11.3 Half-life and reaction order

Half-life is a central concept in chemical kinetics that quantifies how long it takes for a reactant's concentration to drop to half its initial value. The relationship between half-life and reaction order gives you a powerful diagnostic tool: by observing how half-life changes (or doesn't) with concentration, you can identify the reaction order, extract rate constants, and predict how a reaction evolves over time.

Half-life and Reaction Kinetics

Definition and Relationship to Rate Constant and Reaction Order

The half-life ($t_{1/2}$) is the time required for a reactant's concentration to fall to exactly half of its starting value. What makes half-life so useful in kinetics is that its dependence on initial concentration differs for each reaction order, giving you a direct way to distinguish between them.

Here are the three key half-life expressions:

• Zero-order: $t_{1/2} = \frac{[A]_0}{2k}$
• First-order: $t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$
• Second-order (single reactant): $t_{1/2} = \frac{1}{k[A]_0}$

Notice the pattern in how $[A]_0$ appears in each equation:

• Zero-order: $t_{1/2}$ is directly proportional to $[A]_0$. Higher starting concentration means a longer half-life because the reaction consumes reactant at a fixed rate regardless of how much is present.
• First-order: $t_{1/2}$ is independent of $[A]_0$. The half-life stays constant no matter the starting concentration. Radioactive decay is the classic example.
• Second-order: $t_{1/2}$ is inversely proportional to $[A]_0$. Higher starting concentration means a shorter half-life because the rate accelerates with more reactant available (e.g., dimerization of cyclopentadiene).

This concentration dependence is the single most important diagnostic feature of half-life in kinetics.

Calculating Half-life

First-order Reactions

Use $t_{1/2} = \frac{\ln(2)}{k}$. You only need the rate constant.

Example: A first-order reaction has $k = 0.05 \text{ s}^{-1}$.

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

This half-life stays at 13.9 s whether you start with 2.0 M or 0.001 M of reactant.

Zero-order Reactions

Use $t_{1/2} = \frac{[A]_0}{2k}$. You need both the rate constant and the initial concentration.

Example: A zero-order reaction has $k = 0.02 \text{ M s}^{-1}$ and $[A]_0 = 1.0 \text{ M}$.

$t_{1/2} = \frac{1.0 \text{ M}}{2 \times 0.02 \text{ M s}^{-1}} = 25 \text{ s}$

If you doubled $[A]_0$ to 2.0 M, the half-life would double to 50 s. This makes sense: a zero-order reaction depletes reactant at a constant rate (the slope of $[A]$ vs. time is just $-k$), so more reactant simply takes proportionally longer to get through.

Second-order Reactions

Use $t_{1/2} = \frac{1}{k[A]_0}$ for a reaction that is second-order in a single reactant. (For two reactants at equal initial concentrations, the same equation applies.)

Example: A second-order reaction has $k = 0.1 \text{ M}^{-1}\text{s}^{-1}$ and $[A]_0 = 0.5 \text{ M}$.

$t_{1/2} = \frac{1}{0.1 \text{ M}^{-1}\text{s}^{-1} \times 0.5 \text{ M}} = 20 \text{ s}$

If you doubled $[A]_0$ to 1.0 M, the half-life would halve to 10 s.

Determining Reaction Order

Using Half-life Data

If you're given half-life values at different initial concentrations, the pattern tells you the order:

• Half-life stays constant as $[A]_0$ changes → first-order
• Half-life doubles when $[A]_0$ doubles → zero-order
• Half-life halves when $[A]_0$ doubles → second-order

Example: A reaction has the following data:

When the concentration is halved, the half-life doubles. That's the signature of a zero-order reaction ($t_{1/2} \propto [A]_0$). You can verify: $t_{1/2} = \frac{[A]_0}{2k}$, so halving $[A]_0$ should halve $t_{1/2}$... but wait. Here the half-life increases as concentration decreases. That's actually the signature of a second-order reaction ($t_{1/2} \propto \frac{1}{[A]_0}$), because halving $[A]_0$ doubles $\frac{1}{[A]_0}$, which doubles $t_{1/2}$.

This is a common point of confusion. Keep the proportionalities straight:

• Zero-order: $t_{1/2}$ and $[A]_0$ move in the same direction
• Second-order: $t_{1/2}$ and $[A]_0$ move in opposite directions

So the data above (concentration halved, half-life doubled) indicates second-order.

