Question

A reaction is second order with respect to a reaction. How is the rate of reaction affected if the concentration of the reactant is (a) doubled, (b) reduced to $\dfrac{1}{2}$?

Hint: The rate law may be a mathematical relationship obtained by comparing reaction rates with reactant concentrations. The reaction order is the sum of the concentration term exponents during a rate law equation.

Reaction rate refers to the rate or speed at which the reactant forms the product in the chemical reaction. It provides some insight into the timeline to complete the reaction.
For example, wood combustion has a high reaction rate since the process is fast and rusting of iron has a low reaction rate as the process is slow.
Second order reactions can be defined as chemical reactions where the sum of the exponents in the chemical reaction’s corresponding rate law is equal to two.
The rate of such a reaction can be written either as $R=k{{[A]}^{2}}$ or $R=k[A][B]$
Let the concentration of the reactant be $[A]$= a.
Rate of second order reaction is, $R=k{{[A]}^{2}}$=$k{{a}^{2}}$ .
(a) if the concentration of the reactant is doubled, i.e. $[A]$ = 2a, then the rate of the reaction would be ${R}'=k{{[2a]}^{2}}$
=$4k{{a}^{2}}$
= 4R
Therefore, the rate of the reaction would be increased by 4 times.

(b) if the concentration of the reactant is reduced to half, i.e.$[A]$ = $\dfrac{1}{2}$a, then the rate of the reaction would be ${R}''=k{{[\dfrac{1}{2a}]}^{2}}$
=$\dfrac{1}{4}k{{a}^{2}}$
= $\dfrac{1}{4}$ R.
Therefore, the rate of the reaction would be reduced to $\dfrac{1}{4}$th .

Note: Always remember, the chemical reaction rate is directly proportional to the concentration of the reactants.
This implies that the chemical reaction rate increases with increase in concentration and decreases with the decrease in the concentration of reactants.
Rate of reaction depends mainly on the nature of reactants, temperature, pressure, concentration, etc.