Which plot yields a straight line for first-order elimination?

Study for the Pharmaceutics Drug Disposition Test. Prepare with flashcards and multiple choice questions, each answer has hints and explanations. Get set for your exam!

Multiple Choice

Which plot yields a straight line for first-order elimination?

Explanation:
First-order elimination produces exponential decay, meaning the rate of loss is proportional to the current concentration. Mathematically, dC/dt = -kC, and integrating gives ln C = -k t + ln C0. Because of this, plotting the logarithm of concentration against time yields a straight line with slope -k (or -k/2.303 if using base-10 log). This makes the semi-log plot of log C versus time the standard way to visualize first-order elimination and to determine the elimination rate constant from the slope. Plotting concentration versus time gives a curved, exponential decline rather than a straight line. Plotting time versus concentration isn’t linear either, since t = (ln C0 - ln C)/k, which isn’t a straight line when time is graphed against concentration. A double-log plot of C versus time likewise won’t be linear for first-order decay, because the relationship is exponential in time, not a power-law.

First-order elimination produces exponential decay, meaning the rate of loss is proportional to the current concentration. Mathematically, dC/dt = -kC, and integrating gives ln C = -k t + ln C0. Because of this, plotting the logarithm of concentration against time yields a straight line with slope -k (or -k/2.303 if using base-10 log). This makes the semi-log plot of log C versus time the standard way to visualize first-order elimination and to determine the elimination rate constant from the slope.

Plotting concentration versus time gives a curved, exponential decline rather than a straight line. Plotting time versus concentration isn’t linear either, since t = (ln C0 - ln C)/k, which isn’t a straight line when time is graphed against concentration. A double-log plot of C versus time likewise won’t be linear for first-order decay, because the relationship is exponential in time, not a power-law.

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