Half-Life of Reactions
Real-Life Connection: From Nuclear Power to Medicine
Why is plutonium-239 so dangerous for thousands of years? Why does iodine-131 used in thyroid treatment become safe in weeks? Why do some medicines need to be taken every 6 hours while others last 24 hours?
The answer: Half-life!
Nuclear waste management:
- Pu-239: t₁/₂ = 24,000 years → radioactive for millennia
- I-131: t₁/₂ = 8 days → decays quickly (used in medical treatment)
Pharmaceutical design:
- Aspirin: t₁/₂ = 2-3 hours → frequent dosing
- Long-acting drugs: t₁/₂ = 24+ hours → once-daily dosing
Archaeological dating:
- C-14: t₁/₂ = 5,730 years → date artifacts up to 50,000 years old
- U-238: t₁/₂ = 4.5 billion years → date geological samples
Understanding half-life helps us predict reaction timescales, design safe nuclear facilities, and develop effective medications!
What is Half-Life?
Definition
Half-life (t₁/₂) is the time required for the concentration of a reactant to decrease to half of its initial value.
$$\boxed{t_{1/2} = \text{time when } [A]_t = \frac{[A]_0}{2}}$$Key concept: After each half-life, concentration reduces by 50%
General Pattern
| Number of half-lives | Remaining fraction | % Remaining | % Reacted |
|---|---|---|---|
| 0 | 1 | 100% | 0% |
| 1 | 1/2 | 50% | 50% |
| 2 | 1/4 | 25% | 75% |
| 3 | 1/8 | 12.5% | 87.5% |
| 4 | 1/16 | 6.25% | 93.75% |
| n | (1/2)ⁿ | 100/(2ⁿ)% | [100-100/2ⁿ]% |
Formula:
$$\boxed{\frac{[A]_t}{[A]_0} = \left(\frac{1}{2}\right)^n}$$where n = number of half-lives elapsed
Half-Life for Zero Order Reactions
Derivation
Integrated equation:
$$[A]_t = [A]_0 - kt$$At t = t₁/₂:
$$[A]_t = \frac{[A]_0}{2}$$ $$\frac{[A]_0}{2} = [A]_0 - kt_{1/2}$$ $$kt_{1/2} = [A]_0 - \frac{[A]_0}{2} = \frac{[A]_0}{2}$$ $$\boxed{t_{1/2} = \frac{[A]_0}{2k}}$$Key Characteristics
- Depends on initial concentration [A]₀
- Directly proportional to [A]₀
- If [A]₀ doubles, t₁/₂ doubles
- Each successive half-life is shorter (takes less time to halve smaller amounts)
Successive Half-Lives
For zero order:
First half-life: [A]₀ → [A]₀/2
$$t_1 = \frac{[A]_0}{2k}$$Second half-life: [A]₀/2 → [A]₀/4
$$t_2 = \frac{[A]_0/2}{2k} = \frac{[A]_0}{4k}$$Third half-life: [A]₀/4 → [A]₀/8
$$t_3 = \frac{[A]_0/4}{2k} = \frac{[A]_0}{8k}$$Pattern: t₁ : t₂ : t₃ = 1 : 1/2 : 1/4
Each successive half-life is half the previous one!
Example
Photochemical reaction with k = 0.01 M/s
Initial concentration [A]₀ = 0.4 M
First half-life:
$$t_{1/2} = \frac{0.4}{2 \times 0.01} = 20 \text{ s}$$If [A]₀ = 0.8 M:
$$t_{1/2} = \frac{0.8}{2 \times 0.01} = 40 \text{ s}$$Doubling [A]₀ doubles t₁/₂!
