What relationship defines the rate for a second-order reaction when concentration is doubled?

For a second-order reaction, the rate is directly proportional to the square of the concentration of the reactants. This means that if you double the concentration of a reactant, the rate of the reaction will increase by a factor equal to the square of that change in concentration.

To illustrate this, let's say the rate of the reaction can be described by the equation ( \text{Rate} = k[A]^2 ), where ( [A] ) represents the concentration of the reactant and ( k ) is the rate constant. If the concentration ( [A] ) is doubled, the new concentration is ( 2[A] ). Plugging this new concentration into the rate equation gives:

[

\text{New Rate} = k(2[A])^2 = k(4[A]^2) = 4(k[A]^2).

]

This shows that the new rate is four times the original rate, thus confirming that when the concentration is doubled in a second-order reaction, the rate quadruples. This is why the correct answer is that the rate quadruples when concentration is doubled.