56.
The time required for 60% completion
of a first order reaction is 50 min. The
time required for 93.6% completion of
the same reaction will be
(A). 100 min
(B) 83.8 min
(C) 50 min
(D) 150 min
Answers
Answered by
0
Explanation:
k=
t
2.303
log
a−x
a
The reaction is 60% complete in 20 minutes.
k=
20
2.303
log
100−60
100
.....(1)
The reaction is 84% complete
k=
t
2.303
log
100−84
100
.....(2)
But (1) = (2),
2.303
log
100−60
100
=
t
2.303
log
100−84
100
20
1
log
100−60
100
=
t
1
log
100−84
100
t=
log
100−60
100
20
log
100−84
100
t=
0.3979
20
×0.79588
t=40 minutes.
The reaction will take 40 minutes to be 84% complete.
Answered by
1
Answer:
D
Explanation:
For first order reaction, reaction rate (k)
k = 1/t ln(a /a-x)
for 60% completion,
k= 1/50 ln (100/100-60)
k= 1/50 ln (10/4)
for 93.6% completion,
t= 1/k ln (100/100-93.6)
t= 1/k ln(100/6.4)
[putting the value of k]
t=50 {ln(100/6.4) / ln(10/4)}
t=150
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