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. 73Zero First Order Kinetics
8 MCQs4 revision cards9-step worked example
Source: NCERT Chemical KineticsPYQ coverage: NEET 2020, 2022, 2025, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Zero First Order Kinetics, explained for NEET
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 74–76) 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.
Can you answer these Zero First Order Kinetics 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:
Show answer and why every option is right or wrong
Answer: B. For first-order reactions, t₁/₂ = 0.693/k. The expression contains no [A]₀ term, so the half-life is independent of initial concentration (NCERT Class 12 Chemistry Chapter 3, page 74).
Why A is wrong: A describes the zero-order half-life relationship (t₁/₂ = [A]₀/(2k)), not first-order (trap: swapping zero-order and first-order half-life formulas).
Why C is wrong: C describes the second-order half-life relationship (t₁/₂ = 1/(k[A]₀)), not first-order (trap: wrong order formula).
Why D is wrong: D does not match any standard reaction order half-life expression. No common kinetic order yields t₁/₂ ∝ [A]₀² (trap: fabricated relationship).
MCQ 2Easy RecallPractice
The units of the rate constant for a zero-order reaction are:
Show answer and why every option is right or wrong
Answer: A. For zero-order, rate = k. Since rate has units mol L⁻¹ s⁻¹, k must have the same units (NCERT Class 12 Chemistry Chapter 3, page 71).
Why B is wrong: B (s⁻¹) is the unit of the first-order rate constant, not zero-order (trap: confusing the two orders' k-units).
Why C is wrong: C (L mol⁻¹ s⁻¹) is the unit of a second-order rate constant, not zero-order (trap: wrong order).
Why D is wrong: D is not a standard rate constant unit for any common reaction order (trap: fabricated unit combination).
MCQ 3Easy RecallPractice
For a zero-order reaction, a plot of [A] versus time gives:
Show answer and why every option is right or wrong
Answer: B. The zero-order integrated rate law is [A] = [A]₀ − kt, which is linear in t with slope −k and intercept [A]₀ (NCERT Class 12 Chemistry Chapter 3, page 72).
Why A is wrong: A describes the [A] vs t plot for a first-order reaction (exponential decay appears as a curve with decreasing slope), not zero-order (trap: swapping graph shapes between orders).
Why C is wrong: C has the wrong sign. Since concentration decreases with time, the slope must be negative (−k), not positive (trap: sign error in the integrated law).
Why D is wrong: D describes the shape of [A] vs t for first-order kinetics, not zero-order. Zero-order decay is linear, not exponential (trap: applying first-order behaviour to zero-order).
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?
Show answer and why every option is right or wrong
Answer: D. t₁/₂ = 0.693 / k = 0.693 / (6.93 × 10⁻³) = 100 s (NCERT Class 12 Chemistry Chapter 3, page 74).
Why A is wrong: A (10 s) results from dividing 0.0693 by k instead of 0.693, a factor-of-ten arithmetic slip (trap: decimal place error in the 0.693 numerator).
Why B is wrong: B (50 s) could arise from using the zero-order formula t₁/₂ = [A]₀/(2k) with some assumed [A]₀ value, rather than the first-order formula (trap: wrong order formula).
Why C is wrong: C (1000 s) results from using 6.93 instead of 0.693 in the numerator, or equivalently from a power-of-ten error in k (trap: magnitude error).
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?
Show answer and why every option is right or wrong
Answer: C. t₁/₂ = [A]₀ / (2k) = 0.40 / (2 × 2.0 × 10⁻²) = 0.40 / 0.040 = 10.0 s (NCERT Class 12 Chemistry Chapter 3, page 76).
Why A is wrong: A (5.0 s) doubles the factor of 2 in the denominator: [A]₀/(4k) = 0.40/0.080 = 5.0 s.
Why B is wrong: B (20.0 s) forgets the factor of 2: [A]₀/k = 0.40/0.020 = 20.0 s.
