30% of 1st order reaction completes inc200/ days . how much time will require for 80% completion , log5=0.699 log7=0.84 log 10=1
Answers
Answer:
For the first-order reaction, the relationship between the half-life period and the rate constant is
k=
t
1/2
0.693
=
30
0.693
=0.0231min
−1
.
The rate law expression is k=
t
2.303
log
(a−x)
a
.
Substitute values in the above expression.
0.0231=
70
2.303
log
(a−x)
a
log
(a−x)
a
=0.7021
(a−x)
a
=5.036
a
a−x
=
5.036
1
=0.2
Thus, the percentage of the reactant remaining after 70 min will be 20%.
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Answer:
For a first order reaction,
t= 2.3/k log a/a-x
[ where t= time taken, k= rate constant, a= initial reactant concentration, a-x= reactant remaining after the given time ]
Now in the first step they have mentioned 30% completion in 200 days.
We take a= 100, a-x= 100-30= 70 ( remaining amount )
Therefore, 200= 2.3/k log 100/70
Or 200= 2.3/ k (log 100- log 70)
Or 200= 2.3/k [log 100- (log 7 + log 10)] [ Please note that log(ab)= log a + log b ]
Based on the given data, 200= 2.3/k [2- 0.84- 1]
Or 200= 2.3/k [0.16]
Or 200/ 0.16= 2.3/k
Now we must find out time t' required for 80 % completion. ( a= 100, a-x= 20)
t'= 2.3/k log 100/20
t'= 200/0.16 [log 100- log 20]
t'= 200/0.16 [2- 1- 0.3] [ using log (ab)]
t'= 200 [0.7]/ 0.16
t'= 875 days