Half-Life and First-Order Reactions
A half-life is the time required for the concentration or amount of a reacting substance to decrease to one-half of its initial value. In practice, half-life is especially useful for describing first-order reactions, in which the reaction rate is directly proportional to the concentration of the reactant. Instead, its concentration approaches zero exponentially. That's why because the rate slows as the concentration falls, a first-order reaction never theoretically reaches zero concentration at a finite time. Understanding this relationship is essential in chemistry, pharmacology, environmental science, nuclear physics, and many other fields.
What Is a Half-Life?
The half-life of a substance, written as (t_{1/2}), is the time needed for its quantity to fall from an initial value to half of that value. Here's one way to look at it: if a reactant begins at a concentration of (0.40\ \text{mol L}^{-1}) and has a half-life of 25 minutes, its concentration will be:
- (0.20\ \text{mol L}^{-1}) after 25 minutes
- (0.10\ \text{mol L}^{-1}) after 50 minutes
- (0.05\ \text{mol L}^{-1}) after 75 minutes
- (0.025\ \text{mol L}^{-1}) after 100 minutes
The important feature of a first-order reaction is that the time required for each 50% reduction remains constant, even though the absolute amount consumed during each half-life becomes smaller.
After the first half-life, half of the original amount remains. After the second, one-quarter remains. After the third, one-eighth remains.
[ [A]_t = [A]_0\left(\frac{1}{2}\right)^n ]
or equivalently:
[ [A]_t = [A]_0 2^{-n} ]
This relationship makes first-order reactions relatively easy to predict and analyze.
The Rate Law of a First-Order Reaction
For a first-order reaction involving reactant (A), the rate law is:
[ \text{Rate} = k[A] ]
Here:
- ([A]) is the concentration of reactant (A)
- (k) is the rate constant
- (\text{Rate}) is the rate at which (A) is consumed
The rate constant has units of inverse time, such as (\text{s}^{-1}), (\text{min}^{-1}), or (\text{h}^{-1}). Its unit reflects the fact that a first-order rate constant does not depend on the concentration of the reactant It's one of those things that adds up..
For a simple first-order reaction:
[ A \rightarrow \text{products} ]
the rate of disappearance of (A) is written as:
[ -\frac{d[A]}{dt} = k[A] ]
The negative sign indicates that the concentration of (A) decreases with time.
Unlike reactions of other orders, the half-life of a first-order reaction is independent of the initial concentration. If two samples of the same reactant have different starting concentrations but the same rate constant at the same temperature, they will have the same half-life.
Deriving the Half-Life Equation
The first-order rate equation can be separated and integrated:
[ \frac{d[A]}{[A]} = -k,dt ]
Integrating from the initial concentration ([A]_0) at (t=0) to the concentration ([A]_t) at time (t) gives:
[ \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt ]
Rearranging this equation produces the common integrated rate law:
[ \ln[A]_t = \ln[A]_0 - kt ]
An equivalent form is:
[ \ln\left(\frac{[A]_0}{[A]_t}\right) = kt ]
At the half-life, the concentration has fallen to one-half of its initial value:
[ [A]_t = \frac{[A]_0}{2} ]
Substituting this into the integrated rate law gives:
[ \ln\left(\frac{[A]_0}{[A]0/2}\right) = kt{1/2} ]
Therefore:
[ \ln(2) = kt_{1/2} ]
Since (\ln(2)\approx 0.693):
[ \boxed{t_{1/2} = \frac{\ln 2}{
The half‑life expression therefore becomes
[ t_{1/2}= \frac{\ln 2}{k};, ]
so that, numerically,
[ t_{1/2}= \frac{0.693}{k}. ]
Because the rate constant (k) carries units of inverse time (e.(\text{min}^{-1})), the half‑life also has the same time unit. g. This simple inverse relationship makes it straightforward to convert between a measured half‑life and the underlying kinetic constant Small thing, real impact. Simple as that..
Determining (k) and (t_{1/
Determining (k) and (t_{1/2})
In practice the rate constant (k) is obtained from experimental concentration data. Because the integrated rate law for a first‑order process is linear in the natural logarithm, a simple plot of (\ln [A]) versus time yields a straight line whose slope equals (-k).
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Concentration‑time method
- Collect ([A]) measurements at regular intervals (e.g., every minute).
- Compute (\ln [A]) for each point.
- Fit a straight line to the ((\text{time},\ln [A])) data; the slope (m) is (-k).
- The rate constant follows directly: (k = -m).
-
Half‑life method
- Determine the time required for the concentration to fall to one‑half of its initial value, (t_{1/2}).
- Insert this value into the fundamental relation (k = \dfrac{\ln 2}{t_{1/2}}).
- Because the half‑life is constant for a given temperature, a single measurement is sufficient to calculate (k).
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Units and consistency
- The rate constant carries units of inverse time (s(^{-1}), min(^{-1}), h(^{-1}), …).
- When (k) is derived from a half‑life, the resulting unit matches the time unit used for (t_{1/2}) (e.g., if (t_{1/2}) is expressed in minutes, (k) will be min(^{-1})).
Illustrative example
Suppose a sample of reactant A shows a concentration of 0.80 M at (t = 0) and 0.40 M at (t = 15) min.
- Compute (\ln 0.80 = -0.223) and (\ln 0.40 = -0.916).
- Slope (m = \dfrac{-0.916 - (-0.223)}{15 - 0} = \dfrac{-0.693}{15} = -0.0462) min(^{-1}).
- Hence (k = 0.0462) min(^{-1}).
The corresponding half‑life is (t_{1/2} = \dfrac{0.693}{0.0462} \approx 15) min, confirming the consistency of the two approaches.
Conclusion
First‑order reactions are distinguished by an exponential decline of reactant concentration with time, a characteristic that simplifies both theoretical analysis and experimental determination. The integrated rate law (\ln [A] = \ln [A]0 - kt) provides a linear relationship that can be exploited to extract the rate constant (k) from concentration versus time data, while the half‑life expression (t{1/2} = \dfrac{\ln 2}{k}) offers a rapid, concentration‑independent route to the same constant. Because the half‑life does not depend on the initial amount, the kinetic parameters of a first‑order process remain constant under identical temperature conditions, making these reactions especially predictable in fields ranging from pharmaceutical elimination to radioactive decay. Mastery of the straightforward mathematical relationships and the practical techniques described above equips chemists to accurately characterize and control first‑order transformations.