Half-Life for First Order Reactions
Derivation
Integrated equation:
$$\ln\frac{[A]_0}{[A]_t} = kt$$At t = t₁/₂:
$$[A]_t = \frac{[A]_0}{2}$$ $$\ln\frac{[A]_0}{[A]_0/2} = kt_{1/2}$$ $$\ln 2 = kt_{1/2}$$ $$\boxed{t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}}$$Key Characteristics
- Independent of initial concentration [A]₀
- Constant throughout the reaction
- Characteristic property of the substance
- Each successive half-life takes the same time
Why First Order is Special
Only for first order reactions:
- Half-life is constant
- Independent of how much you start with
- Each half-life period reduces concentration by exactly 50%
This makes first order kinetics ideal for:
- Radioactive dating (predictable decay)
- Drug dosing (predictable elimination)
- Process control (consistent behavior)
Successive Half-Lives
For first order, all half-lives are equal:
t₁ = t₂ = t₃ = … = 0.693/k
[A]
|
100%|___
| |___
50% | | |___
| | | |___
25% | | | |
|___|___|___|___
0 t₁/₂ 2t₁/₂ 3t₁/₂
Relationship to Rate Constant
$$k = \frac{0.693}{t_{1/2}}$$Large k → Small t₁/₂ → Fast reaction Small k → Large t₁/₂ → Slow reaction
Examples
C-14 decay:
- t₁/₂ = 5,730 years
- k = 0.693/5730 = 1.21 × 10⁻⁴ year⁻¹
I-131 decay:
- t₁/₂ = 8 days
- k = 0.693/8 = 0.0866 day⁻¹
Drug elimination:
- Caffeine: t₁/₂ = 5 hours
- Aspirin: t₁/₂ = 2-3 hours
Half-Life for Second Order Reactions
Derivation (Type I: Rate = k[A]²)
Integrated equation:
$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$At t = t₁/₂:
$$[A]_t = \frac{[A]_0}{2}$$ $$\frac{1}{[A]_0/2} = \frac{1}{[A]_0} + kt_{1/2}$$ $$\frac{2}{[A]_0} = \frac{1}{[A]_0} + kt_{1/2}$$ $$kt_{1/2} = \frac{2}{[A]_0} - \frac{1}{[A]_0} = \frac{1}{[A]_0}$$ $$\boxed{t_{1/2} = \frac{1}{k[A]_0}}$$Key Characteristics
- Inversely proportional to [A]₀
- If [A]₀ doubles, t₁/₂ halves
- Each successive half-life is twice the previous one
- Reaction slows down significantly over time
Successive Half-Lives
For second order:
First half-life: [A]₀ → [A]₀/2
$$t_1 = \frac{1}{k[A]_0}$$Second half-life: [A]₀/2 → [A]₀/4
$$t_2 = \frac{1}{k[A]_0/2} = \frac{2}{k[A]_0} = 2t_1$$Third half-life: [A]₀/4 → [A]₀/8
$$t_3 = \frac{1}{k[A]_0/4} = \frac{4}{k[A]_0} = 4t_1$$Pattern: t₁ : t₂ : t₃ = 1 : 2 : 4
Each successive half-life doubles!
Example
NO₂ decomposition with k = 0.5 M⁻¹s⁻¹
Initial concentration [A]₀ = 0.2 M
First half-life:
$$t_{1/2} = \frac{1}{0.5 \times 0.2} = 10 \text{ s}$$Second half-life:
$$t_{1/2} = \frac{1}{0.5 \times 0.1} = 20 \text{ s}$$Total time for 75% completion = 10 + 20 = 30 s
Comparison Table: Half-Life for Different Orders
| Order | Formula | Dependence on [A]₀ | Successive half-lives | Pattern |
|---|---|---|---|---|
| 0 | t₁/₂ = [A]₀/(2k) | Directly proportional | Decrease | 1, 1/2, 1/4, … |
| 1 | t₁/₂ = 0.693/k | Independent | Constant | 1, 1, 1, … |
| 2 | t₁/₂ = 1/(k[A]₀) | Inversely proportional | Increase | 1, 2, 4, … |
Visual Comparison
Zero Order:
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First Order:
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Second Order:
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Memory Tricks
“ZIP” for Half-life Patterns
Zero order: Decreasing half-lives (faster and faster) I (one/first) order: Identical half-lives (constant) Pair (two/second) order: Progressively increasing (doubling)
“FIC” for First Order
Fixed half-life Independent of concentration Constant throughout
“SID” for Second Order
Slowly increasing half-lives Inversely proportional to [A]₀ Doubling each time
Mathematical Pattern
- 0 order: t₁/₂ has [A]₀ in numerator (increases with [A]₀)
- 1 order: t₁/₂ has no [A]₀ term (independent)
- 2 order: t₁/₂ has [A]₀ in denominator (decreases with [A]₀)
Determining Reaction Order from Half-Life
Method: Vary [A]₀ and Measure t₁/₂
| Observation | Order |
|---|---|
| t₁/₂ increases as [A]₀ increases | Zero |
| t₁/₂ constant (independent of [A]₀) | First |
| t₁/₂ decreases as [A]₀ increases | Second |
Mathematical Test
Measure t₁/₂ at two different [A]₀:
$$\frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \left(\frac{[A]_0^{(1)}}{[A]_0^{(2)}}\right)^{1-n}$$where n = order
- If ratio = 1 → n = 1 (first order)
- If ratio > 1 when [A]₀⁽¹⁾ > [A]₀⁽²⁾ → n = 0 (zero order)
- If ratio < 1 when [A]₀⁽¹⁾ > [A]₀⁽²⁾ → n = 2 (second order)
Common JEE Mistakes
Mistake 1: Using Wrong Formula
Wrong: Using t₁/₂ = 0.693/k for all reactions
Correct: This formula is only for first order!
