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

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

• A. -K
• B. -2.303K
• C. -K/2.303
• D. -1/(2.303K)

Answer: (C)

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.