What is the linear form of the first-order kinetics equation?

What is the linear form of the first-order kinetics equation?

• A. t = (1/K) ln(C0/C)
• B. dC/dt = -K[C]
• C. C = C0 e^{-Kt}
• D. log C = log C0 - Kt/2.303

Answer: (D)

Linearizing first-order kinetics turns the exponential decay into a straight line by taking the logarithm of the concentration. Starting from the rate law dC/dt = -k C, separating variables gives dC/C = -k dt. Integrating leads to ln C = -k t + ln C0. If you use common logarithms, log C = log C0 - (k/2.303) t. This means plotting log C against time yields a straight line with slope -k/2.303 and intercept log C0. The option that expresses this straight-line relationship using common logs is log C = log C0 - K t/2.303, which matches the linear form. The other forms correspond to the original differential equation (dC/dt = -k C) or the exponential decay solution (C = C0 e^{-k t}), or a rearranged form for t rather than a direct log vs. time plot, and thus do not represent the linear form used for plotting concentration on a log scale.

Linearizing first-order kinetics turns the exponential decay into a straight line by taking the logarithm of the concentration. Starting from the rate law dC/dt = -k C, separating variables gives dC/C = -k dt. Integrating leads to ln C = -k t + ln C0. If you use common logarithms, log C = log C0 - (k/2.303) t. This means plotting log C against time yields a straight line with slope -k/2.303 and intercept log C0. The option that expresses this straight-line relationship using common logs is log C = log C0 - K t/2.303, which matches the linear form. The other forms correspond to the original differential equation (dC/dt = -k C) or the exponential decay solution (C = C0 e^{-k t}), or a rearranged form for t rather than a direct log vs. time plot, and thus do not represent the linear form used for plotting concentration on a log scale.