Why do we need an integrated equation?
The rate law gives the speed at one moment. But in a lab we want to know: how much is left after 10 minutes? For that we integrate (add up tiny changes over time). The result is the integrated rate equation.
Zero-order reactions
Rate = −d[R]/dt = k[R]⁰ = k. So the concentration falls by the same amount every second.
Derivation, step by step:
- d[R] = −k dt
- Integrate: [R] = −kt + I
- At t = 0, [R] = [R]₀, so I = [R]₀
- [R] = [R]₀ − kt, or k = ([R]₀ − [R])/t
Graph of [R] against t: a straight line, slope −k, intercept [R]₀. Examples: breakdown of NH₃ on a hot platinum surface at high pressure; some enzyme reactions when the enzyme is fully busy.
First-order reactions
Rate = −d[R]/dt = k[R].
Derivation, step by step:
- d[R]/[R] = −k dt
- Integrate: ln[R] = −kt + I
- At t = 0: I = ln[R]₀
- ln[R] = ln[R]₀ − kt, so k = (1/t) ln([R]₀/[R])
- Change to log base 10: k = (2.303/t) log([R]₀/[R])
- Also [R] = [R]₀ e^(−kt)
Graphs: [R] against t is a falling curve; ln[R] (or log[R]) against t is a straight line with slope −k (or −k/2.303). Examples: radioactive decay, breakdown of N₂O₅, hydrogenation of ethene.
Gas-phase first-order reaction
For A(g) → B(g) + C(g) at constant volume, if the starting pressure is p₀ and the total pressure at time t is pₜ, then the pressure of A left is 2p₀ − pₜ. So
k = (2.303/t) log[p₀/(2p₀ − pₜ)]
Half-life
Half-life (t½) is the time for the concentration to fall to half its starting value.
- Zero order: put [R] = [R]₀/2 in [R] = [R]₀ − kt → t½ = [R]₀/2k. More starting amount → longer half-life.
- First order: put [R] = [R]₀/2 in k = (2.303/t) log([R]₀/[R]) → t½ = 2.303 log 2 / k → t½ = 0.693/k. It does not depend on [R]₀.
After n half-lives of a first-order reaction, the amount left = [R]₀ × (½)ⁿ.
Pseudo first-order reactions
Some reactions have two reactants but one is in huge excess, so its concentration hardly changes. Examples: acid hydrolysis of ethyl acetate and the inversion of cane sugar in lots of water. Rate = k′[ester]; they follow the first-order equation.
Try it: predict, then check
In the 3D, pick first order. Predict how many blocks will be left after 3 half-lives (answer: 16 × ⅛ = 2). Now switch to zero order and double [R]₀: predict the half-life before you read it. At home: fill a bottle with water, make a small hole at the bottom and time how long each half takes to drain. The water drains faster when the bottle is full, a lot like first order.
Key formulas and definitions
- Zero order: [R] = [R]₀ − kt; t½ = [R]₀/2k; unit of k = mol L⁻¹ s⁻¹
- First order: k = (2.303/t) log([R]₀/[R]); [R] = [R]₀e^(−kt)
- First order: t½ = 0.693/k; unit of k = s⁻¹
- Amount left after n half-lives = [R]₀/2ⁿ
- Gas phase: k = (2.303/t) log[p₀/(2p₀ − pₜ)]
Worked examples
1. A zero-order reaction has k = 0.02 mol L⁻¹ s⁻¹ and [R]₀ = 1.0 mol/L. Find [R] after 20 s and the half-life.
[R] = 1.0 − 0.02 × 20 = 0.6 mol/L. t½ = [R]₀/2k = 1.0/0.04 = 25 s.
2. A first-order reaction has k = 0.0693 min⁻¹. Find its half-life.
t½ = 0.693/k = 0.693/0.0693 = 10 min.
3. The half-life of a first-order reaction is 30 min. How much of 80 g is left after 2 hours?
2 h = 120 min = 4 half-lives. Left = 80/2⁴ = 80/16 = 5 g.
4. In a first-order reaction, [R] falls from 0.8 to 0.2 mol/L in 40 min. Find k.
k = (2.303/40) log(0.8/0.2) = (2.303/40) × log 4 = 0.0576 × 0.602 ≈ 0.0347 min⁻¹.
5. k = 0.0231 min⁻¹ for a first-order reaction. How long does it take to be 75% complete?
75% done → [R] = [R]₀/4. t = (2.303/0.0231) log 4 = 99.7 × 0.602 ≈ 60 min (two half-lives of 30 min).
6. Show that for a first-order reaction, the time for 99.9% completion is about 10 times the half-life.
99.9% done → [R]₀/[R] = 1000. t = (2.303/k) log 1000 = 6.909/k. t½ = 0.693/k. Ratio = 6.909/0.693 ≈ 10.
7. A first-order reaction is 20% complete in 10 min. Find k and the time for 50% completion.
[R]₀/[R] = 100/80 = 1.25. k = (2.303/10) log 1.25 = 0.2303 × 0.0969 ≈ 0.0223 min⁻¹. t½ = 0.693/0.0223 ≈ 31 min.
8. Gas A → B + C (first order, constant volume). p₀ = 0.5 atm; after 100 s, total pressure pₜ = 0.6 atm. Find k.
Pressure of A left = 2p₀ − pₜ = 1.0 − 0.6 = 0.4 atm. k = (2.303/100) log(0.5/0.4) = 0.02303 × 0.0969 ≈ 2.23 × 10⁻³ s⁻¹.
Common mistakes
- Using t½ = 0.693/k for a zero-order reaction. That formula is only for first order.
- Writing log([R]/[R]₀) instead of log([R]₀/[R]); the answer then comes out negative.
- Mixing up ln and log: ln x = 2.303 log x.
- Forgetting to match time units: if k is in min⁻¹, time must be in minutes.