NEETAceToday's NEET Daily

Kinetics of iodide ions with hydrogen peroxide at room temperature: NEET MCQs with answers

50 questions on this topic of Principles Related to Practical Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Principles Related to Practical Chemistry freeAll Principles Related to Practical Chemistry questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. In the kinetic experiment of H₂O₂ with iodide ions using the iodine clock method, which of the following expressions correctly represents the relationship between the initial reaction rate and the time (t) taken for the blue colour to appear, given that a constant amount of S₂O₃²⁻ is used across all runs?

  1. Rate ∝ 1/t
  2. Rate ∝ t
  3. Rate ∝ t²
  4. Rate ∝ 1/t²
Show answer and explanation

Answer: (A) Rate ∝ 1/t

In the iodine clock reaction, a constant number of moles of I₂ (equivalent to the fixed amount of Na₂S₂O₃ initially added) is produced before the blue starch-iodine complex appears. Since the amount of product formed during this fixed interval is constant (Δ[I₂] = constant), the initial rate is given by Δ[I₂]/t, which is inversely proportional to time (Rate ∝ 1/t).

medium

2. What will happen if the volume of Na₂S₂O₃ solution added to the reaction mixture is increased by 50% while keeping all other reagent volumes and concentrations constant?

  1. The time required for the blue colour to appear increases by 50%.
  2. The rate of the slow reaction increases by 50%.
  3. The blue colour appears immediately upon mixing.
  4. The rate constant of the reaction decreases by 50%.
Show answer and explanation

Answer: (A) The time required for the blue colour to appear increases by 50%.

The rate of the slow reaction depends only on [H₂O₂] and [I⁻], so the rate remains unchanged. However, a larger amount of S₂O₃²⁻ means more I₂ must be generated before thiosulphate is exhausted; hence, the time interval Δt taken for the blue colour to appear increases proportionally by 50%.

hard

3. In a quantitative clock study of H₂O₂ and I⁻, a student forgets to add dilute H₂SO₄, keeping the solution neutral (pH ≈ 7). What is the expected consequence on the reaction kinetics compared to the acidified medium?

  1. The rate of iodine formation decreases significantly because H⁺ participates in accelerating the rate-determining step.
  2. The reaction is instantly explosive due to rapid catalytic disproportionation of H₂O₂.
  3. The blue colour appears faster because S₂O₃²⁻ decomposes spontaneously under neutral conditions.
  4. Starch hydrolyzes rapidly in neutral medium, completely preventing colour complexation.
Show answer and explanation

Answer: (A) The rate of iodine formation decreases significantly because H⁺ participates in accelerating the rate-determining step.

The reaction between H₂O₂ and I⁻ proceeds much more rapidly in an acidic medium than in neutral or alkaline media. In neutral medium, the rate of oxidation of I⁻ by H₂O₂ is considerably slower; therefore, it takes significantly longer for enough iodine to be generated to consume the thiosulphate and show the blue colour.

medium

4. During the kinetic investigation of the reaction between H₂O₂ and KI, why is the total volume of the reaction mixture kept constant across different runs by adding distilled water?

  1. To ensure the ionic strength and depth of solution remain identical for visual detection while keeping individual concentrations directly proportional to added volumes.
  2. To keep the temperature of the reaction mixture strictly constant via thermal buffering.
  3. To prevent the precipitation of potassium sulfate formed during the reaction.
  4. To dilute the starch indicator so it does not coagulate in acidic medium.
Show answer and explanation

Answer: (A) To ensure the ionic strength and depth of solution remain identical for visual detection while keeping individual concentrations directly proportional to added volumes.

Keeping the total volume constant ensures that the concentration of each reactant is directly proportional to the volume of stock solution pipetted into the flask. It also keeps the liquid depth identical, ensuring reliable visual detection of the blue colour.

easy

5. Why is fresh starch solution preferred over an old starch solution in the iodine clock experiment?

  1. Old starch solution permanently precipitates I⁻ ions.
  2. Old starch decomposes into glucose which reduces H₂O₂.
  3. Old starch reacts rapidly with sodium thiosulphate instead of iodine.
  4. Old starch undergoes bacterial degradation or hydrolysis, losing its sensitivity to I₂.
Show answer and explanation

Answer: (D) Old starch undergoes bacterial degradation or hydrolysis, losing its sensitivity to I₂.