Interpreting Concentration-Time Graphs

You can also identify reaction order by finding which linearized plot gives a straight line:

• Plot $[A]$ vs. time → straight line means zero-order (slope = $-k$)
• Plot $\ln[A]$ vs. time → straight line means first-order (slope = $-k$)
• Plot $\frac{1}{[A]}$ vs. time → straight line means second-order (slope = $+k$)

Only one of these three plots will be linear for a given data set. That's the one that matches the reaction order.

Concentration Effects on Half-life

First-order Reactions

The half-life doesn't change with concentration. Whether you start with 1.0 M or 0.01 M, $t_{1/2}$ is the same. This is because the integrated rate law is exponential: the fraction remaining depends only on time and $k$, not on how much you started with.

Zero-order Reactions

The half-life scales linearly with $[A]_0$. Doubling the initial concentration doubles the half-life. This follows from the fact that a zero-order reaction has a constant rate of consumption. If you have twice as much reactant but the reaction eats through it at the same fixed rate, it takes twice as long to get halfway.

Example: If $t_{1/2} = 20$ min at $[A]_0 = 1.0$ M, then at $[A]_0 = 2.0$ M, $t_{1/2} = 40$ min.

Second-order Reactions

The half-life is inversely proportional to $[A]_0$. Doubling the initial concentration halves the half-life. The reaction rate depends on $[A]^2$, so at higher concentrations the reaction proceeds much faster, burning through the first half of reactant more quickly.

Example: If $t_{1/2} = 30$ min at $[A]_0 = 0.1$ M, then at $[A]_0 = 0.2$ M, $t_{1/2} = 15$ min.

Problem Solving with Half-life

Calculating Half-life from a Rate Constant

Pick the formula that matches the reaction order and plug in.

Example: Find $t_{1/2}$ for a first-order reaction with $k = 0.02 \text{ s}^{-1}$.

$t_{1/2} = \frac{\ln(2)}{0.02 \text{ s}^{-1}} = 34.7 \text{ s}$

Determining Rate Constants from Half-life

Rearrange the appropriate half-life equation to isolate $k$.

Example: A zero-order reaction has $t_{1/2} = 15$ min and $[A]_0 = 0.5$ M. Find $k$.

1. Start with $t_{1/2} = \frac{[A]_0}{2k}$
2. Rearrange: $k = \frac{[A]_0}{2 \, t_{1/2}}$
3. Substitute: $k = \frac{0.5 \text{ M}}{2 \times 15 \text{ min}} = 0.0167 \text{ M min}^{-1}$

Time to Reach a Specific Concentration

For first-order reactions, you can count half-lives if the target concentration is a convenient power of $\frac{1}{2}$.

Example: A first-order reaction has $t_{1/2} = 20$ min. How long to reach 25% of $[A]_0$?

1. 25% of $[A]_0$ means the concentration has been halved twice: $100\% \rightarrow 50\% \rightarrow 25\%$
2. That's 2 half-lives: $2 \times 20 \text{ min} = 40 \text{ min}$

For targets that aren't neat fractions, use the integrated rate law directly instead of counting half-lives.

Complex Problem: Second-order Concentration After a Given Time

Example: A second-order reaction has $k = 0.05 \text{ M}^{-1}\text{s}^{-1}$ and $[A]_0 = 0.8 \text{ M}$. Find $[A]$ after 30 s.

Important: You cannot simply raise $\frac{1}{2}$ to a fractional number of half-lives for a second-order reaction. The $[A] = [A]_0 \times \left(\frac{1}{2}\right)^n$ shortcut only works for first-order kinetics because only first-order decay is exponential. For second-order reactions, use the integrated rate law:

1. Write the second-order integrated rate law: $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$
2. Substitute: $\frac{1}{[A]} = \frac{1}{0.8 \text{ M}} + (0.05 \text{ M}^{-1}\text{s}^{-1})(30 \text{ s})$
3. Calculate: $\frac{1}{[A]} = 1.25 \text{ M}^{-1} + 1.50 \text{ M}^{-1} = 2.75 \text{ M}^{-1}$
4. Invert: $[A] = \frac{1}{2.75 \text{ M}^{-1}} = 0.364 \text{ M}$

This is the correct approach. The half-life is still useful here for quick estimates ($t_{1/2} = 25$ s, so 30 s is a bit more than one half-life, meaning $[A]$ should be somewhat less than 0.4 M), but the integrated rate law gives you the exact answer.