Formula
First-order kinetics
r = k[A]; integrated: ln[A] = ln[A]₀ - kt. Half-life t_½ = (ln 2)/k = 0.693/k (independent of [A]₀).
-- NCERT Class 12 Chemistry, Ch. 3, p. 73Half Life
8 MCQs6 revision cards9-step worked example
Source: NCERT Chemical KineticsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Half Life, explained for NEET
The trap that costs marks on half-life questions is simple: you apply the first-order half-life formula to a zero-order reaction, or vice versa. Both formulas contain k and produce a time — so under exam pressure, the wrong one feels right until the answer doesn't match any option.
The core distinction. For a first-order reaction, half-life is independent of initial concentration:
t₁/₂ = 0.693 / k
For a zero-order reaction, half-life depends directly on the initial concentration:
t₁/₂ = [A]₀ / (2k)
This means: double the starting concentration of a zero-order reactant and its half-life doubles. Do the same for a first-order reactant and the half-life stays unchanged. NCERT Class 12 Chemistry Chapter 3 (page 76) derives both expressions from their respective integrated rate laws.
Why this matters on NEET. Questions on first-order half-life appear regularly (observed in 2022, 2023, 2024 papers). The standard pattern: given k or t₁/₂, find the other — or find the fraction remaining after n half-lives. The distractor that catches students is the zero-order formula plugged into a first-order problem (or vice versa). The numbers work out to a plausible-looking wrong answer.
Watch-out. When a problem states the order explicitly, use the matching formula. When a problem says "half-life is independent of concentration," that is the fingerprint of first-order kinetics — do not reach for the zero-order expression. When half-life changes with concentration, think zero-order (or second-order: t₁/₂ = 1/(k[A]₀), which is less frequent on NEET but occasionally tested).
Can you answer these Half Life MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The half-life of a first-order reaction is given by which expression?
Show answer and why every option is right or wrong
Answer: D. For a first-order reaction, the integrated rate law ln([A]₀/[A]) = kt yields t₁/₂ = 0.693/k when [A] = [A]₀/2 (NCERT Class 12 Chemistry Chapter 3, page 76). The half-life is independent of initial concentration.
Why A is wrong: A gives the zero-order half-life formula t₁/₂ = [A]₀/(2k). This depends on initial concentration — a hallmark of zero-order, not first-order kinetics (trap: zero-order vs first-order formula swap).
Why B is wrong: B rearranges the zero-order formula incorrectly, placing 2 in the denominator of [A]₀ instead of multiplying k by 2. No standard kinetic order uses this expression.
Why C is wrong: C gives the second-order half-life formula t₁/₂ = 1/(k[A]₀). This also depends on [A]₀, which contradicts the concentration-independent half-life of first-order reactions.
MCQ 2Easy RecallPractice
The half-life of a zero-order reaction depends on:
Show answer and why every option is right or wrong
Answer: C. For zero-order kinetics, t₁/₂ = [A]₀/(2k). The half-life explicitly contains [A]₀, so it depends on the initial concentration of the reactant (NCERT Class 12 Chemistry Chapter 3, page 76).
Why A is wrong: A is incomplete. While temperature affects k (and thus half-life indirectly), the zero-order half-life formula t₁/₂ = [A]₀/(2k) shows direct dependence on [A]₀ — the more specific and correct answer (trap: confusing first-order independence with zero-order).
Why B is wrong: B describes first-order kinetics where t₁/₂ = 0.693/k depends only on k. Zero-order half-life requires both [A]₀ and k (trap: applying first-order half-life logic to zero-order).
Why D is wrong: D is irrelevant. A catalyst changes the rate constant k but is not a variable in the half-life formula. The question asks what the half-life depends on as a mathematical function.
MCQ 3Easy RecallPractice
For which order of reaction is the half-life independent of the initial concentration of the reactant?
Show answer and why every option is right or wrong
Answer: A. First-order half-life t₁/₂ = 0.693/k contains no [A]₀ term, making it independent of initial concentration (NCERT Class 12 Chemistry Chapter 3, page 76).
Why B is wrong: B is incorrect. Zero-order half-life is t₁/₂ = [A]₀/(2k), which directly depends on [A]₀ — halve the starting concentration and the half-life halves (trap: zero-order vs first-order half-life dependence).
Why C is wrong: C is incorrect. Second-order half-life is t₁/₂ = 1/(k[A]₀), which is inversely proportional to [A]₀.
Why D is wrong: D is incorrect. Third-order half-life is t₁/₂ = 3/(2k[A]₀²), showing an even stronger dependence on initial concentration.
