Chapter 13.5: Half Lives and Radioactive Decay Kinetics
 Page ID
 42112
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Prince George's Community College 

Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry 
Learning Objective
 To know how to use halflives to describe the rates of firstorder reactions.
HalfLives
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to onehalf its initial value. This period of time is called the halflifeThe period of time it takes for the concentration of a reactant to decrease to onehalf its initial value. of the reaction, written as t_{1/2}. Thus the halflife of a reaction is the time required for the reactant concentration to decrease from [A]_{0} to [A]_{0}/2. If two reactions have the same order, the faster reaction will have a shorter halflife, and the slower reaction will have a longer halflife.
The halflife of a firstorder reaction under a given set of reaction conditions is a constant. This is not true for zeroth and secondorder reactions. The halflife of a firstorder reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a firstorder reaction (Equation 13.5.1) to produce the following equation:
\( ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]} = kt \tag{13.5.1} \)
Substituting [A]_{0}/2 for [A] and t_{1/2} for t (to indicate a halflife) into Equation 13.5.1 gives
\( ln \dfrac{\left [ A \right ]_{0}}{\left [ A \right ]_{0}/2} = ln 2=kt_{1/2} \tag{13.5.2} \)
The natural logarithm of 2 (to three decimal places) is 0.693. Substituting this value into the equation, we obtain the expression for the halflife of a firstorder reaction:
\( t_{1/2}=\dfrac{0.693}{k} \tag{13.5.3} \)
Thus, for a firstorder reaction, each successive halflife is the same length of time, as shown in Figure 13.5.1 , and is independent of [A].
Figure 13.5.1 The HalfLife of a FirstOrder Reaction This plot shows the concentration of the reactant in a firstorder reaction as a function of time and identifies a series of halflives, intervals in which the reactant concentration decreases by a factor of 2. In a firstorder reaction, every halflife is the same length of time.
If we know the rate constant for a firstorder reaction, then we can use halflives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of HalfLives  Percentage of Reactant Remaining  

1  \( \dfrac{100 \%}{2}=50\% \)  \( \dfrac{1}{2}\left (100 \% \right )= 50 \% \) 
2  \( \dfrac{50 \%}{2} =25 \%\)  \( \dfrac{1}{2} \left ( \dfrac{1}{2} \right )\left (100 \% \right )= 25 \% \) 
3  \( \dfrac{25 \%}{2}=12.5 \% \)  \( \dfrac{1}{2} \left ( \dfrac{1}{2} \right ) \left ( \dfrac{1}{2} \right ) \left (100 \% \right )= 12.5 \% \) 
n  \( \dfrac{100 \%}{2^{n}} \)  \( \left ( \dfrac{1}{2} \right )^{n} \left (100 \% \right )= \left ( \dfrac{1}{2} \right )^{n} \% \) 
As you can see from this table, the amount of reactant left after n halflives of a firstorder reaction is (1/2)^{n} times the initial concentration.
Note the Pattern
For a firstorder reaction, the concentration of the reactant decreases by a constant with each halflife and is independent of [A].
Example 13.5.1
The anticancer drug cisplatin hydrolyzes in water with a rate constant of 1.5 × 10^{−3} min^{−1} at pH 7.0 and 25°C. Calculate the halflife for the hydrolysis reaction under these conditions. If a freshly prepared solution of cisplatin has a concentration of 0.053 M, what will be the concentration of cisplatin after 5 halflives? after 10 halflives? What is the percent completion of the reaction after 5 halflives? after 10 halflives?
Given: rate constant, initial concentration, and number of halflives
Asked for: halflife, final concentrations, and percent completion
Strategy:
A Use Equation 13.5.3 to calculate the halflife of the reaction.
B Multiply the initial concentration by 1/2 to the power corresponding to the number of halflives to obtain the remaining concentrations after those halflives.
C Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution:
A We can calculate the halflife of the reaction using Equation 13.5.3:
\( t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5 \times 10^{3}\;min^{1}}=4.6 \times 10^{2}\;min \)
Thus it takes almost 8 h for half of the cisplatin to hydrolyze.
B After 5 halflives (about 38 h), the remaining concentration of cisplatin will be as follows:
\( \dfrac{0.053\;M}{2^{5}}=\dfrac{0.053\;M}{32}=0.0017\;M \)
After 10 halflives (77 h), the remaining concentration of cisplatin will be as follows:
\( \dfrac{0.053\;M}{2^{10}}=\dfrac{0.053\;M}{1024}=5.2 \times 10^{5}\;M \)
C The percent completion after 5 halflives will be as follows:
\( percent\;completion=\dfrac{\left (0.053\;M0.0017\;M \right )100}{0.53\;M}=97\% \)
The percent completion after 10 halflives will be as follows:
\( percent\;completion=\dfrac{\left (0.053\;M5.2 \times 10^{5}\;M \right )100}{0.53\;M}=100\% \)
Thus a firstorder chemical reaction is 97% complete after 5 halflives and 100% complete after 10 halflives.
Exercise
In Example 4 you found that ethyl chloride decomposes to ethylene and HCl in a firstorder reaction that has a rate constant of 1.6 × 10^{−6} s^{−1} at 650°C. What is the halflife for the reaction under these conditions? If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 halflives?
Answer: 4.3 × 10^{5} s = 120 h = 5.0 days; 4.8 × 10^{−3} M
Radioactive Decay Rates
As you learned in Chapter 1 radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how halflives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decayThe decrease in the number of a radioisotope’s nuclei per unit time. of the sample, which is also called its activity (A)The decrease in the number of a radioisotope’s nuclei per unit time: A=−dN/dt as the decrease in the number of the radioisotope’s nuclei per unit time:
\( A=\dfrac{d N}{d t} \tag{13.5.4}\)
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
\( A=kN \tag{13.5.5}\)
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s^{−1}, yr^{−1}) and a characteristic value for each radioactive isotope. If we combine Equation 13.5.4 and Equation 13.5.5, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
\( \dfrac{d N}{d t}=kN \tag{13.5.6}\)
Equation 13.5.6 is the same as the equation for the reaction rate of a firstorder reaction (Equation 13.3.2), except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a firstorder process and can be described in terms of either the differential rate law (Equation 13.5.7) or the integrated rate law:
\( N=N_{0}e^{kt} \tag{13.5.7}\)
Because radioactive decay is a firstorder process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the halflife of the isotope. The halflife tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter halflives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer halflives. The halflives of several isotopes are listed in Table 13.5.1, along with some of their applications.
Table 13.5.1 HalfLives and Applications of Some Radioactive Isotopes
Radioactive Isotope  HalfLife  Typical Uses 

