D15.6 Half-Life of a Reaction
The half-life (t½) 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}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-0722a564dbb9e58143ca1894127e79b1_l3.png "Rendered by QuickLaTeX.com")
When t = t½:
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}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-0032df4d1467aaa83e91cc3383d31b5f_l3.png "Rendered by QuickLaTeX.com")
The half-life of a first-order reaction is inversely proportional to the rate constant k: a larger k (a faster reaction, all else equal) has a shorter half-life; a smaller k (a slower reaction, all else equal) 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](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-e380acfc268dbc86e4af6e30fc484d79_l3.png "Rendered by QuickLaTeX.com")
When t = t½, [A]t½ = 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}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-3f82860e0d0489ada41de4db1571af2b_l3.png "Rendered by QuickLaTeX.com")
For a second-order reaction, t½ is inversely proportional to the rate constant and the concentration of the reactant. Therefore, t½ is not constant throughout the reaction. As the reaction proceeds, each successive half-life increases 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}]](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-7766fed1b40da8f8fd939f562f7c8e5a_l3.png "Rendered by QuickLaTeX.com")
When t = t½, [A]t½ = 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}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-a935d076ebd7bac0e7ab77b46488f23d_l3.png "Rendered by QuickLaTeX.com")
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 t½ decreases as the reaction progresses and the reactant concentration decreases.
Activity: Half-life and Order from Concentration-Time Data
Some dyes can be decomposed by sunlight. The figure below shows the decomposition of an aqueous solution of a green dye in sunlight at 40 °C. Without making a plot/graph of the data, determine the order of the reaction and then calculate its rate constant.
Write in your notebook, then left-click here for an explanation.
The data show that it takes 6 hours for the 1.000 M solution to decrease to 0.500 M. It takes another 6 hours for the concentration to reach 0.250 M, and each successive halving also takes 6 hours. Therefore the half-life for the dye decomposition reaction is 6 hours or 2.16 × 104 s. Because the half-life is independent of dye concentration, the reaction must be first-order. (Half-lives for other reaction orders vary with concentration.)
To find the rate constant at the given conditions:
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