Integrated Rate Laws & Half-Life
Predict a concentration at any moment — and discover why some reactions have a fixed, concentration-proof half-life.
⏱️ About 20 min
A rate law tells you the speed right now. But right now keeps changing as the reactant runs out. Integrated rate laws solve that: give them a starting concentration and a clock, and they tell you exactly how much is left — the same maths that dates ancient bones by carbon-14.
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The big idea: Integrating the rate law turns 'how fast' into 'how much is left at time t'. For a first-order reaction the concentration decays exponentially, and its half-life t½ = 0.693/k is a constant — it does not depend on how much you started with.
From rate to amount-remaining
The rate law relates rate to concentration at an instant. Using calculus to sum up all those instants (integrating) gives an integrated rate law: a formula for the concentration at any time t. Each order has its own form.
\[ \text{First order:}\quad \ln[\text{A}]_t = \ln[\text{A}]_0 - kt \]
Equivalently [A]ₜ = [A]₀·e^(−kt): first-order reactants decay exponentially.
\[ \text{Second order:}\quad \frac{1}{[\text{A}]_t} = \frac{1}{[\text{A}]_0} + kt \qquad \text{Zero order:}\quad [\text{A}]_t = [\text{A}]_0 - kt \]
Three orders, three straight-line forms. Whichever plot of the data comes out linear tells you the order.
✨ The linear-plot trick
Each integrated law is the equation of a straight line. Plot ln[A] vs t and get a line → first order (slope = −k). Plot 1/[A] vs t and get a line → second order (slope = +k). Plot [A] vs t and get a line → zero order. The graph that straightens out reveals the order.
Half-life: the time to fall by half
The half-life (t½) is the time for a reactant to drop to half its current amount. For a first-order reaction it has a strikingly simple form — and a surprising property.
\[ t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k} \]
First-order half-life. Notice what is missing: there is no concentration term.
🔑 First-order half-life is concentration-independent
Because [A]₀ cancels out when you set [A]ₜ = ½[A]₀, the first-order half-life depends only on k — not on how much you start with. Whether you have 10 mol or 0.001 mol, each half-life takes the same time. That constant t½ is the fingerprint of first-order kinetics, and it is exactly why radioactive decay has a fixed half-life.
📝 Worked example: A first-order reaction has k = 0.0693 s⁻¹. What is its half-life, and does starting with more reactant change it?
- Use t½ = 0.693 / k.
- = 0.693 / 0.0693 = 10 s.
- The formula contains no concentration term, so starting with more (or less) reactant does not change t½ — it stays 10 s.
✓ t½ = 10 s, and it is the same no matter the starting concentration.
📝 Worked example: If one half-life has passed, then a second, then a third, what fraction of the original reactant is left?
- Each half-life leaves half of what was there: ½, then ½ of ½ = ¼, then ½ of ¼ = ⅛.
- After n half-lives the fraction remaining is (½)ⁿ.
- After 3 half-lives: (½)³ = 1/8 = 0.125.
✓ One-eighth (12.5%) remains after three half-lives.
✏️ Practice: A first-order reaction has k = 0.030 s⁻¹. What is its half-life in seconds? (Use t½ = 0.693/k.)
s
Solution
- t½ = 0.693 / k.
- = 0.693 / 0.030.
- = 23.1 s. (Independent of the starting concentration.)
✏️ Practice: A first-order reaction starts at [A]₀ = 0.80 M with k = 0.15 min⁻¹. What is [A] after 10 minutes? (Use [A]ₜ = [A]₀·e^(−kt).)
M
Solution
- Compute the exponent: −kt = −(0.15)(10) = −1.5.
- [A]ₜ = 0.80 × e^(−1.5) = 0.80 × 0.223.
- = 0.18 M (about 0.179 M). Just under one-quarter is left — roughly two half-lives (t½ ≈ 4.6 min).
Check your understanding
1. For a first-order reaction, if you double the starting concentration, the half-life…
First-order half-life t½ = 0.693/k contains no concentration term, so it is unchanged by the starting amount. This is the signature of first-order kinetics.
2. You plot ln[A] against time and get a straight line. The reaction is…
ln[A] vs t linear ⇒ first order, with slope = −k. (A linear 1/[A] vs t would be second order; a linear [A] vs t would be zero order.)
3. After 4 half-lives, what fraction of a first-order reactant remains?
Each half-life leaves half: (½)⁴ = 1/16. The reactant approaches zero but never reaches it exactly.
✅ Key takeaways
- Integrated rate laws give concentration at any time t for zero, first and second order.
- First order: ln[A]ₜ = ln[A]₀ − kt, i.e. [A] decays exponentially.
- The order shows up as which plot is straight: [A], ln[A], or 1/[A] vs t.
- First-order half-life t½ = 0.693/k is CONSTANT — independent of concentration.
- After n half-lives, the fraction remaining is (½)ⁿ.
➡️ We can now predict how fast a reaction empties its reactants. But WHY does k have the value it does — and why does heating things up speed them so dramatically? For that we zoom in to the colliding molecules themselves.
Want to test yourself on this?
Try the Chemistry practice test →