The rate of a first order reaction at a constant temperature

The correct answer is Option (2) → Decreases as the reaction proceeds

For a first order reaction the rate is proportional to the concentration of the reactant:

    \( \displaystyle \text{Rate} = k[\mathrm{R}] \)

At constant temperature k does not change. As the reaction proceeds, the reactant is used up and [R] falls, so the rate falls with it. [R] decays exponentially, \( [\mathrm{R}] = [\mathrm{R}]_0 e^{-kt} \), so the rate also decreases steadily with time: after one half-life it is half the initial rate, after two half-lives a quarter.

Why the other options are incorrect

• Option (1) Increases as the reaction proceeds — Incorrect. The rate could rise only if [R] rose or a product speeded up the reaction (autocatalysis); neither happens in a simple first order reaction.

• Option (3) May increase or decrease as the reaction proceeds — Incorrect. With k fixed, the rate depends on [R] alone, which can only fall, so the change has one direction only.

• Option (4) Remains constant as the reaction proceeds — Incorrect. A constant rate is the mark of a zero order reaction (Rate = k), whose rate does not depend on concentration.