- Check order first
- Use appropriate formula
Mistake 2: Successive Half-Lives
Wrong: “All reactions have equal successive half-lives”
Correct: Only first order has equal successive half-lives!
- Zero order: decreasing
- Second order: increasing
Mistake 3: n Half-Lives Calculation
Wrong: After 3 half-lives, 3/4 of reactant remains
Correct: After n half-lives, (1/2)ⁿ remains
- After 3 half-lives: (1/2)³ = 1/8 remains (7/8 reacted)
Mistake 4: Units
Wrong: Treating t₁/₂ values in different units as comparable
Correct: Always convert to same time unit before comparing!
Practice Problems
Level 1: JEE Main Foundation
Problem 1: A first order reaction has k = 0.0693 min⁻¹. Calculate half-life.
Solution:
$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0693} = 10 \text{ min}$$Problem 2: In a first order reaction, 75% of reactant decomposes in 60 minutes. Find half-life.
Solution:
75% decomposed → 25% remains
$$\frac{[A]_t}{[A]_0} = 0.25 = \left(\frac{1}{2}\right)^n$$ $$\left(\frac{1}{2}\right)^2 = 0.25$$So n = 2 half-lives
$$2 \times t_{1/2} = 60 \text{ min}$$ $$t_{1/2} = 30 \text{ min}$$Level 2: JEE Main/Advanced
Problem 3: A radioactive sample has half-life of 10 days. Calculate: (a) Time for 87.5% decay (b) % remaining after 30 days (c) Rate constant
Solution:
(a) 87.5% decay → 12.5% remains = 1/8 of original
$$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$So 3 half-lives needed
$$t = 3 \times 10 = 30 \text{ days}$$(b) 30 days = 3 half-lives
$$\text{Remaining} = \left(\frac{1}{2}\right)^3 \times 100\% = 12.5\%$$(c)
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ day}^{-1}$$Problem 4: For a second order reaction with k = 0.2 M⁻¹min⁻¹, calculate time for 75% completion if [A]₀ = 0.1 M.
Solution:
75% completion → [A]ₜ = 0.25[A]₀ = 0.025 M
Method 1 - Direct calculation:
$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$ $$\frac{1}{0.025} = \frac{1}{0.1} + 0.2t$$ $$40 = 10 + 0.2t$$ $$t = 150 \text{ min}$$Method 2 - Using half-lives:
First half-life:
$$t_1 = \frac{1}{k[A]_0} = \frac{1}{0.2 \times 0.1} = 50 \text{ min}$$Second half-life (when [A] = 0.05 M):
$$t_2 = \frac{1}{k \times 0.05} = \frac{1}{0.2 \times 0.05} = 100 \text{ min}$$Total = 50 + 100 = 150 min
Level 3: JEE Advanced
Problem 5: A substance decomposes by first order kinetics. If 60% decomposes in 60 minutes, calculate: (a) Half-life (b) Time for 90% decomposition (c) Rate constant (d) Time for 99% decomposition
Solution:
(a) 60% decomposed → 40% remains
$$\ln\frac{100}{40} = kt$$ $$\ln 2.5 = k \times 60$$ $$0.916 = 60k$$ $$k = 0.01527 \text{ min}^{-1}$$ $$t_{1/2} = \frac{0.693}{0.01527} = 45.4 \text{ min}$$(b) For 90% decomposition (10% remains):
$$\ln\frac{100}{10} = kt$$ $$\ln 10 = 0.01527 \times t$$ $$2.303 = 0.01527t$$ $$t = 150.8 \text{ min}$$(c) k = 0.01527 min⁻¹ (from part a)
(d) For 99% decomposition (1% remains):
$$\ln\frac{100}{1} = 0.01527t$$ $$\ln 100 = 0.01527t$$ $$4.606 = 0.01527t$$ $$t = 301.6 \text{ min}$$Alternative for parts b and d using half-lives:
(b) 90% decomposition:
$$t = \frac{2.303}{k} = \frac{2.303}{0.01527} \approx 3.32 \times t_{1/2} = 150.8 \text{ min}$$(d) 99% decomposition:
$$t = \frac{2 \times 2.303}{k} \approx 6.64 \times t_{1/2} = 301.5 \text{ min}$$Problem 6: Two substances A and B decompose by first order kinetics. Half-life of A is twice that of B. Calculate the ratio of time taken for 75% decomposition of A to 87.5% decomposition of B.
