What happens to the rate of a reaction when the concentration of a first-order reactant is halved?

In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one reactant. This relationship can be expressed by the rate law equation:

[ \text{Rate} = k[\text{A}] ]

where ( k ) is the rate constant and [A] is the concentration of the reactant.

When the concentration of the first-order reactant is halved, the new concentration can be represented as ( \frac{1}{2}[\text{A}] ). Substituting this new concentration into the rate law, the new rate can be described as:

[ \text{New Rate} = k\left(\frac{1}{2}[\text{A}]\right) = \frac{1}{2}k[\text{A}] ]

This indicates that the new rate of the reaction is indeed half of the original rate. Therefore, halving the concentration results directly in halving the rate of the reaction, confirming that the correct answer is that the rate is halved.

This principle highlights the foundational concept in kinetics where the dependence of the reaction rate on concentration is characterized differently for zero-order and second-order reactions, but for first-order reactions, such linear