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

Options

A. 44.1 s
B. 54.1 s
C. 24.1 s
D. 34.1 s

Answer

C

Explanation

For a first order reaction, A `rarr` Products and for concentration of the reactant at two different times,
`k=(2.303)/(t_2-t_1)log([A]_1)/([A]_2)`
`therefore k= (2.303)/(t_2-t_1)log(("rate")_1)/(("rate"_2)` (`since" rate `prop[A])
`k=(2.303)/((20-10))log ((0.04)/(0.03))=0.0287` `sec^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.0287" "sec^(-1))`=24.14 sec