Which equation best represents the relationship between ke and t1/2 in a first-order elimination process?

Which equation best represents the relationship between ke and t1/2 in a first-order elimination process?

• A. Ke = slope of ln(concentration) vs time; t1/2 = 0.693/ke.
• B. Ke = slope of concentration vs time; t1/2 = ke/0.693.
• C. Ke = intercept; t1/2 = 1/ke.
• D. Ke = slope of ln(concentration) vs time; t1/2 = 2.303/ke.

Answer: (A)

In a first-order elimination, concentration falls exponentially: C = C0 e^{-k t}. Taking natural logs gives ln C = ln C0 - k t, so the plot of ln(concentration) versus time is a straight line with slope -k. The rate constant ke is the magnitude of that slope, and the half-life follows from ln 2 = k t1/2, giving t1/2 = 0.693 / ke. So the best relationship is ke equals the slope of ln(concentration) versus time, and t1/2 equals 0.693 divided by ke. The other options misstate which plot yields ke, or mix up the constants (using 2.303 instead of 0.693 corresponds to a base-10 log, not natural log).

In a first-order elimination, concentration falls exponentially: C = C0 e^{-k t}. Taking natural logs gives ln C = ln C0 - k t, so the plot of ln(concentration) versus time is a straight line with slope -k. The rate constant ke is the magnitude of that slope, and the half-life follows from ln 2 = k t1/2, giving t1/2 = 0.693 / ke. So the best relationship is ke equals the slope of ln(concentration) versus time, and t1/2 equals 0.693 divided by ke. The other options misstate which plot yields ke, or mix up the constants (using 2.303 instead of 0.693 corresponds to a base-10 log, not natural log).