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

Options

A. 34 min
B. 45 min
C. 24 min
D. 60 min

Answer

C

Explanation

For a first order reaction , rate is directly proportional to the concentration of a single reactant reised to the first power:
Rate `porp` concentration
Suppose `C_(10)` and `C_(20)` are the concentration of the reactants at the end of `10` min and `20` min respectively. Thus
Rate at `10` min, `(Rate)_(10)propC_(10)`
Rate at `20` min, `(Rate)_(20propC_(20)`
or `((Rate)_(10))/((Rate)_(20))=(C_(10))/(C_(20))`
`(0.04molL^(-1)s^(-1))/(0.03molL^(-1)s^(-1))=(C_(10))/(C_(20))`
Thus, `(C_(10))/(C_(20))=(4)/(3)`
When initial concentration of the reactant is not given, we can write the integrated rate law as
`k=(2.303)/(t_(2)-t_(1))log([R]_(1))/([R]_(2))`
In the present situation
`k=(2.303)/(t_(20)-t_(10))log(C_(10))/(C_(20))`
`=(2.303)/((20-10)min)log(4)/(3)`
`=(2.303)/(10min)(log4-log3)`
`=(2.303)/(10mion)(2xxx0.3010-0.4770)`
`=(2.303)/(10min)(0.125)`
`=2.879xx10^(-2)min^(-1)`
Half life of a first order reaction is given as
`t_(1//2)=(0.693)/(k)=(0.693)/(2.879xx10^(-2)min^(-1))`
`=24.07min`