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 from 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

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 left 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 reactat will go on getting half, i.e., `t_(1//2)` a will be constant. So, `4 mol L^(-1) overset("20 min")rarr 2 mol 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.