Integrated rate laws relate concentration to time directly, allowing us to:

• Predict reactant concentration at any time
• Determine reaction order from experimental data
• Calculate rate constants from concentration-time data

These equations are derived from differential rate laws using calculus.

Zero Order Reactions

Rate law: Rate = k

Integrated form:

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

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

Test for zero order: Plot [A] vs t

• Not linear (curved)

Test for first order: Plot ln[A] vs t

• Points: (0, 0.000), (10, -0.462), (20, -0.777), (30, -1.022), (40, -1.238)
• Check linearity: Δ(ln[A])/Δt ≈ -0.031 per second (constant!)
• Linear! → First order

Test for second order: Plot 1/[A] vs t

• Not as linear as ln[A] plot

Conclusion: First order

Calculate k from slope:

slope = Δ(ln[A])/Δt = (-1.238 - 0.000)/(40 - 0) = -0.031 s⁻¹

For first order: slope = -k k = 0.031 s⁻¹

Verification: Calculate t₁/₂ = 0.693/k = 0.693/0.031 = 22.4 s Check data: At t ≈ 22 s, [A] should be ≈ 0.50 M (half of 1.00) Interpolating between t=20 (0.46 M) and t=30 (0.36 M) gives ≈ 0.48 M ✓

Answer: First order, k = 0.031 s⁻¹

Answer: 18.5 s

Verification: After 18.5 s, [A] should be 0.050 M

1/[A]₁₈.₅ = 10 + (0.54)(18.5) = 10 + 10 = 20 M⁻¹ [A]₁₈.₅ = 1/20 = 0.050 M ✓

(c) Second half-life

Key point: For second order, half-life increases!

After first t₁/₂: [A] = 0.050 M (new starting point)

Second t₁/₂ = 1/(k × 0.050) = 1/((0.54)(0.050)) Second t₁/₂ = 1/0.027 = 37.0 s

Answer: 37.0 s

Important: Second half-life (37 s) is twice the first (18.5 s)

• For second order: each successive half-life doubles
• Total time for 2 half-lives: 18.5 + 37.0 = 55.5 s
• After 55.5 s: [A] = 0.025 M (1/4 of original)

Contrast with first order:

• First order: all half-lives equal
• Second order: each half-life = 2 × previous
• Zero order: each half-life = 1/2 × previous