Reaction: H2O2 + 2I⁻ + 2H⁺ → I2 + 2H2O. Rate = k[H2O2][I⁻] (overall second order). Use clock reaction (with thiosulfate + starch) — appearance of blue colour times the reaction. Vary [I⁻] and [H2O2] separately to extract rate orders by initial-rates method.
-- NCERT Class 12 Chemistry, Ch. 3, p. 116Kinetic Iodide H2o2
8 MCQs2 revision cards9-step worked example
Source: NCERT Unit 20PYQ coverage: NEET 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap first: In the iodide–hydrogen peroxide kinetic study, aspirants frequently overshoot the end-point by adding thiosulfate past the starch-iodine colour discharge — confusing a transient fade with the true equivalence. This single error invalidates the entire rate measurement because the recorded volume no longer corresponds to the iodine actually liberated in that time interval.
The reaction system: Hydrogen peroxide oxidises iodide in acidic medium:
H₂O₂ + 2 KI + H₂SO₄ → I₂ + K₂SO₄ + 2 H₂O
The liberated iodine is titrated against standard sodium thiosulfate:
I₂ + 2 Na₂S₂O₃ → 2 NaI + Na₂S₄O₆
Starch indicator turns blue-black with free I₂; the end-point is the first persistent discharge of this blue-black colour (NCERT Class 12 Chemistry Chapter 3, page 116).
Kinetic measurement: By withdrawing aliquots at fixed time intervals and titrating liberated I₂, you obtain [I₂] vs time data. The volume of thiosulfate consumed is directly proportional to [I₂] formed — and hence to the extent of reaction at that instant. The rate law is determined by varying initial [H₂O₂] or [KI] across separate runs.
Indicator and end-point discipline: Starch is added near the end-point (not at the start — early addition causes adsorption of I₂ on starch granules, giving sluggish colour change). The end-point is the first persistent colourless state lasting ≥30 seconds. If colour returns on standing, the titration has not overshot — the solution was simply not mixed.
Watch-out for NEET: Questions test whether you can identify the correct indicator, state the end-point criterion, or calculate molarity of H₂O₂ from thiosulfate volume using stoichiometric ratios. The 2:1 mole ratio (2 mol Na₂S₂O₃ per mol I₂) and the 1:1 ratio (1 mol H₂O₂ per mol I₂) are the conversion links 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
In the kinetic study of iodide–H₂O₂ reaction, what indicator is used and when is it added?
MCQ 2Easy RecallPractice
The end-point of the thiosulfate–iodine titration in this experiment is identified by:
MCQ 3Easy RecallPractice
In the reaction H₂O₂ + 2 KI + H₂SO₄ → I₂ + K₂SO₄ + 2 H₂O, the mole ratio of H₂O₂ to I₂ liberated is:
MCQ 4Direct ApplicationPractice
20.0 mL of an H₂O₂ solution liberates I₂ that requires 25.0 mL of 0.10 N Na₂S₂O₃. The normality of H₂O₂ is:
MCQ 5Direct ApplicationPractice
In the iodometric back-titration, 1 mol I₂ reacts with Na₂S₂O₃ in the mole ratio:
MCQ 6Direct ApplicationPractice
A student adds starch indicator at the very beginning of the thiosulfate titration. The most likely consequence is:
MCQ 7CalculationPractice
In the kinetic study, aliquots are withdrawn at 5, 10, 15, and 20 minutes. If the thiosulfate volumes consumed are 8.0, 14.5, 19.8, and 24.0 mL respectively, the rate of I₂ liberation is best described as:
MCQ 8Concept TrapPractice
A student performing the iodide–H₂O₂ kinetic experiment records the end-point when the blue-black colour fades momentarily on swirling but returns within 10 seconds. The student's titre value will be:
Quick recall before you leave
Worked Example
- 1
Given
In a kinetic run, a 10.0 mL aliquot of the reaction mixture (containing liberated I₂) is titrated with 0.050 M Na₂S₂O₃. The burette reading at end-point is 18.0 mL.
- 2
Required
Concentration of I₂ in the aliquot (in mol/L).
- 3
Concept
The titration reaction I₂ + 2 Na₂S₂O₃ → 2 NaI + Na₂S₄O₆ shows that 1 mol I₂ ≡ 2 mol Na₂S₂O₃. Using the molarity-stoichiometry relationship with n(I₂) = 1 and n(Na₂S₂O₃) = 2.
