1. The rate of a first order reaction is 0.04 mol/l/sec
at 10 minutes and
0.03 mol L-1s-1 at 20 minutes after initiation. Find the half life of the reaction ?
Answers
Answered by
0
Explanation:
The rate law expression for the first order reaction is =K×[A].
Substitute values in the above expression.
0.04=K[A]
10
......(1)
0.03=K[A]
20
......(2)
Divide equation (1) by equation (2).
[A]
20
[A]
10
=
0.03
0.04
=
3
4
At 10 min, the expression for the rate constant will be t=
K
2.303
log
[A]
20
[A]
10
.
10=
K
2.303
log
3
4
.
Hence, the rate constant K=
10
2.303
log
3
4
=0.0288min
−1
The half life period is t
1/2
=
K
0.693
=
0.0288
0.693
=24.06min
−1
Similar questions