Why D is wrong: D (34.65 s) results from using the first-order formula t₁/₂ = 0.693/k = 0.693/0.02 = 34.65 s, ignoring that the reaction is zero-order (trap: applying first-order half-life to a zero-order reaction).
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:
Show answer and why every option is right or wrong
Answer: D. 75% consumed means 25% remains, so [A] = [A]₀/4. For first-order: kt = ln([A]₀/[A]) = ln 4 = 2 ln 2. So t = 2 × t₁/₂ = 32 min, giving t₁/₂ = 16 min (NCERT Class 12 Chemistry Chapter 3, page 74).
Why A is wrong: A (8 min) results from dividing 32 by 4 (the fraction denominator), which has no basis in the kinetic equations (trap: numerically dividing time by the dilution factor instead of using the logarithmic relationship).
Why B is wrong: B (32 min) treats the total time as the half-life, ignoring that 75% consumption corresponds to two half-lives, not one (trap: confusing total reaction time with half-life).
Why C is wrong: C (24 min) might come from subtracting 8 from 32 or some other arithmetic guess rather than applying the first-order integrated law (trap: arbitrary arithmetic without applying the correct formula).
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?
Show answer and why every option is right or wrong
Answer: A. First t₁/₂ = [A]₀/(2k) = 0.10/(2 × 5.0 × 10⁻³) = 10.0 min. After the first half-life, [A] = 0.050 mol L⁻¹. The second half-life uses this new concentration: t₁/₂(2nd) = 0.050/(2 × 5.0 × 10⁻³) = 5.0 min (NCERT Class 12 Chemistry Chapter 3, page 76).
Why B is wrong: B (10.0 min) assumes the half-life stays constant, which is true for first-order reactions but NOT for zero-order. For zero-order, t₁/₂ depends on the current concentration, which has halved (trap: applying first-order constant half-life logic to zero-order).
Why C is wrong: C (2.5 min) results from halving the second half-life again, as if calculating a third half-life instead of the second (trap: one extra halving step).
Why D is wrong: D (20.0 min) doubles the first half-life, implying concentration increased, which contradicts the reaction consuming reactant (trap: arithmetic inversion).
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?
Show answer and why every option is right or wrong
Answer: C. A constant half-life independent of initial concentration is the defining characteristic of first-order kinetics, where t₁/₂ = 0.693/k (NCERT Class 12 Chemistry Chapter 3, page 74). Zero-order and second-order half-lives both depend on [A]₀.
Why A is wrong: A is wrong because zero-order half-life is t₁/₂ = [A]₀/(2k), which changes when [A]₀ changes. A constant half-life rules out zero-order (trap: confusing zero-order and first-order half-life dependence on concentration).
Why B is wrong: B is wrong because second-order half-life is t₁/₂ = 1/(k[A]₀), which is inversely proportional to [A]₀ and therefore not constant (trap: wrong order assignment).
Why D is wrong: D is wrong because the concentration-independence of half-life uniquely identifies first-order kinetics. Half-life behaviour is a standard diagnostic for reaction order (trap: unnecessary caution — this is a well-defined diagnostic).
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Zero First Order Kinetics: quick recall before you leave
How do you solve a Zero First Order Kinetics question? A 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.)
---
What to remember before solving Zero First Order Kinetics 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 Zero First Order Kinetics 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 Zero First Order Kinetics?
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.
Zero First Order Kinetics questions from past NEET papers
4 questions from NEET 2020, 2022, 2025, 2026. Answers verified against NTA official keys.
NEET 2026
11.5 min
20.75 min
31.0 min
42.0 min
NTA Answer: Option 2(final)
NEET 2025
121.0 s
269.3 s
323.1 s
4210 s
NTA Answer: Option 2(final)
NEET 2022
10.2303
21.3818
30.9212
40.4606
NTA Answer: Option 3(final)
NEET 2020
11000 s
2100 s
3200 s
4500 s
NTA Answer: Option 4(final)
How does NEET ask about Zero First Order Kinetics?
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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