A first order reaction takes 40 min for 30% decomposition. Calculate
t1/2.
(Given : log10 2 = 0.3010, log10 3 = 0.4771, log10 5 = 0.6990 , log10 7 = 0.8450)
Answers
Answered by
0
Answer:
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])
−
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.
Answered by
1
Sorry for handwriting (^^)
hope u understand it❤️
Attachments:
Similar questions