hydrogen3 (tritium)  12.32 yr  biochemical tracer 
carbon11  20.33 min  positron emission tomography (biomedical imaging) 
carbon14  5.70 × 10^{3} yr  dating of artifacts 
sodium24  14.951 h  cardiovascular system tracer 
phosphorus32  14.26 days  biochemical tracer 
potassium40  1.248 × 10^{9} yr  dating of rocks 
iron59  44.495 days  red blood cell lifetime tracer 
cobalt60  5.2712 yr  radiation therapy for cancer 
technetium99m*  6.006 h  biomedical imaging 
iodine131  8.0207 days  thyroid studies tracer 
radium226  1.600 × 10^{3} yr  radiation therapy for cancer 
uranium238  4.468 × 10^{9} yr  dating of rocks and Earth’s crust 
americium241  432.2 yr  smoke detectors 
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope. 
Note the Pattern
Radioactive decay is a firstorder process.
Radioisotope Dating Techniques
In our earlier discussion, we used the halflife of a firstorder reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow firstorder kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the halflives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon14 dating. The carbon14 isotope, created continuously in the upper regions of Earth’s atmosphere by nuclear reactions initiated by energetic cosmic rays, reacts with atmospheric oxygen or ozone to form ^{14}CO_{2}. As a result, the CO_{2} that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of ^{14}CO_{2} molecules as well as nonradioactive ^{12}CO_{2} and ^{13}CO_{2}. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon14 nuclei in its tissues decay to nitrogen14 nuclei by a radioactive process known as beta decay, which releases lowenergy electrons (β particles) that can be detected and measured:
\( ^{14}C\rightarrow ^{14}N + \beta ^{} \tag{13.5.8} \)
The halflife for this reaction is 5700 ± 30 yr.
The ^{14}C/^{12}C ratio in living organisms is 1.3 × 10^{−12}, with a decay rate of 15 dpm/g of carbon (Figure 13.5.2). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11. Using this method implicitly assumes that the ^{14}CO_{2}/^{12}CO_{2} ratio in the atmosphere is constant, which is not strictly correct because the intensity of the cosmic ray flux varies somewhat. Moreover the method is not useful for samples that were alive after 1945 because of the relatively large amount of ^{14}C created by nuclear bomb explosions in the atmosphere. Other methods, such as dating by counting treerings and comparing the treering dates to the ^{14}C/^{12}C ratio in a ring, have been used to calibrate the dates obtained by radiocarbon dating. All radiocarbon dates reported are now corrected for minor changes in the ^{14}CO_{2}/^{12}CO_{2} ratio over time.
Figure 13.5.2 Radiocarbon Dating A plot of the specific activity of ^{14}C versus age for a number of archaeological samples shows an inverse linear relationship between ^{14}C content (a log scale) and age (a linear scale).
Example 13.5.2
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the ^{14}C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation 13.5.5 to calculate N_{0}/N. Then substitute the value for the halflife of ^{14}C into Equation 13.5.3 to find the rate constant for the reaction.
B Using the values obtained for N_{0}/N and the rate constant, solve Equation 13.5.7 to obtain the elapsed time.
Solution:
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the halflife, so we can use the integrated rate law for a firstorder nuclear reaction (Equation 13.5.7) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\( ln \dfrac{N}{N_{0}}=kt \)
\( \dfrac{ln \left (N_{0}/N \right )}{k}=t \)
A From Equation 13.5.5, we know that A = kN. We can therefore use the initial and final activities (A_{0} = 15 dpm and A = 8.0 dpm) to calculate N_{0}/N:
\( \dfrac{A_{0}}{A}= \dfrac{\cancel{k}N_{0}}{\cancel{k}N}=\dfrac{N_{0}}{N}=\dfrac{15.}{8.0} \)
Now we need only calculate the rate constant for the reaction from its halflife (5730 yr) using Equation 13.5.3:
\( t_{1/2}=\dfrac{0.693}{k} \)
This equation can be rearranged as follows:
\( k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\;yr}=1.22 \times 10^{4} \; yr^{1} \)
B Substituting into the equation for t,
\( t=\dfrac{ln\left ( N_{0}/N \right )}{k}=\dfrac{ln\left ( 15./8.0 \right )}{1.22 \times 10^{4} \; yr^{1}}=5.2 \times 10^{3} \; yr \)
From our calculations, the man died 5200 yr ago.
Exercise
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a ^{14}C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer: 30,000 yr
Summary
The halflife of a reaction is the time required for the reactant concentration to decrease to onehalf its initial value. The halflife of a firstorder reaction is a constant that is related to the rate constant for the reaction: t_{1/2} = 0.693/k.
Radioactive decay reactions are firstorder reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time.
Key Takeaways
 The halflife of a firstorder reaction is independent of the concentration of the reactants.
 The halflives of radioactive isotopes can be used to date objects.
Key Equations
halflife of firstorder reaction
Equation 13.5.3 \( t_{1/2}=\dfrac{0.693}{k} \)
radioactive decay
Equation 13.5.5: \( A=kN \)
Conceptual Problems

