4.4 Half-life and its applications

Half-Life and Its Applications

Half-life measures the time it takes for a reactant's concentration to drop to half its starting value. For first-order reactions, this time stays constant no matter how much reactant you begin with, which makes half-life a powerful tool for predicting how a reaction progresses over time.

Definition and Significance of Half-Life

Half-life ($t_{1/2}$) is the time required for a reactant concentration to decrease to half its initial value. What makes it especially useful for first-order reactions is that it remains constant throughout the reaction.

• A shorter half-life means a faster reaction. Uranium-235 decays with a half-life of about 704 million years, which sounds long, but is short compared to uranium-238's 4.5 billion-year half-life.
• A longer half-life means the reactant sticks around longer, reacting more slowly.
• Because half-life is independent of initial concentration for first-order reactions, you can predict reaction progress without knowing how much reactant you started with.

This makes half-life practical for real problems: figuring out how long a drug stays active in your body, or determining the age of an archaeological sample.

Half-Life and Rate Constant Relationship

The connection between half-life and the rate constant $k$ comes directly from the first-order integrated rate law. Here's the derivation:

1. Start with the first-order integrated rate law: $\ln\frac{[A]_t}{[A]_0} = -kt$

2. At the half-life, the concentration has dropped to half, so $[A]_t = \frac{1}{2}[A]_0$. Substitute that in: $\ln\frac{\frac{1}{2}[A]_0}{[A]_0} = -kt_{1/2}$

3. The $[A]_0$ terms cancel, leaving: $\ln\frac{1}{2} = -kt_{1/2}$

4. Since $\ln\frac{1}{2} = -\ln 2$, solve for $t_{1/2}$:

$t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$

Notice that $[A]_0$ canceled out entirely in step 3. That's why first-order half-life doesn't depend on initial concentration. The math confirms it directly.

Calculation of First-Order Reaction Half-Life

When you know the rate constant, plug straight into the formula:

• Example: If $k = 0.05 \text{ s}^{-1}$, then

$t_{1/2} = \frac{0.693}{0.05 \text{ s}^{-1}} \approx 13.9 \text{ s}$

When you know concentrations at two times, work backward to find $k$ first, then calculate half-life:

1. Suppose $[A]_0 = 1.0 \text{ M}$ and $[A]_t = 0.25 \text{ M}$ at $t = 20 \text{ s}$.
2. Plug into the integrated rate law: $\ln\frac{0.25}{1.0} = -k(20 \text{ s})$ $-1.386 = -k(20 \text{ s})$ $k = 0.0693 \text{ s}^{-1}$
3. Now find half-life: $t_{1/2} = \frac{0.693}{0.0693 \text{ s}^{-1}} = 10.0 \text{ s}$

A quick check confirms this: going from 1.0 M to 0.25 M means the concentration halved twice (1.0 → 0.50 → 0.25), so two half-lives passed in 20 s, giving $t_{1/2} = 10.0 \text{ s}$.

Applications of Half-Life

Radioactive decay follows first-order kinetics, so every radioisotope has a characteristic, constant half-life. The amount remaining after $n$ half-lives is:

$A_t = A_0\left(\frac{1}{2}\right)^n$

• $A_0$ = initial amount of the radioisotope
• $A_t$ = amount remaining after time $t$
• $n$ = number of half-lives elapsed ($n = t / t_{1/2}$, and doesn't have to be a whole number)

Carbon-14 dating uses this directly. Carbon-14 has a half-life of about 5,730 years. By measuring how much carbon-14 remains in an organic sample compared to what it started with, you can calculate how many half-lives have passed and determine the sample's age.

Drug elimination in the body also commonly follows first-order kinetics. The half-life tells clinicians how long a drug stays effective and how often to dose it.

• From a known half-life, find the elimination rate constant: $k = \frac{\ln 2}{t_{1/2}}$
• Then predict the drug concentration at any time: $[D]_t = [D]_0 e^{-kt}$

For example, caffeine has a half-life of roughly 5 hours in most adults. If you drink coffee containing 200 mg of caffeine, about 100 mg remains after 5 hours, 50 mg after 10 hours, and so on. This is why drinking coffee late in the afternoon can still affect your sleep.