MCQ 4Direct ApplicationPractice
A first-order reaction has a rate constant of 3.465 × 10⁻² s⁻¹. What is the half-life of this reaction?
Show answer and why every option is right or wrong
Answer: D. t₁/₂ = 0.693 / k = 0.693 / (3.465 × 10⁻²) = 20 s (NCERT Class 12 Chemistry Chapter 3, page 76).
Why A is wrong: A results from using 0.693/(2k) = 0.693/(2 × 3.465 × 10⁻²) = 10 s, which incorrectly introduces an extra factor of 2 — possibly from confusing with the zero-order formula structure (trap: formula swap).
Why B is wrong: B results from using [A]₀/(2k) with an assumed [A]₀ value, applying the zero-order formula to a first-order reaction (trap: zero-order vs first-order formula swap).
Why C is wrong: C results from using 1/k = 1/(3.465 × 10⁻²) ≈ 28.9 s, rounded to 30 — this gives the mean lifetime, not the half-life.
MCQ 5Direct ApplicationPractice
A zero-order reaction has a rate constant k = 0.010 mol L⁻¹ s⁻¹. If the initial concentration is 0.40 mol L⁻¹, the half-life is:
Show answer and why every option is right or wrong
Answer: B. For zero-order: t₁/₂ = [A]₀/(2k) = 0.40/(2 × 0.010) = 0.40/0.020 = 20 s (NCERT Class 12 Chemistry Chapter 3, page 76).
Why A is wrong: A results from putting 4k in the denominator (0.40/0.040 = 10 s), doubling the factor of 2 instead of using 2k.
Why C is wrong: C results from using [A]₀/k = 0.40/0.010 = 40 s, which gives the total time for the reaction to go to completion (zero-order), not the half-life.
Why D is wrong: D results from applying the first-order formula: 0.693/0.010 = 69.3 s, ignoring that this is a zero-order reaction (trap: zero-order vs first-order formula swap).
MCQ 6Direct ApplicationPractice
A first-order reaction has a half-life of 10 minutes. How much of the original reactant remains after 30 minutes?
Show answer and why every option is right or wrong
Answer: B. After n half-lives, the fraction remaining is (1/2)ⁿ. Here n = 30/10 = 3 half-lives. Fraction remaining = (1/2)³ = 1/8 (NCERT Class 12 Chemistry Chapter 3, page 76). First-order half-life is constant, so each 10-minute interval halves the concentration.
Why A is wrong: A corresponds to only 1 half-life (10 min), not 3. The question specifies 30 minutes, which is 3 half-lives.
Why C is wrong: C corresponds to 2 half-lives (20 min). After 30 min = 3 half-lives, one more halving occurs.
Why D is wrong: D corresponds to 4 half-lives (40 min), overshooting the given time of 30 minutes.
MCQ 7CalculationPractice
For a zero-order reaction with k = 5.0 × 10⁻³ mol L⁻¹ s⁻¹, the half-life is found to be 50 s. If the initial concentration is doubled, what is the new half-life?
Show answer and why every option is right or wrong
Answer: C. Step 1: From the original data, [A]₀ = 2k × t₁/₂ = 2 × 5.0 × 10⁻³ × 50 = 0.50 mol L⁻¹. Step 2: New [A]₀ = 2 × 0.50 = 1.0 mol L⁻¹. Step 3: New t₁/₂ = [A]₀,new/(2k) = 1.0/(2 × 5.0 × 10⁻³) = 100 s (NCERT Class 12 Chemistry Chapter 3, page 76). Zero-order half-life is directly proportional to initial concentration.
Why A is wrong: A assumes half-life halves when concentration doubles. That would describe second-order kinetics (t₁/₂ = 1/(k[A]₀)), not zero-order.
Why B is wrong: B assumes half-life stays constant when concentration changes — this is the first-order behavior (t₁/₂ = 0.693/k), not zero-order (trap: applying first-order logic to zero-order reaction).
Why D is wrong: D assumes half-life quadruples when concentration doubles. No standard kinetic order produces this relationship.
MCQ 8CalculationPractice
A first-order reaction has a half-life of 20 minutes. What is the time required for 75% of the reactant to decompose?
Show answer and why every option is right or wrong
Answer: A. Step 1: 75% decomposed means 25% remains, i.e., [A] = [A]₀/4 = [A]₀ × (1/2)². Step 2: This corresponds to n = 2 half-lives. Step 3: t = 2 × 20 = 40 min. Alternatively, k = 0.693/20 min⁻¹, then t = (1/k) × ln([A]₀/[A]) = (20/0.693) × ln(4) = (20/0.693) × 1.386 = 40 min (NCERT Class 12 Chemistry Chapter 3, page 76).
