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. 14Zero First Order Kinetics
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 zero- and first-order kinetics questions is formula swapping: applying the first-order half-life formula to a zero-order reaction, or vice versa. The two formulas look superficially similar but encode fundamentally different concentration-time behaviours.
Zero-order reactions have a rate that does not depend on reactant concentration: rate = k. The integrated rate law is linear:
[A] = [A]₀ − kt
Half-life: t₁/₂ = [A]₀ / (2k). This depends on initial concentration — halve [A]₀ and the half-life halves too.
First-order reactions have rate = k[A]. The integrated rate law is logarithmic:
ln([A]₀ / [A]) = kt
Half-life: t₁/₂ = 0.693 / k. This is independent of [A]₀ — every successive half-life is the same duration regardless of starting concentration.
The NCERT Class 12 Chemistry Chapter 3 (pages 14–16) derives both integrated laws and their half-life expressions. The key diagnostic: if a problem states the half-life changes when you change [A]₀, the reaction is NOT first-order.
Concentration-time graph diagnostic:
- Zero-order → straight-line plot of [A] vs t (slope = −k).
- First-order → straight-line plot of ln[A] vs t (slope = −k).
Plotting [A] vs t for a first-order reaction gives a curve, not a line — a common confusion in graph-based questions.
Watch-out: When a question gives successive half-lives and each one is shorter than the last, the reaction is zero-order (because [A]₀ keeps shrinking, so t₁/₂ = [A]₀/(2k) shrinks). If successive half-lives are identical, the reaction is first-order. NEET uses this diagnostic regularly.
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
For a first-order reaction, the half-life is:
MCQ 2Easy RecallPractice
The units of the rate constant for a zero-order reaction are:
MCQ 3Easy RecallPractice
For a zero-order reaction, a plot of [A] versus time gives:
MCQ 4Direct ApplicationPractice
A first-order reaction has a rate constant k = 6.93 × 10⁻³ s⁻¹. What is the half-life of this reaction?
MCQ 5Direct ApplicationPractice
A zero-order reaction has k = 2.0 × 10⁻² mol L⁻¹ s⁻¹ and [A]₀ = 0.40 mol L⁻¹. What is the half-life?
MCQ 6Direct ApplicationPractice
For a first-order reaction, if 75% of the reactant is consumed in 32 minutes, the half-life of the reaction is:
MCQ 7CalculationPractice
A zero-order reaction has an initial concentration [A]₀ = 0.10 mol L⁻¹ and k = 5.0 × 10⁻³ mol L⁻¹ min⁻¹. After the first half-life elapses, what is the second half-life?
MCQ 8Concept TrapPractice
A reaction's half-life is observed to remain constant at 20 minutes regardless of how much reactant is initially present. Which statement is correct?
Quick recall before you leave
Worked Example
Pattern: First-order half-life application (NEET pattern: first order half life — observed in NEET 2022, 2023, 2024)
- 1
Given
- Reaction order: first-order - t₁/₂ = 40 min - Target: [A] = 0.125 [A]₀ (i.e., 12.5% remaining)
- 2
Required
Time *t* for concentration to reach 12.5% of [A]₀.
- 3
Concept
For first-order kinetics, each half-life reduces the concentration by half. Since the half-life is constant (independent of [A]₀), we can count how many half-lives bring [A]₀ down to the target fraction.
- 4
Formula
ln([A]₀/[A]) = kt, where k = 0.693/t₁/₂ Alternatively: after *n* half-lives, [A] = [A]₀ / 2ⁿ.
- 5
Substitution
[A]₀ / 2ⁿ = 0.125 [A]₀ 1/2ⁿ = 1/8 2ⁿ = 8 n = 3
- 6
Calculation
t = n × t₁/₂ = 3 × 40 = 120 min Note on exact values: the number 3 (half-life count) and the fraction 1/8 are exact integers/fractions. The half-life of 40 min is a given exact value. These do not limit significant figures in the answer.
- 7
Final answer
t = 120 min (or 2.0 hours)
- 8
Common trap
Applying the zero-order formula here would give a different (incorrect) answer because t₁/₂ would change with each successive period. If you mistakenly used [A] = [A]₀ − kt with constant k, you would get a linear decay to 12.5%, reaching it at a different time. The diagnostic: the problem states "first-order," so t₁/₂ is constant and the 2ⁿ method applies.
- 9
Similar NEET-style question
"A first-order reaction is 87.5% complete in 60 minutes. Calculate the half-life of the reaction." (Answer: 87.5% complete → 12.5% remaining → 3 half-lives → t₁/₂ = 60/3 = 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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