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First-order reactions

We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:

ln [ A ] 0 [ A ] = k t t = ln [ A ] 0 [ A ] × 1 k

If we set the time t equal to the half-life, t 1 / 2 , the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when t = t 1 / 2 , [ A ] = 1 2 [ A ] 0 .

Therefore:

t 1 / 2 = ln [ A ] 0 1 2 [ A ] 0 × 1 k = ln 2 × 1 k = 0.693 × 1 k

Thus:

t 1 / 2 = 0.693 k

We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k . A fast reaction (shorter half-life) will have a larger k ; a slow reaction (longer half-life) will have a smaller k .

Calculation of a first-order rate constant using half-life

Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in [link] .

A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, “1.000 M, 0 s, and ( 0 h ).” The second beaker contains a slightly lighter green substance and is labeled below as, “0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).” The third beaker contains an even lighter green substance and is labeled below as, “0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).” The fourth beaker contains a green tinted substance and is labeled below as, “0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).” The fifth beaker contains a colorless substance and is labeled below as, “0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).”
The decomposition of H 2 O 2 ( 2H 2 O 2 2H 2 O + O 2 ) at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H 2 O 2 at the indicated times; H 2 O 2 is actually colorless.

Solution

The half-life for the decomposition of H 2 O 2 is 2.16 × 10 4 s:

t 1 / 2 = 0.693 k k = 0.693 t 1 / 2 = 0.693 2.16 × 10 4 s = 3.21 × 10 −5 s −1

Check your learning

The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d −1 . What is the half-life for this decay?

Answer:

5.02 d.

Got questions? Get instant answers now!

Second-order reactions

We can derive the equation for calculating the half-life of a second order as follows:

1 [ A ] = k t + 1 [ A ] 0

or

1 [ A ] 1 [ A ] 0 = k t

If

t = t 1 / 2

then

[ A ] = 1 2 [ A ] 0

and we can write:

1 1 2 [ A ] 0 1 [ A ] 0 = k t 1 / 2 2 [ A ] 0 1 [ A ] 0 = k t 1 / 2 1 [ A ] 0 = k t 1 / 2

Thus:

t 1 / 2 = 1 k [ A ] 0

For a second-order reaction, t 1 / 2 is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.

Zero-order reactions

We can derive an equation for calculating the half-life of a zero order reaction as follows:

[ A ] = k t + [ A ] 0

When half of the initial amount of reactant has been consumed t = t 1 / 2 and [ A ] = [ A ] 0 2 . Thus:

[ A ] 0 2 = k t 1 / 2 + [ A ] 0 k t 1 / 2 = [ A ] 0 2

and

t 1 / 2 = [ A ] 0 2 k

The half-life of a zero-order reaction increases as the initial concentration increases.

Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in [link] .

Summary of Rate Laws for Zero-, First-, and Second-Order Reactions
Zero-Order First-Order Second-Order
rate law rate = k rate = k [ A ] rate = k [ A ] 2
units of rate constant M s −1 s −1 M −1 s −1
integrated rate law [ A ] = − kt + [ A ] 0 ln[ A ] = − kt + ln[ A ] 0 1 [ A ] = k t + ( 1 [ A ] 0 )
plot needed for linear fit of rate data [ A ] vs. t ln[ A ] vs. t 1 [ A ] vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = +slope
half-life t 1 / 2 = [ A ] 0 2 k t 1 / 2 = 0.693 k t 1 / 2 = 1 [ A ] 0 k

Key concepts and summary

Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.

Practice Key Terms 2

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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