Integrated Rate Equations and Half-Life

• Derive and use integrated rate laws for zero, first, and second order reactions
• Calculate concentration at any time t from initial conditions
• Determine half-life for each order and understand its concentration dependence
• Use graphical methods to identify reaction order from experimental data
• Connect radioactive decay to first-order kinetics

Level 1 — Foundations

Why integrate? The differential rate law (rate = k[A]^n) tells you the instantaneous rate. Integrating gives [A] as a function of t — essential for practical calculations.

Key insight: First-order t₁/₂ is independent of [A]₀ — only true for first order. Zero and second order t₁/₂ depend on initial concentration.

Level 2 — JEE Depth

Zero Order Derivation
−d[A]/dt = k → d[A] = −k dt → integrating: [A] = [A]₀ − kt
At t = t₁/₂: [A] = [A]₀/2 → [A]₀/2 = [A]₀ − kt₁/₂ → t₁/₂ = [A]₀/(2k)
Units of k: mol L⁻¹ s⁻¹

First Order Derivation
−d[A]/dt = k[A] → d[A]/[A] = −k dt → ln[A] = ln[A]₀ − kt
Also: [A] = [A]₀ e^(−kt)
At t₁/₂: [A]₀/2 = [A]₀ e^(−kt₁/₂) → ln 2 = kt₁/₂ → t₁/₂ = 0.693/k
Time for [A] to fall to [A]₀/n: t = ln(n)/k
Units of k: s⁻¹

Second Order Derivation
−d[A]/dt = k[A]² → d[A]/[A]² = −k dt → 1/[A] = 1/[A]₀ + kt
At t₁/₂: 2/[A]₀ = 1/[A]₀ + kt₁/₂ → t₁/₂ = 1/(k[A]₀)
Units of k: L mol⁻¹ s⁻¹

Graphical Method for Identifying Order
Plot each candidate graph; the one giving a straight line reveals the order:

• [A] vs t → straight line? → zero order (slope = −k)
• ln[A] vs t → straight line? → first order (slope = −k)
• 1/[A] vs t → straight line? → second order (slope = +k)

Radioactive Decay — Always First Order
N = N₀ e^(−λt), where λ (decay constant) = k; t₁/₂ = 0.693/λ
JEE Advanced: fraction remaining after n half-lives = (1/2)^n

Useful First-Order Fraction Relationships

JEE Traps

• For second-order, t₁/₂ doubles when [A]₀ halves — opposite of zero order
• ln 2 = 0.693, not 0.632 (0.632 is for 1 − e⁻¹, i.e., when kt = 1)
• Plotting wrong graph: ensure you use natural log (ln), not log₁₀, unless you account for the 2.303 factor

Example 1: First-Order Calculation

Given: k = 0.0693 min⁻¹, [A]₀ = 1 M, find [A] after 20 min and t₁/₂

Step 1: Half-life
  t₁/₂ = 0.693/k = 0.693/0.0693 = 10 min

Step 2: [A] after 20 min (= 2 half-lives)
  Method A (using powers of half):
    After 1st t₁/₂ (10 min): [A] = 0.5 M
    After 2nd t₁/₂ (20 min): [A] = 0.25 M

  Method B (using integrated law):
    ln[A] = ln(1) − (0.0693)(20)
    ln[A] = 0 − 1.386 = −1.386
    [A] = e^(−1.386) = 0.25 M ✓

Answer: t₁/₂ = 10 min, [A] at t = 20 min is 0.25 M

Example 2: Finding Order and k from % Decomposition

Given: 75% of substance decomposes in 100 min (assume first order, verify)

Step 1: If 75% decomposed, [A]/[A]₀ = 0.25 → [A] = 0.25[A]₀

Step 2: Apply first-order integrated law
  ln[A] = ln[A]₀ − kt
  ln(0.25[A]₀) = ln[A]₀ − k(100)
  ln(0.25) = −k(100)
  −1.386 = −100k
  k = 0.01386 min⁻¹

Step 3: Calculate t₁/₂ to verify consistency
  t₁/₂ = 0.693/0.01386 = 50 min

Step 4: Verify: 75% decomposed = 2 half-lives → t = 2 × 50 = 100 min ✓

Answer: k = 0.01386 min⁻¹, t₁/₂ = 50 min, consistent with first order

JEE check: 75% decomposed → (1/2)² = 0.25 remaining → 2 half-lives
  so t₁/₂ = 100/2 = 50 min → k = 0.693/50 = 0.01386 min⁻¹ (same, faster method)

• Q1: A first-order reaction has t₁/₂ = 30 s. What fraction remains after 2 minutes?
• Q2: For a zero-order reaction, [A]₀ = 0.5 M and k = 0.01 mol/L/s. When does [A] reach zero?
• Q3: A plot of 1/[A] vs t gives a straight line with slope 0.05 L mol⁻¹ min⁻¹. What is the order and k?
• Q4: 87.5% of a radioactive element decays in 3 hours. What is its half-life?
• Stretch: Q5: A second-order reaction has [A]₀ = 2 M and k = 0.1 L/mol/s. How long until [A] = 0.5 M? How does t₁/₂ at [A]₀ = 1 M compare to [A]₀ = 2 M?

NCERT Chapter 3 link: Sections 3.4–3.6 cover integrated rate equations for zero, first, and second order, with graphical analysis. NCERT worked examples include calculation of t₁/₂ and concentration at time t for first-order reactions.

Exam connections: JEE asks for: (a) matching order to graph shape, (b) finding k from given data using integrated law, (c) calculating concentration after n half-lives, (d) radioactive decay as first-order kinetics, (e) comparing half-lives of different orders when [A]₀ changes. "Which graph gives a straight line?" is a perennial JEE MCQ.

Study strategy: Derive all three integrated rate laws from scratch at least twice — derivation cements the formula. Draw the three graphs side-by-side. Practise converting between % decomposed and [A]/[A]₀ ratio as a standalone mental skill. For JEE Advanced, be ready to combine first-order integrated law with Arrhenius equation in a single problem.

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