D34.6 Half-Life of a Reaction

John Moore; Jia Zhou; Etienne Garand

D34.6 Half-Life of a Reaction

The half-life (t_½) of a reaction is the time required for the concentration of a reactant to be reduced to half of its initial value. In each successive half-life, the remaining concentration of the reactant is again halved. The half-life of a reaction can be derived from the integrated rate law. Hence, there is a general equation for half-life for zeroth-order, first-order, and second-order reaction.

First-Order Reaction

The integrated rate law gives:

$\begin{array}{rcl} kt &=& \text{ln}\left(\dfrac{[\text{A}]_0}{[\text{A}]_t}\right) \\[1em] t &=& \text{ln}\left(\dfrac{[\text{A}]_0}{[\text{A}]_t}\right) \times \dfrac{1}{k} \end{array}$

When t = t_½:

[A]_{t_½} = ½[A]₀

Therefore:

$\begin{array}{rcl} t_{_{1/2}} &=& \text{ln}\left(\dfrac{[\text{A}]_0}{\frac{1}{2}[\text{A}]_0}\right) \times \dfrac{1}{k} \\[1.5em] t_{_{1/2}} &=& \text{ln}(2) \times \dfrac{1}{k} \\[1em] t_{_{1/2}} &=& \dfrac{0.693}{k} \end{array}$

The half-life of a first-order reaction is inversely proportional to the rate constant k: a larger k (a faster reaction) has a shorter half-life; a smaller k (a slower reaction) has a longer half-life. Moreover, the half-life is conveniently independent of the concentration of the reactant. Therefore, you do not need to know the initial concentration to calculate the rate constant from the half-life, or vice versa.

Exercise: Half-Life, Rate, and Concentration

Second-Order Reactions

The integrated rate law is:

$\dfrac{1}{[\text{A}]_t} - \dfrac{1}{[\text{A}]_0} = kt$

When t = t_½, [A]_{t_½} = ½[A]₀, therefore:

$\begin{array}{rcl} \dfrac{1}{\frac{1}{2}[\text{A}]_0} - \dfrac{1}{[\text{A}]_0} &=& kt_{_{1/2}} \\[1em] \dfrac{2}{[\text{A}]_0} - \dfrac{1}{[\text{A}]_0} &=& kt_{_{1/2}} \\[1em] \dfrac{1}{[\text{A}]_0} &=& kt_{_{1/2}} \\[1em] t_{_{1/2}} &=& \dfrac{1}{k[\text{A}]_0} \end{array}$

For a second-order reaction, t_½ is inversely proportional to the rate constant and the concentration of the reactant. Therefore, t_½ is not constant throughout the reaction. Each successive half-life increases as the reaction proceeds due to decreasing concentration of reactant. Consequently, unlike the situation with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration relating to that half-life is known.

Zeroth-Order Reactions

For a zeroth-order reaction:

$[\text{A}]_t = -kt + [\text{A}]$

When t = t_½, [A]_{t_½} = ½[A]₀, therefore:

$\begin{array}{rcl} \dfrac{[\text{A}]_0}{2} & = & -kt_{_{1/2}} + [\text{A}]_0 \\[1em] kt_{_{1/2}} &=& \dfrac{[\text{A}]_0}{2} \\[1em] t_{_{1/2}} &=& \dfrac{[\text{A}]_0}{2k} \end{array}$

The half-life of a zeroth-order reaction is inversely proportional to the rate constant and directly proportional to the concentration of the reactant. Therefore, each successive t_½ decreases as the reaction progresses and the reactant concentration decreases.

Activity: Half-life and Order from Concentration-Time Data

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Chemistry 109 Fall 2021 Copyright © by John Moore; Jia Zhou; and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.