Integrated Rate Laws
Relate concentration and time for common reaction orders
Understanding Integrated Rate Laws
Integrated rate laws are mathematical equations that relate reactant concentrations to time, derived by integrating differential rate laws. While differential rate laws (Rate = k[A]n) describe instantaneous rates, integrated rate laws allow chemists to calculate concentrations at any time point or determine how long a reaction takes to reach a certain completion level. This makes them invaluable for practical applications—from predicting medication dosages in pharmacokinetics to calculating radioactive decay timescales in nuclear chemistry.
Each reaction order (zero, first, second) yields a unique integrated form with distinct mathematical behavior and graphical signature. Understanding these patterns is essential for experimental kinetics: by plotting concentration data appropriately (linear vs. time, ln vs. time, or 1/concentration vs. time), chemists can identify reaction order and extract rate constants from slopes. This graphical method, pioneered by physical chemists in the early 20th century, remains standard practice in research laboratories worldwide.
Integrated rate laws also reveal fundamental insights about half-lives. First-order reactions have constant half-lives independent of initial concentration, explaining radioactive decay consistency and first-pass drug metabolism. Zero-order half-lives depend linearly on initial concentration, while second-order half-lives depend inversely. These relationships are crucial for drug design, environmental remediation, and industrial process optimization where predicting time-dependent behavior is essential.
Zero-Order Reactions
[A]t = [A]0 − kt
Linear decrease with time
Differential Form:
Rate = k (independent of [A])
Linear Plot:
[A] vs t → straight line, slope = −k
Half-Life:
t1/2 = [A]0/(2k) (depends on [A]0!)
Units of k:
M/s or mol·Lâ»Â¹Â·sâ»Â¹
When Does Zero-Order Occur?
- Surface reactions: When reactant adsorption saturates catalyst surface
- Enzyme kinetics: At high substrate concentrations (Vmax regime in Michaelis-Menten)
- Photochemical reactions: Limited by photon flux, not reactant concentration
- Example: Alcohol elimination by liver enzymes (~constant 0.015% BAC/hour once saturated)
First-Order Reactions
ln[A]t = ln[A]0 − kt
or [A]t = [A]0 e−kt (exponential decay)
Differential Form:
Rate = k[A]
Linear Plot:
ln[A] vs t → straight line, slope = −k
Half-Life:
t1/2 = 0.693/k = ln2/k (constant!)
Units of k:
sâ»Â¹ or minâ»Â¹ or hrâ»Â¹
Most Common Reaction Order:
- Radioactive decay: All isotopes follow first-order kinetics (C-14 dating, medical isotopes)
- Drug elimination: Many drugs have first-order metabolism (constant percentage removed per hour)
- Unimolecular reactions: A → products (molecular rearrangements, decompositions)
- Example: Nâ‚‚Oâ‚… → 2NOâ‚‚ + ½Oâ‚‚ has k = 6.2 × 10â»â´ sâ»Â¹ at 65°C, t1/2 = 1118 s
Second-Order Reactions
1/[A]t = 1/[A]0 + kt
Reciprocal concentration increases linearly
Differential Form:
Rate = k[A]² or k[A][B]
Linear Plot:
1/[A] vs t → straight line, slope = k
Half-Life:
t1/2 = 1/(k[A]0) (inversely proportional to [A]0)
Units of k:
L·molâ»Â¹Â·sâ»Â¹ or Mâ»Â¹Â·sâ»Â¹
Common Second-Order Cases:
- Bimolecular reactions: 2A → products or A + B → products (most elementary steps)
- Gas-phase collisions: 2NOâ‚‚ → 2NO + Oâ‚‚ (k = 0.54 Mâ»Â¹Â·sâ»Â¹ at 300°C)
- Radical recombination: 2R• → R-R (termination steps in polymerization)
- Practical Note: Higher initial concentration → shorter half-life (reacts faster when concentrated)
Detailed Example: First-Order Kinetics
Problem: Cyclopropane isomerizes to propene (first-order). If [cyclopropane]0 = 0.100 M, k = 6.0 × 10â»â´ sâ»Â¹, find [cyclopropane] at t = 1000 s.
Method 1: Using ln[A] = ln[A]0 − kt
ln[A]1000 = ln(0.100) − (6.0 × 10â»â´)(1000)
ln[A]1000 = −2.3026 − 0.60
ln[A]1000 = −2.9026
[A]1000 = e−2.9026 = 0.0549 M
Method 2: Using [A] = [A]0 e−kt
[A]1000 = 0.100 × e−0.60
[A]1000 = 0.100 × 0.5488 = 0.0549 M ✓
Answer: [cyclopropane] = 0.0549 M after 1000 seconds
About 55% reacted (45% remaining). Calculate half-life: t1/2 = 0.693/k = 1155 s, so we're close to one half-life.
Comparison Table: Reaction Orders
| Property | Zero-Order | First-Order | Second-Order |
|---|---|---|---|
| Differential | Rate = k | Rate = k[A] | Rate = k[A]² |
| Integrated | [A] = [A]0 − kt | ln[A] = ln[A]0 − kt | 1/[A] = 1/[A]0 + kt |
| Linear Plot | [A] vs t | ln[A] vs t | 1/[A] vs t |
| Slope | −k | −k | +k |
| Half-Life | [A]0/(2k) | 0.693/k | 1/(k[A]0) |
| t1/2 depends on | [A]0 | Independent of [A]0 | 1/[A]0 |
| k units | M/s | sâ»Â¹ | Mâ»Â¹Â·sâ»Â¹ |
Applications & Real-World Examples
1. Pharmacokinetics
Most drugs follow first-order elimination. If a drug has t1/2 = 6 hours and initial plasma concentration = 100 μg/mL, after 12 hours (2 half-lives): C = 100 × (1/2)² = 25 μg/mL. This predictability guides dosing schedules.
2. Radiocarbon Dating
C-14 decay is first-order with t1/2 = 5730 years. Ancient artifacts with 25% of original C-14 are ~11,460 years old (2 half-lives). The constant half-life makes age determination reliable.
3. Atmospheric Ozone Depletion
CFC decomposition in stratosphere follows first-order kinetics with k ~ 10â»âµ sâ»Â¹, giving t1/2 ~ 20 hours for key reactions. Integrated rate laws predict ozone hole recovery timescales (decades).
4. Industrial Reactors
Batch reactor design uses integrated rate laws to calculate required reaction time for desired conversion. For 95% conversion of second-order reaction: t = (1/(k[A]0)) × 19 (solving 1/[A] = 20/[A]0).
Common Mistakes & Tips
Wrong Linear Plot
Plot all three ([A] vs t, ln[A] vs t, 1/[A] vs t) and see which is linear! The straight-line plot reveals reaction order. Non-linearity in all three suggests complex kinetics or experimental error.
Unit Inconsistency
Check k units match reaction order! First-order k must have timeâ»Â¹ units (sâ»Â¹, minâ»Â¹), second-order needs Mâ»Â¹Â·timeâ»Â¹. Mismatched units mean wrong order or calculation error.
Natural Log vs Logâ‚â‚€
Integrated rate laws use ln (natural logarithm, base e), NOT logâ‚â‚€! Using logâ‚â‚€([A]) vs t gives wrong slope. Remember: ln = 2.303 × logâ‚â‚€.
Pro Tip: Half-Life Method
Quick diagnostic: If successive half-lives are constant → first-order. If t1/2 increases → second-order. If t1/2 proportional to [A]0 → zero-order. This checks order without making plots!