Question

A first-order reaction: A $\to$ Products and a second order reaction: 2R $\to$ Products both have a half time of 20 min when they are carried out taking 4 mol $L^{-1}$ of their respective reactants. The number of a mole per litre of A and R remaining unreached after 60 min from the start of the reaction, respectively, will be:

• 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 1

Answer: (C)

0.5 and 1 M

In the case of first order reaction $t_{1/2}$ will remain constant independent of initial concentration. So,
$4mol{L}^{-1}\stackrel{20min}{\to }2mol{L}^{-1}\stackrel{20min}{\to }1mol{L}^{-1}\stackrel{20min}{\to }0.5mol{L}^{-1}.$
That is, after 60 min 0.5 mol $L^{-1}$ of A will be left unreached.
In the case of second order reaction, $t_{1/2}$ in 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,
$4mol{L}^{-1}\stackrel{20min}{\to }2mol{L}^{-1}\stackrel{40min}{\to }2mol{L}^{-1}.$
That is, after 60 min, the concentration of R remaining unreached will be 1 mol $L^{-1}$.