Characteristics:

• Linear decrease in [A] over time
• Plot [A] vs t → straight line, slope = -k
• Half-life depends on [A]₀: $t_{1/2} = \frac{[A]_0}{2k}$

Example: Surface-catalyzed reactions where surface is saturated

First Order Reactions

Rate law: Rate = k[A]

Integrated form:

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

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

Characteristics:

• Exponential decay
• Plot ln[A] vs t → straight line, slope = -k
• Constant half-life: $t_{1/2} = \frac{0.693}{k}$ (independent of [A]₀)

Common examples: Radioactive decay, many decompositions

Second Order Reactions

Rate law: Rate = k[A]²

Integrated form:

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

Characteristics:

• Plot 1/[A] vs t → straight line, slope = k
• Half-life increases as reaction proceeds: $t_{1/2} = \frac{1}{k[A]_0}$

Example: Gas-phase dimerizations

Comparison Table

Determining Order Graphically

Procedure:

1. Plot [A] vs t (zero order check)
2. Plot ln[A] vs t (first order check)
3. Plot 1/[A] vs t (second order check)
4. Whichever is linear indicates the order

From slope:

• Zero/first order: k = -slope
• Second order: k = +slope

Half-Life

Half-life (t₁/₂): Time for [A] to decrease to half its value

Key insight: Only first order has constant t₁/₂

• After 1 half-life: [A] = [A]₀/2
• After 2 half-lives: [A] = [A]₀/4
• After n half-lives: [A] = [A]₀/2ⁿ

Applications:

• Radioactive dating (¹⁴C, t₁/₂ = 5730 years)
• Drug metabolism (pharmacokinetics)
• Chemical stability predictions

(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 ✓