Integrated Rate Laws and Half-Life

Master integrated rate laws for zero, first, and second order reactions, calculate concentrations over time, and understand half-life concepts.

Integrated Rate Laws and Half-Life

Overview

Integrated rate laws: Relate concentration to time directly

  • Derived from differential rate laws using calculus
  • Allow prediction of [A] at any time t
  • Used to determine reaction order graphically

Zero Order Reactions

Rate law: Rate = k

Integrated form:

[A]t=[A]0−kt[A]_t = [A]_0 - kt

Characteristics:

  • Linear decrease in [A] over time
  • Plot [A] vs t → straight line, slope = -k
  • Half-life depends on [A]₀: t1/2=[A]02kt_{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\ln[A]_t = \ln[A]_0 - kt

Or: [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}

Characteristics:

  • Exponential decay
  • Plot ln[A] vs t → straight line, slope = -k
  • Constant half-life: t1/2=0.693kt_{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:

1[A]t=1[A]0+kt\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: t1/2=1k[A]0t_{1/2} = \frac{1}{k[A]_0}

Example: Gas-phase dimerizations

Comparison Table

| Order | Integrated Law | Linear Plot | Slope | Half-life | k Units | |-------|----------------|-------------|-------|-----------|---------| | 0 | [A] = [A]₀ - kt | [A] vs t | -k | [A]₀/(2k) | M·s⁻¹ | | 1 | ln[A] = ln[A]₀ - kt | ln[A] vs t | -k | 0.693/k | s⁻¹ | | 2 | 1/[A] = 1/[A]₀ + kt | 1/[A] vs t | +k | 1/(k[A]₀) | M⁻¹·s⁻¹ |

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

📚 Practice Problems

1Problem 1easy

❓ Question:

The decomposition of N₂O₅ is first order with k = 5.0 × 10⁻⁴ s⁻¹. If [N₂O₅]₀ = 0.200 M, find: (a) [N₂O₅] after 100 seconds, (b) time to reach 0.050 M, (c) half-life.

💡 Show Solution

Given:

  • First order: k = 5.0 × 10⁻⁴ s⁻¹
  • [N₂O₅]₀ = 0.200 M

First order integrated law: ln[A]_t = ln[A]₀ - kt

(a) Find [N₂O₅] after t = 100 s

ln[N₂O₅]₁₀₀ = ln(0.200) - (5.0 × 10⁻⁴)(100) ln[N₂O₅]₁₀₀ = -1.609 - 0.050 = -1.659

[N₂O₅]₁₀₀ = e⁻¹·⁶⁵⁹ = 0.190 M

Answer: 0.190 M

(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 ✓

2Problem 2medium

❓ Question:

Given concentration vs time data, determine the order and rate constant: t(s): 0, 10, 20, 30, 40 | A: 1.00, 0.63, 0.46, 0.36, 0.29

💡 Show Solution

Data: | t (s) | [A] (M) | ln[A] | 1/[A] (M⁻¹) | |-------|---------|-------|-------------| | 0 | 1.00 | 0.000 | 1.00 | | 10 | 0.63 | -0.462| 1.59 | | 20 | 0.46 | -0.777| 2.17 | | 30 | 0.36 | -1.022| 2.78 | | 40 | 0.29 | -1.238| 3.45 |

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

3Problem 3hard

❓ Question:

A second-order reaction has k = 0.54 M⁻¹·s⁻¹ and [A]₀ = 0.10 M. (a) What is [A] after 2.0 s? (b) What is the first half-life? (c) What is the second half-life?

💡 Show Solution

Given:

  • Second order: k = 0.54 M⁻¹·s⁻¹
  • [A]₀ = 0.10 M

Second order integrated law: 1/[A]_t = 1/[A]₀ + kt

(a) [A] after 2.0 s

1/[A]₂ = 1/0.10 + (0.54)(2.0) 1/[A]₂ = 10 + 1.08 = 11.08 M⁻¹

[A]₂ = 1/11.08 = 0.0903 M

Answer: 0.090 M

(b) First half-life

For second order: t₁/₂ = 1/(k[A]₀)

t₁/₂ = 1/((0.54)(0.10)) = 1/0.054 = 18.5 s

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
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