Analyze chemical kinetics for zero, first, and second order reactions
Reaction Order: Determines how reaction rate depends on concentration
Zero order: Rate = k | First order: Rate = k[A] | Second order: Rate = k[A]²
M (mol/L)
M (mol/L)
seconds, minutes, hours, etc.
| Order | Rate Law | Integrated Form | Half-Life | k Units |
|---|---|---|---|---|
| 0 | Rate = k | [A] = [A]₀ - kt | [A]₀/(2k) | M/s |
| 1 | Rate = k[A] | ln[A] = ln[A]₀ - kt | ln(2)/k | s⁻¹ |
| 2 | Rate = k[A]² | 1/[A] = 1/[A]₀ + kt | 1/(k[A]₀) | M⁻¹s⁻¹ |
Reaction order describes how the rate of a chemical reaction depends on the concentration of reactants. It is determined experimentally and reveals the mechanism by which reactions occur.
Rate = k[A]ⁿ
Rate Law: Rate = k
Integrated Law: [A]ₜ = [A]₀ - kt
Half-Life: t₁/₂ = [A]₀/(2k) (depends on initial concentration)
Linear Plot: [A] vs time gives straight line
Rate is independent of concentration. Often occurs when reaction occurs on a saturated catalyst surface or with enzyme saturation.
Example: Decomposition of N₂O on platinum surface
Rate Law: Rate = k[A]
Integrated Law: ln[A]ₜ = ln[A]₀ - kt
Half-Life: t₁/₂ = ln(2)/k ≈ 0.693/k (constant, independent of concentration)
Linear Plot: ln[A] vs time gives straight line with slope = -k
Rate is directly proportional to concentration. Most common reaction order.
Examples: Radioactive decay, sucrose hydrolysis, N₂O₅ decomposition
Rate Law: Rate = k[A]²
Integrated Law: 1/[A]ₜ = 1/[A]₀ + kt
Half-Life: t₁/₂ = 1/(k[A]₀) (inversely proportional to initial concentration)
Linear Plot: 1/[A] vs time gives straight line with slope = k
Rate is proportional to concentration squared. Common in gas phase reactions.
Examples: NO₂ decomposition (2NO₂ → 2NO + O₂), many dimerization reactions
Plot concentration data in different forms:
Measure half-lives at different initial concentrations:
Compare initial rates at different initial concentrations:
If [A]₀ doubles and rate doubles → First order
If [A]₀ doubles and rate quadruples → Second order
Problem: Carbon-14 has a half-life of 5730 years. If a sample initially contains 100 g of ¹⁴C, how much remains after 10,000 years?
Solution:
Step 1: Calculate k from half-life
k = ln(2)/t₁/₂ = 0.693/5730 = 1.21×10⁻⁴ year⁻¹
Step 2: Use integrated first-order law
ln[A]ₜ = ln[A]₀ - kt
ln[A]ₜ = ln(100) - (1.21×10⁻⁴)(10000)
ln[A]ₜ = 4.605 - 1.21 = 3.395
[A]ₜ = e³·³⁹⁵ = 29.8 g
Answer: 29.8 g of ¹⁴C remains after 10,000 years
This represents approximately 1.75 half-lives (10000/5730 ≈ 1.75)
The rate constant k varies with temperature according to the Arrhenius equation:
k = A·e^(-Ea/RT)
Higher temperatures increase k, making reactions faster. A rule of thumb: reaction rate doubles for every 10°C temperature increase.