What do chemists mean by the halflife of a reaction?

If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their halflives compare?
Numerical Problems

Halflives for the reaction A + B → C were calculated at three values of [A]_{0}, and [B] was the same in all cases. The data are listed in the following table:
[A]_{0} (M) t_{½} (s) 0.50 420 0.75 280 1.0 210 Does this reaction follow firstorder kinetics? On what do you base your answer?

Ethyl2nitrobenzoate (NO_{2}C_{6}H_{4}CO_{2}C_{2}H_{5}) hydrolyzes under basic conditions. A plot of [NO_{2}C_{6}H_{4}CO_{2}C_{2}H_{5}] versus t was used to calculate t_{½}, with the following results:
[NO_{2}C_{6}H_{4}CO_{2}C_{2}H_{5}] (M/cm^{3}) t_{½} (s) 0.050 240 0.040 300 0.030 400 Is this a firstorder reaction? Explain your reasoning.

Azomethane (CH_{3}N_{2}CH_{3}) decomposes at 600 K to C_{2}H_{6} and N_{2}. The decomposition is first order in azomethane. Calculate t_{½} from the data in the following table:
Time (s) P_{Azomethane} (atm) 0 8.2 × 10^{−2} 2000 3.99 × 10^{−2} 4000 1.94 × 10^{−2} How long will it take for the decomposition to be 99.9% complete?

The firstorder decomposition of hydrogen peroxide has a halflife of 10.7 h at 20°C. What is the rate constant (expressed in s^{−1}) for this reaction? If you started with a solution that was 7.5 × 10^{−3} M H_{2}O_{2}, what would be the initial rate of decomposition (M/s)? What would be the concentration of H_{2}O_{2} after 3.3 h?
Answers

No; the reaction is second order in A because the halflife decreases with increasing reactant concentration according to t_{1/2} = 1/k[A_{0}].

t_{1/2} = 1.92 × 10^{3} s or 1920 s; 19100 s or 5.32 hrs.
Contributors
 Anonymous
Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)