A first order reaction takes place in 40 min for 30% completion.calculate the rate constant
Answers
Answered by
5
Answer:
Explanation:
given the time is (t) = 40min
the first order reaction rate constant is -
k = 2.303/t ㏒ [R°]/[R]
=2.303/40 ㏒ 100/100-30 { [R°]= 100 and [R] = 100-30}
= 0.0576 ㏒ 100/70
= 0.0576 ㏒10/7
= 0.0576 ( ㏒ 10 - log7)
= 0.0576 ( 1 - 0.845)
=0.0576 × 0.155
= 0.00892 min^-1
Answered by
0
Answer:
77.7 min
Explanation:
Let a M be the initial concentration. After 40 minutes, the concentration is a−
100
30a
=0.70a.
k=
t
2.303
log
[A]
[A]
o
k=
40
2.303
log
0.70a
a
k=8.92×10
−3
/min
The half life period, t
1/2
=
k
0.693
=
8.92×10
−3
/min
0.693
=77.7 min.
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