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. 14Half Life
8 MCQs2 revision cards9-step worked example
Source: NCERT Organic Chemistry — Basic PrinciplesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 (pages 14 and 16) 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).
Practice 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?
MCQ 2Easy RecallPractice
The half-life of a zero-order reaction depends on:
MCQ 3Easy RecallPractice
For which order of reaction is the half-life independent of the initial concentration of the reactant?
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?
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:
MCQ 6Direct ApplicationPractice
A first-order reaction has a half-life of 10 minutes. How much of the original reactant remains after 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?
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?
Quick recall before you leave
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. ---
Before solving, remember these
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. 16Formulas
Arrhenius equation
Temperature dependence of rate constant. Higher Ea → more T-sensitive rate.
| Symbol | Quantity | SI Unit |
|---|---|---|
| A | frequency factor | same as k |
| Ea | activation energy | J/mol |
| R | gas constant | J/mol/K |
| T | temp | K |
Valid when
- T in kelvins
- Most reactions in modest T range
Arrhenius for two temperatures
Compare rate constants at two temperatures to find Ea.
| Symbol | Quantity | SI Unit |
|---|---|---|
| k1, k2 | rate constants | same units |
| T1, T2 | temperatures | K |
| Ea | activation energy | J/mol |
Valid when
- A constant across temperature range
- T in kelvins
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)
Exam Traps & Common Mistakes
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: sign error
Correction
ln(k2/k1) = (Ea/R) × (1/T1 - 1/T2). At higher T2, k2 > k1, so ln(k2/k1) > 0. Need 1/T1 > 1/T2.
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).
Root cause: concept gap
Correction
Rate law from EXPERIMENT, not stoichiometry. Order = sum of exponents in rate = k[A]^x[B]^y. Molecularity follows mechanism.
Past Year Questions
10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1[Co(NH 3 ) 5 Cl]Cl
2[Co(NH 3 ) 3 Cl 3 ]
3[Co(NH ) Cl ]
4[Co(NH ) ]Cl 3 4 2 3 6 3 Answer (2, 3)
NTA Answer: Option multi:2,3(final)
NEET 2025
110 minutes
22 minutes
34 minutes
45 minutes
NTA Answer: Option 1(final)
NEET 2024Revised key
Activation energy of any chemical reaction can be calculated if one knows the value of
1rate constant at standard temperature
2probability of collision
3orientation of reactant molecules during collision
4rate constant at two different temperatures
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
Which reaction is NOT a redox reaction?
1Zn + CuSO → ZnSO + Cu 4 4
22KClO 3 + I 2 → 2KIO 3 + Cl 2
3H + Cl → 2HCl 2 2
4BaCl + Na SO → BaSO + 2NaCl 2 2 4 4
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
138.04 kJ/mol
2380.4 kJ/mol
33.80 kJ/mol
43804 kJ/mol
NTA Answer: Option 1(revised_final)
NEET 2023
Which one is an example of heterogenous catalysis?
1Hydrolysis of sugar catalysed by H+ ions
2Decomposition of ozone in presence of nitrogen monoxide
3Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron
4Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen
NTA Answer: Option 3(final)
NEET 2023
1Increase by a factor of six
2Increase by a factor of nine
3Increase by a factor of three
4Decrease by a factor of nine
NTA Answer: Option 2(final)
NEET 2022
1Statement I is incorrect but Statement II is correct.
2Both Statement I and Statement II are correct.
3Both Statement I and Statement II are incorrect.
4Statement I is correct but Statement II is incorrect.
NTA Answer: Option 4(final)
NEET 2021
1Zone refining
2Electrolysis
3Chromatography
4Distillation
NTA Answer: Option 4(final)
NEET 2021
1–83 kJ mol–1
241.5 kJ mol–1
383.0 kJ mol–1
4166 kJ
NTA Answer: Option 2(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Find Ea from rate-constant ratio at two temperatures using ln(k2/k1) = Ea/R · (1/T1 − 1/T2).
Multi StepMedium
Common distractors
wrong sign of 1 over t
Mixes T1 and T2 in subtraction
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
Test yourself on this topic with real past-paper questions:
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