Key characteristics:

• Concentration decreases linearly with time
• Plot [A] vs t gives straight line with slope = -k
• Half-life depends on initial concentration

Half-life formula:

$t_{1/2} = \frac{[A]_0}{2k}$

Example: Surface-catalyzed reactions where catalyst surface is saturated

First Order Reactions

Rate law: Rate = k[A]

Integrated form:

$\ln[A]_t = \ln[A]_0 - kt$

Alternative form:

$[A]_t = [A]_0 e^{-kt}$

Key characteristics:

• Concentration decreases exponentially
• Plot ln[A] vs t gives straight line with slope = -k
• Half-life is constant (independent of concentration)

Half-life formula:

$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$

Common examples: Radioactive decay, many decomposition reactions

Second Order Reactions

Rate law: Rate = k[A]²

Integrated form:

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Key characteristics:

• Plot 1/[A] vs t gives straight line with slope = +k
• Half-life increases as reaction proceeds

Half-life formula:

$t_{1/2} = \frac{1}{k[A]_0}$

Example: Gas-phase dimerization reactions (2A → A₂)

Comparison at a Glance (Mobile‑Friendly)

Rather than a cramped table, here’s a stacked, easy‑scan summary for each key property:

Rate law

• Zero order: Rate = k
• First order: Rate = k[A]
• Second order: Rate = k[A]²

Integrated law

• Zero order: $[A]_t = [A]_0 - kt$
• First order: $\ln[A]_t = \ln[A]_0 - kt$ (equivalently, $[A]_t = [A]_0 e^{-kt}$)
• Second order: $\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Linear plot for straight line

• Zero order: [A] vs t (slope = −k)
• First order: ln[A] vs t (slope = −k)
• Second order: 1/[A] vs t (slope = +k)

Half‑life

• Zero order: $t_{1/2} = \dfrac{[A]_0}{2k}$ (depends on [A]₀)
• First order: $t_{1/2} = \dfrac{0.693}{k}$ (constant)
• Second order: $t_{1/2} = \dfrac{1}{k[A]_0}$ (depends on [A]₀)

Units of k

• Zero order: M·s⁻¹
• First order: s⁻¹
• Second order: M⁻¹·s⁻¹

Determining Reaction Order Graphically

Method:

1. Collect data: Measure [A] at various times
2. Make three plots:
   • [A] vs t
   • ln[A] vs t
   • 1/[A] vs t
3. Identify the linear plot - this reveals the order:
   • [A] vs t is linear → Zero order
   • ln[A] vs t is linear → First order
   • 1/[A] vs t is linear → Second order

Extracting the rate constant:

• Zero/First order: k = -slope (note negative!)
• Second order: k = +slope (positive)

Half-Life Concepts

Definition: Time required for concentration to decrease to half its initial value

After n half-lives:

$[A] = \frac{[A]_0}{2^n}$

Examples:

• After 1 half-life: [A] = [A]₀/2 = 50% remaining
• After 2 half-lives: [A] = [A]₀/4 = 25% remaining
• After 3 half-lives: [A] = [A]₀/8 = 12.5% remaining

Key insight: Only first-order reactions have a constant half-life!

Practice Example

Problem: A reaction has the following data:

Determine the order and rate constant.

Solution:

Calculate ln[A] and 1/[A]:

Plot ln[A] vs t → Linear! (slope ≈ -0.01)

Conclusion: First order reaction with k = 0.01 s⁻¹

Half-life: t₁/₂ = 0.693/0.01 = 69.3 seconds

Real-World Applications

Radioactive Dating:

• Carbon-14 dating: t₁/₂ = 5,730 years (first order)
• Used to date organic materials up to ~50,000 years old

Pharmacology:

• Drug metabolism follows first-order kinetics
• Constant half-life allows predictable dosing schedules
• Example: Aspirin t₁/₂ ≈ 2-3 hours

Environmental Chemistry:

• Pollutant degradation rates
• Predicting cleanup timelines

Chemical Engineering:

• Reactor design and optimization
• Determining optimal reaction times

(b) Time to reach 0.050 M

Rearrange: t = (1/k)ln([A]₀/[A]_t)

t = (1/(5.0 × 10⁻⁴))ln(0.200/0.050) t = 2000 × ln(4) = 2000 × 1.386 t = 2772 s = 46.2 min

Answer: 2772 s or 46 min

(c) Half-life

For first order: t₁/₂ = 0.693/k

t₁/₂ = 0.693/(5.0 × 10⁻⁴) = 1386 s = 23.1 min

Answer: 1386 s or 23 min

Check: In part (b), 0.200 M → 0.050 M is a factor of 4 = 2² So time = 2 half-lives = 2(1386) = 2772 s ✓