In the linear form of first-order kinetics, what is the slope of log C versus time?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

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

In the linear form of first-order kinetics, what is the slope of log C versus time?

Explanation:
In first-order kinetics, concentration decays exponentially: C = C0 e^(−k t). Taking natural logs gives ln C = ln C0 − k t, so plotting ln C versus time yields a straight line with slope −k. If you plot common logarithm (base 10) of C instead, use log10 C = (ln C)/ln 10 = log10 C0 − (k/ln 10) t. Since ln 10 ≈ 2.303, the slope becomes −k/2.303. Therefore, the slope of log C versus time is −K/2.303.

In first-order kinetics, concentration decays exponentially: C = C0 e^(−k t). Taking natural logs gives ln C = ln C0 − k t, so plotting ln C versus time yields a straight line with slope −k.

If you plot common logarithm (base 10) of C instead, use log10 C = (ln C)/ln 10 = log10 C0 − (k/ln 10) t. Since ln 10 ≈ 2.303, the slope becomes −k/2.303.

Therefore, the slope of log C versus time is −K/2.303.

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy