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D32.6 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}

When t = t½:

[A]t½ = 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, 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

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}

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}]

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}

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.

Five beakers are shown, each with a green liquid in it. The green color gets lighter from left to right. From left to right the beakers are labeled: beaker 1, 1.000 M, 0 s, (0 h); beaker 2, 0.500 M, \(2.16\times10^4\ \text{s}\), (6 h); beaker 3, 0.250 M, \(4.32\times10^4\ \text{s}\), (12 h); beaker 4, 0.125 M, \(6.48\times10^4\ \text{s}\), (18 h); beaker 5, 0.0625 M, \(8.64\times10^4\ \text{s}\), (24 h).

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:

t_{1/2} = \dfrac{0.693}{k}

k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{2.16 \times 10^4\ \text{s}} = 3.21 \times 10^{-5}\ \text{s}^{-1}

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