If the concentration of a reactant in a first-order reaction is halved, what happens to the reaction rate?

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Multiple Choice

If the concentration of a reactant in a first-order reaction is halved, what happens to the reaction rate?

Explanation:
In a first-order reaction, the rate depends linearly on the concentration: rate = k[A]. If the concentration is halved, the rate becomes k([A]/2) = (1/2)k[A], so the rate drops to half of its original value. That’s why halving the reactant concentration causes the rate to halve. The other possibilities—doubling, staying the same, or quadrupling—would require different dependencies on concentration, not a simple first-order relationship.

In a first-order reaction, the rate depends linearly on the concentration: rate = k[A]. If the concentration is halved, the rate becomes k([A]/2) = (1/2)k[A], so the rate drops to half of its original value. That’s why halving the reactant concentration causes the rate to halve. The other possibilities—doubling, staying the same, or quadrupling—would require different dependencies on concentration, not a simple first-order relationship.

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