- 4
Formula
M(I₂) × V(I₂) / 1 = M(Na₂S₂O₃) × V(Na₂S₂O₃) / 2
- 5
Substitution
M(I₂) × 10.0 / 1 = 0.050 × 18.0 / 2
- 6
Calculation
M(I₂) × 10.0 = 0.45 M(I₂) = 0.45 / 10.0 = 0.045 mol/L Note on exact constants: the stoichiometric coefficients 1 and 2 are exact integers from the balanced equation and do not limit significant figures.
- 7
Final answer
[I₂] = 4.5 × 10⁻² mol/L (2 significant figures, limited by 0.050 M).
- 8
Common trap
Forgetting the factor of 2 — i.e., using M(I₂) = M(thio) × V(thio) / V(I₂) directly without dividing by the stoichiometric coefficient. This gives 0.090 mol/L (double the true value). This is the most common computational error in iodometric problems.
- 9
Similar NEET-style question
"25.0 mL of a solution containing dissolved I₂ is titrated against 0.020 M Na₂S₂O₃. If 30.0 mL of thiosulfate is consumed, what is the molarity of the I₂ solution?" (Answer: M(I₂) = 0.020 × 30.0 / (2 × 25.0) = 0.012 M) ---
Before solving, remember these
Formulas
Molarity-stoichiometry titration
Use when normality is awkward (e.g., diprotic acids). Stoichiometric coefficients from balanced equation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| V | volume | L |
| n | coefficient | - |
Valid when
- Balanced equation known
- Same end-point
Normality equation in titration
Equivalents of acid = equivalents of base at end-point. Or for redox: equivalents of oxidant = equivalents of reductant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | normality | eq/L |
| V | volume | mL or L |
Valid when
- Same titration end-point
- Equivalent factors known
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: Inorganic Exception
Cations like Pb²⁺ precipitate in BOTH Group I (with HCl) and Group II (with H2S) — assigning to only one group misses the redundancy.
When it triggers
Cation that appears in two analytical groups, e.g. Pb²⁺ (Group I + Group II) or Hg²⁺ vs Hg2²⁺.
How to avoid
Apply confirmatory tests for each candidate group; do not assume mutual exclusivity.
Category: Overthinking
Continuing to add titrant past the first persistent colour change because the colour seemed to fade after a swirl.
When it triggers
Question describes 'colour faded after swirling' or 'persistent colour' — distinguishes transient vs end-point.
How to avoid
End-point = first PERSISTENT colour change (lasts ≥30 s). Transient fades back to original on swirling.
Category: Similar Terms
Phenolphthalein (pH 8.2–10) and methyl orange (pH 3.1–4.4) only mark equivalence when the eq-pt pH falls within their range; using the wrong indicator gives an end-point that disagrees with the actual equivalence point.
When it triggers
Titration prompt mentions a specific weak/strong combination but asks which indicator is suitable.
How to avoid
Match the indicator's pH-change range to the equivalence-point pH: phenolphthalein for eq-pt > 7, methyl orange for eq-pt < 7.
Root cause: concept gap
Correction
Pb²⁺ appears in both Group I (PbCl2 white ppt, soluble in hot water) and Group II (PbS black ppt). Confirm with hot-water solubility test.
Root cause: concept gap
Correction
Strong-acid + strong-base: any indicator (eq pt = 7). Weak-acid + strong-base: phenolphthalein (eq pt > 7). Strong-acid + weak-base: methyl orange (eq pt < 7).
Root cause: rushed under time pressure
Correction
Slow drop-wise addition near the end-point; first persistent colour change is the end-point. Re-do if overshot.
Past Year Questions
3 questions from NEET 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1A-III, B-II, C-I, D-IV
2A-III, B-IV, C-II, D-I
3A-III, B-IV, C-I, D-II
4A-III, B-II, C-IV, D-I
NTA Answer: Option 3(final)
NEET 2024Revised key
1dilute hydrochloric acid
2concentrated sulphuric acid
3dilute nitric acid
4dilute sulphuric acid
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
1B, A, D, C, E
2B, C, A, D, E
3E, C, D, B, A
4E, A, B, C, D
NTA Answer: Option 1(revised_final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Practical chemistry — titrimetric end-point selection, indicator–pH matching, qualitative-analysis cation/anion identification, organic functional-group tests.
Direct ApplicationMedium
Common distractors
wrong zero error direction
Sign convention is easily flipped.
Sources
NCERT refs: Class 12 Chemistry Chapter 3, p.116
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