Question

The half-life period of a second order reaction is:
(A) Proportional to the initial concentration of reactants
(B) Independent of the initial concentration of reactants
(C) Inversely proportional to the initial concentration of reactants
(D) Inversely proportional to the square of initial concentration of the reactants

Answer 1

Hint: The half life period of any reaction can be understood as the time that is needed for the concentration of the reactants to be reduced to exactly half of the original value of their concentrations.

Complete Step-by-Step answer:
The integrated form of the second order rate law can be represented as follows:
$${\displaystyle \frac{d[A]}{dt}}=-k[A{]}^2$$
This equation can be simplified further and be represented as follows:
$${\displaystyle \frac{1}{[A]}}={\displaystyle \frac{1}{{[A]}_0}}+kt$$
$${\displaystyle \frac{1}{[A]}}-{\displaystyle \frac{1}{{[A]}_0}}=kt$$
Now, since $$[A{]}_{t_{1/2}}={\displaystyle \frac{1}{2}}[A{]}_0$$;
When $$t=t_{1/2}$$, the given simplified rate equation transforms to:
\$${\displaystyle \frac{1}{{\displaystyle \frac{1}{2}}{[A]}_0}}-{\displaystyle \frac{1}{{[A]}_0}}=k{t}_{1/2}$$
Hence,
$${\displaystyle \frac{2-1}{{[A]}_0}}=k{t}_{1/2}$$
$${\displaystyle \frac{1}{{[A]}_0}}=k{t}_{1/2}$$
From this equation above, we have established a relation between the concentration of the reactant and the half-life period, for a second order reaction.
Upon observation, we can deduce that in second order reactions, the half-life period is inversely proportional to the initial concentration of the reactant.

Hence, Option C is the correct option.

Note: This inverse relationship indirectly states that when the initial concentration of the reactant is increased, there is a higher probability of the two reactant molecules interacting to form a product. This equation also implies that since the half-life is longer when the concentrations are low, species decaying according to second-order kinetics may exist for a longer amount of time if their initial concentrations are small.