A first order reaction: `A rarr` Products and a second order reaction: `2R rarr` Products both have half time of `20 min` when they are carried out taking `4 mol L^(-1)` of their respective reactants. The number of mole per litre of `A` and `R` remaining unreacted after `60 min` form the start of the reaction, respectively, will be

Options

A. `1` and `0.5 M`
B. `0.5 M` and negligible
C. `0.5` and `1 M`
D. `1` and `0.25 M`

Answer

C

Explanation

In the case of first order reaction `t_(1//2)` will remain constant independent of initial concentration. So,
`4 mol L^(-1) overset(20 min)rarr 2 mol L^(-1) overset(20 min)rarr 1 mol L^(-1) overset(20 min)rarr 0.5 mol L^(-1)`.
That is, after `60 min 0.5 mol L^(-1)` of `A` will be laft unreacted.
In the case of second order reaction, `t_(1//2)` is inversely proportional to the initial concentration of reactant, i.e., `t_(1//2)` will go on doubling as the concentration of reactant will go on getting half, i.e., `t_(1//2) a` will be constant. So `4mol L^(-1) overset(20 min)rarr 2mol L^(-1) overset(40 min)rarr 1 mol L^(-1)`.
That is, after `60 min`, the concentration of `R` remaining unreacted will be `1 mol L^(-1)`.
Note: `t_(1//2) a = 20 min xx 4 M = 40 min xx 2M = 80 mol L^(-1) min^(-1), a` constant.