Which statement correctly describes the difference between instantaneous rate and average rate for a first-order reaction and how to determine from a rate law?

• A. Instantaneous rate is the slope of [A] versus time at a given moment; average rate is Δ[A]/Δt; for a first-order reaction, rate = k[A], and ln[A] versus t is linear.
• B. Instantaneous rate equals Δ[A]/Δt; average rate equals -d[A]/dt.
• C. For first order, rate = k[A]^2.
• D. Instantaneous rate is the same as the initial rate.

Answer: (C)

For a first-order reaction, the rate at any moment is proportional to the amount of A still present. That means the instantaneous rate of disappearance of A is given by -d[A]/dt = k[A], so the rate at that moment equals k times the current concentration. As [A] decreases over time, the instantaneous rate also decreases.

The average rate over a time interval is the overall change in concentration divided by the time interval: -Δ[A]/Δt (taking the rate of disappearance as positive). This is not fixed; it depends on how much [A] changes during that interval, unlike the instantaneous rate which uses the concentration at a specific moment.

A hallmark of first-order kinetics is the integrated form: ln[A] = -kt + ln[A]0. This means a plot of ln[A] versus time is a straight line with slope -k, confirming the proportionality to [A] and the first-order nature.

The statement claiming the rate for a first-order reaction is proportional to the square of the concentration is not correct, as that would describe a second-order dependence.