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The rate of first order reaction is 0....

The rate of first order reaction is `0.04"mol.L"^(-1).s^(-1)` at 10 seconds and `0.03"mol.L"^(-1)` at 20 seconds after initiation of the reaction . The half-life period of the reaction is -

A

44.1s

B

54.1s

C

24.1s

D

34.1s

Text Solution

Verified by Experts

The correct Answer is:
C
For first order reaction , `r=k[A]_(0)`
After 10s , `r = k [A]_(1) = 0.04`
After 20s , `r=k[A]_(2)=0.03`
`therefore([A]_(2))/([A]_(1))=(0.03)/(0.04)=0.75 " or, " [A]_(2)=0.75[A]_(1)`
`t_(2)-t_(1)=Deltat=(2.303)/(k)log.([A]_(0))/([A]_(2))-(2.303)/(k)log.([A]_(0))/([A]_(1))`
`=(2.303)/(k)log.([A]_(1))/([A]_(2))`
or, `20-10=(2.303)/(k)log.(1)/(0.75) " or, " k = 0.02877s^(-1)`
`therefore t_(1//2)=(0.693)/(k)=(0.693)/(0.02877)=24.08s.`
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