Why B is wrong: B (30 min) would be 1.5 half-lives, leaving ~35% of reactant — more than the 25% target. This error arises from incorrectly averaging one and two half-lives.
Why C is wrong: C (20 min) corresponds to 1 half-life, where only 50% has decomposed — not the required 75%.
Why D is wrong: D (60 min) corresponds to 3 half-lives, where 87.5% has decomposed — overshooting the 75% target.
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Half Life: quick recall before you leave
How do you solve a Half Life question? A worked example
- 1
Given
A first-order reaction has a rate constant k = 1.386 × 10⁻² min⁻¹.
- 2
Required
(a) Find the half-life.
(b) Find the fraction of reactant remaining after 200 min. - 3
Concept
For first-order kinetics, half-life is independent of initial concentration. The fraction remaining after time t is found using the integrated rate law: [A]/[A]₀ = e^(−kt), or equivalently, the number of half-lives elapsed gives (1/2)ⁿ.
- 4
Formula
t₁/₂ = 0.693 / k
Fraction remaining = (1/2)^(t / t₁/₂) - 5
Substitution
(a) t₁/₂ = 0.693 / (1.386 × 10⁻²)
(b) n = t / t₁/₂ = 200 / t₁/₂ - 6
Calculation
(a) t₁/₂ = 0.693 / 0.01386 = 50.0 min
(b) n = 200 / 50.0 = 4.00 half-lives
Fraction remaining = (1/2)⁴ = 1/16 = 0.0625
Note on exact values: the number 200 and the exponent 4 are exact counting values. The value 0.693 is ln 2 rounded to three significant figures — this is the precision-limiting factor. - 7
Final answer
(a) t₁/₂ = 50.0 min
(b) Fraction remaining after 200 min = 1/16 (6.25%)
The answer is reported to 3 significant figures, consistent with the precision of k. - 8
Common trap
Using the zero-order formula t₁/₂ = [A]₀/(2k) would give a concentration-dependent half-life and a completely different remaining fraction. The problem states first-order — use t₁/₂ = 0.693/k. If you catch yourself needing [A]₀ to calculate the half-life of a first-order reaction, you have picked the wrong formula.
- 9
Similar NEET-style question
A first-order reaction is 87.5% complete in 60 minutes. What is the half-life?
Approach: 87.5% complete → 12.5% remaining → (1/2)³ = 1/8. So 3 half-lives = 60 min → t₁/₂ = 20 min.
---
What to remember before solving Half Life questions
Formula
Zero-order kinetics
r = k (independent of [A]); integrated: [A] = [A]₀ - kt. Half-life t_½ = [A]₀/(2k) (depends on [A]₀).
-- NCERT Class 12 Chemistry, Ch. 3, p. 76Which Half Life formulas do you need for NEET?
First-order kinetics
Concentration decays exponentially. Half-life independent of [A]_0.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [A] | conc at time t | mol/L |
| k | rate constant | 1/s |
| t | time | s |
Valid when
- First-order reaction (rate = k[A])
Zero-order kinetics
Concentration decays linearly. Half-life depends on initial concentration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [A]_0 | initial conc | mol/L |
| k | rate constant | mol/L/s |
| t | time | s |
Valid when
- Zero-order reaction (rate = k, no concentration dependence)
Where do students lose marks on Half Life?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Zero-order t_1/2 depends on [A]_0. First-order t_1/2 INDEPENDENT of [A]_0. Student uses wrong formula.
When it triggers
Half-life question with order specified.
How to avoid
1st order: t_1/2 = 0.693/k (constant). Zero order: t_1/2 = [A]_0/(2k) (varies with initial conc). Second order: t_1/2 = 1/(k[A]_0).
Root cause: formula misuse
Correction
First order: t_1/2 = 0.693/k (constant). Zero order: t_1/2 = [A]_0/(2k) (depends on [A]_0). Second order: t_1/2 = 1/(k[A]_0).
Root cause: formula misuse
Correction
Zero-order: t_1/2 = [A]_0/(2k) (depends on initial conc). First-order: t_1/2 = 0.693/k (independent of [A]_0).
More in Chemical Kinetics: 2 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Half Life questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys.
NEET 2025
110 minutes
22 minutes
34 minutes
45 minutes
NTA Answer: Option 1(final)
How does NEET ask about Half Life?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
First-order: t_1/2 = 0.693/k. Apply or invert to find k or remaining concentration.
Direct ApplicationEasy
Common distractors
uses zero order formula
Plugs into [A]_0/(2k) wrong-order formula
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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