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

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Multiple Choice

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

Explanation:
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.

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