R
twice the half life of S
the half life of S
the half life of R
Validate
Solution:
Substance R
2k rate constant
t
1
∕
2
Half life period
Substance S
k rate constant
2
t
1
2
Half life period
T = n ×
t
1
∕
2
where n = number of half life period
Amount of R left =
0.5
(
2
)
T
∕
t
1
∕
2
;
Amount of S left =
0.25
(
2
)
T
∕
2
t
1
∕
2
Equating both
0.5
0.25
=
(
2
)
T
∕
t
1
∕
2
(
2
)
T
∕
2
t
1
∕
2
or 2 =
(
2
)
T
∕
t
1
∕
2
∴ T =
2
t
1
∕
2
.
2
t
1
∕
2
s half life of S and twice the half -life of R
© examsnet.com
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