Solution:
Let t₁/₂(B) = t Then t₁/₂(A) = 2t
For A (75% decomposition = 25% remaining = 2 half-lives):
$$t_A = 2 \times t_{1/2}(A) = 2 \times 2t = 4t$$For B (87.5% decomposition = 12.5% remaining = 3 half-lives):
$$t_B = 3 \times t_{1/2}(B) = 3t$$ $$\frac{t_A}{t_B} = \frac{4t}{3t} = \frac{4}{3}$$Answer: 4:3
Average Life (Mean Life)
Related concept: Average life or mean life (τ)
For first order reactions:
$$\boxed{\tau = \frac{1}{k} = 1.443 \times t_{1/2}}$$Relationship:
$$t_{1/2} = 0.693 \tau$$Physical meaning: Average time a molecule exists before reacting
Used in:
- Radioactive decay (mean lifetime of nucleus)
- Excited state lifetimes in spectroscopy
- Fluorescence decay
Applications of Half-Life
1. Radioactive Dating
C-14 dating:
- Living organisms have constant C-14/C-12 ratio
- After death, C-14 decays (t₁/₂ = 5,730 years)
- Measuring ratio gives age
Example: Dead wood with 1/4 original C-14
- 2 half-lives have passed
- Age = 2 × 5,730 = 11,460 years
2. Drug Dosing
Pharmacokinetics:
- Most drugs follow first-order elimination
- After 5 half-lives: >96% eliminated
- Dosing interval based on t₁/₂
Example: Drug with t₁/₂ = 6 hours
- Dose every 6 hours to maintain therapeutic levels
- Complete elimination in ~30 hours (5 half-lives)
3. Nuclear Medicine
Medical isotopes:
- Short t₁/₂ → quick decay → less radiation exposure
- I-131 (t₁/₂ = 8 days) for thyroid treatment
- Tc-99m (t₁/₂ = 6 hours) for imaging
4. Environmental Science
Pesticide persistence:
- DDT: t₁/₂ ~ 15 years (very persistent)
- Safer alternatives: t₁/₂ ~ days to months
5. Nuclear Waste
Safe disposal:
- Pu-239: t₁/₂ = 24,000 years
- Must be isolated for >10 half-lives (~240,000 years!)
Connection to Other Topics
Link to Integrated Rate Equations
- Half-life derived from integrated rate laws
- See: Integrated Rate Equations
Link to Rate Constant
- k and t₁/₂ are inversely related (for first order)
- See: Rate Law
Link to Order Determination
- Half-life behavior identifies reaction order
- See: Order and Molecularity
Link to Arrhenius Equation
- Temperature affects k, which affects t₁/₂
- See: Arrhenius Equation
JEE Previous Year Questions
JEE Main 2022: The half-life of a first order reaction is 10 minutes. What % of the reactant will remain after 30 minutes?
Solution:
30 min = 3 half-lives
$$\text{Remaining} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125$$% remaining = 12.5%
JEE Advanced 2019: For a first order reaction, time required for 99% completion is n times the time required for 90% completion. Find n.
Solution:
$$t = \frac{2.303}{k}\log\frac{100}{100-\%}$$For 90%:
$$t_{90} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}$$For 99%:
$$t_{99} = \frac{2.303}{k}\log 100 = \frac{2 \times 2.303}{k}$$ $$n = \frac{t_{99}}{t_{90}} = 2$$Answer: n = 2
Quick Revision Points
- Half-life = time for [A] to reduce to [A]₀/2
- Zero order: t₁/₂ ∝ [A]₀ (decreasing successive half-lives)
- First order: t₁/₂ = 0.693/k (constant, independent of [A]₀)
- Second order: t₁/₂ ∝ 1/[A]₀ (increasing successive half-lives)
- After n half-lives: (1/2)ⁿ remains
- First order is unique - only order with constant t₁/₂
- Measuring t₁/₂ at different [A]₀ determines order
- Radioactive decay always first order
Summary
Half-life is a fundamental concept in chemical kinetics with:
- Different formulas for different orders
- Unique behavior for first order (constant t₁/₂)
- Wide applications from archaeology to medicine
- Diagnostic value for determining reaction order
Understanding half-life patterns is essential for:
- Solving JEE kinetics problems
- Interpreting experimental data
- Real-world applications in science and technology
The key is remembering that only first-order reactions have constant half-lives - this single fact solves many JEE problems!
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