Which expression describes the zero-order concentration change over time?

Which expression describes the zero-order concentration change over time?

• A. [A] = [A]0 - kt
• B. [A] = [A]0 e^{-kt}
• C. 1/[A] = kt + 1/[A]0
• D. ln[A] = -kt

Answer: (A)

Zero-order behavior means the reaction consumes A at a constant rate, independent of how much A is present. This leads to the rate equation d[A]/dt = -k. Integrating with the initial condition [A](0) = [A]0 gives [A](t) = [A]0 − kt. So the concentration decreases linearly over time, with slope −k, until the reactant is exhausted. The units of k are concentration per time (for example, M/s), so kt has the same units as [A]. The other forms correspond to different kinetic orders: a first-order process gives exponential decay [A] = [A]0 e^(−kt) (or ln[A] = −kt), while a second-order process gives 1/[A] = kt + 1/[A]0.

Zero-order behavior means the reaction consumes A at a constant rate, independent of how much A is present. This leads to the rate equation d[A]/dt = -k. Integrating with the initial condition A = [A]0 gives A = [A]0 − kt. So the concentration decreases linearly over time, with slope −k, until the reactant is exhausted. The units of k are concentration per time (for example, M/s), so kt has the same units as [A].

The other forms correspond to different kinetic orders: a first-order process gives exponential decay [A] = [A]0 e^(−kt) (or ln[A] = −kt), while a second-order process gives 1/[A] = kt + 1/[A]0.