Why is the half-life for a first-order reaction independent of the initial concentration?

Why is the half-life for a first-order reaction independent of the initial concentration?

• A. t1/2 = ln 2 / k; it depends only on the rate constant k, not on [A]0.
• B. t1/2 = k / ln 2; depends on initial concentration.
• C. t1/2 = ln 2 / (k [A]0); depends on initial concentration.
• D. t1/2 = (ln 2) / (k^2); independent of concentration.

Answer: (A)

In a first-order reaction, the rate is proportional to the instantaneous concentration, so -d[A]/dt = k[A]. Integrating gives [A] = [A]0 e^{-kt}. The half-life t1/2 is the time when [A] = [A]0/2, which leads to [A]0/2 = [A]0 e^{-kt1/2} => 1/2 = e^{-kt1/2} => t1/2 = ln 2 / k. This shows t1/2 depends only on the rate constant k and not on the initial concentration [A]0. Even if you start with a larger or smaller amount, the time to halve remains the same because the decay rate at any moment scales with the current concentration. The other proposed forms do not fit: they either give wrong units, imply dependence on [A]0, or both.

In a first-order reaction, the rate is proportional to the instantaneous concentration, so -d[A]/dt = k[A]. Integrating gives [A] = [A]0 e^{-kt}. The half-life t1/2 is the time when [A] = [A]0/2, which leads to [A]0/2 = [A]0 e^{-kt1/2} => 1/2 = e^{-kt1/2} => t1/2 = ln 2 / k. This shows t1/2 depends only on the rate constant k and not on the initial concentration [A]0. Even if you start with a larger or smaller amount, the time to halve remains the same because the decay rate at any moment scales with the current concentration. The other proposed forms do not fit: they either give wrong units, imply dependence on [A]0, or both.