A zero-order reaction has rate = k. If [A] doubles, how does the rate change?

A zero-order reaction has rate = k. If [A] doubles, how does the rate change?

• A. Rate doubles
• B. Rate halves
• C. Rate does not change
• D. Rate quadruples

Answer: (C)

In a zero-order reaction, the rate is independent of the reactant concentration and is given by rate = k. Because the rate does not depend on [A], doubling the amount of A doesn’t speed things up or slow them down—the rate stays the same at k. This happens when the process is controlled by something else, like a limited number of active sites on a surface or enzyme saturation, so adding more reactant doesn’t change how fast the reaction proceeds. If the reaction were first-order, doubling [A] would double the rate; if it were second-order, doubling [A] would quadruple the rate.

In a zero-order reaction, the rate is independent of the reactant concentration and is given by rate = k. Because the rate does not depend on [A], doubling the amount of A doesn’t speed things up or slow them down—the rate stays the same at k. This happens when the process is controlled by something else, like a limited number of active sites on a surface or enzyme saturation, so adding more reactant doesn’t change how fast the reaction proceeds. If the reaction were first-order, doubling [A] would double the rate; if it were second-order, doubling [A] would quadruple the rate.