Starch solutions tend to ferment or hydrolyze upon standing over long periods, losing their ability to form the intense blue-coloured inclusion complex with triiodide (I₃⁻) ions. Freshly prepared starch ensures sharp and reproducible detection of free iodine.

hard

6. In an experiment, 5.0 mL of 0.04 M KI, 5.0 mL of 0.002 M Na₂S₂O₃, 2.0 mL of starch, and 3.0 mL of dilute H₂SO₄ are mixed. Then 5.0 mL of 0.02 M H₂O₂ is added to start the reaction (Total volume = 20.0 mL). How many moles of I₂ are formed when the blue colour just appears?

  1. 1.0 × 10⁻⁵ mol
  2. 2.0 × 10⁻⁵ mol
  3. 5.0 × 10⁻⁶ mol
  4. 2.5 × 10⁻⁶ mol
Show answer and explanation

Answer: (C) 5.0 × 10⁻⁶ mol

Moles of S₂O₃²⁻ added = (5.0 mL × 0.002 M) = 1.0 × 10⁻⁵ mol. The fast titration reaction is: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻. Therefore, moles of I₂ required to consume all S₂O₃²⁻ = (1/2) × 1.0 × 10⁻⁵ mol = 5.0 × 10⁻⁶ mol. As soon as this amount of I₂ is formed and reacted, the very next trace of I₂ turns the starch blue.

medium

7. If the concentration of iodide ions is increased while other conditions are kept constant, the blue colour appears:

  1. Later
  2. At the same time
  3. Never
  4. Sooner, because the rate increases
Show answer and explanation

Answer: (D) Sooner, because the rate increases

The rate depends on [I⁻]; increasing it speeds up formation of iodine, so the blue colour appears in less time.

NCERT Class 11

medium

8. Which of the following experimental errors would cause the time taken for the blue colour to appear to be recorded as unexpectedly SHORTER than the true value?

  1. Accidentally adding a higher volume of standard Na₂S₂O₃ solution
  2. Carrying out the reaction at a lower temperature than room temperature
  3. Diluting the mixture with excess distilled water
  4. Presence of a trace amount of Fe²⁺ or Fe³⁺ ions in the reaction mixture
Show answer and explanation

Answer: (D) Presence of a trace amount of Fe²⁺ or Fe³⁺ ions in the reaction mixture

Iron ions (Fe²⁺/Fe³⁺) act as efficient homogeneous catalysts for the oxidation of iodide by hydrogen peroxide. Their presence accelerates the reaction, consuming the thiosulphate faster and shortening the time required for the blue colour to appear.

hard

9. Given the following two steps proposed for the mechanism of the reaction between H₂O₂ and I⁻ in acidic solution: Step 1: H₂O₂ + I⁻ → IO⁻ + H₂O (slow) Step 2: IO⁻ + 2H⁺ + I⁻ → I₂ + H₂O (fast) What is the overall order of the reaction with respect to H⁺ ions under initial rate conditions?

  1. 2
  2. 0
  3. 1
  4. 3
Show answer and explanation

Answer: (B) 0

The rate-determining step is Step 1: Rate = k[H₂O₂][I⁻]. The H⁺ ions only participate in the subsequent fast step (Step 2). Therefore, under conditions where H⁺ is in sufficient supply, the rate does not depend on [H⁺], giving an order of zero with respect to H⁺.

medium

10. In the kinetic experiment between H₂O₂ and I⁻ ions, which of the following represents the correct sequence of events occurring before the blue colour appears?

  1. I⁻ directly reduces starch, which is re-oxidized by H₂O₂.
  2. I₂ reacts with starch first, and then S₂O₃²⁻ decolorizes the complex.
  3. S₂O₃²⁻ reduces H₂O₂ to H₂O, preventing I⁻ oxidation temporarily.
  4. I₂ produced is immediately reduced by S₂O₃²⁻ to I⁻ until S₂O₃²⁻ is exhausted.
Show answer and explanation

Answer: (D) I₂ produced is immediately reduced by S₂O₃²⁻ to I⁻ until S₂O₃²⁻ is exhausted.

As I₂ is generated by the slow reaction of H₂O₂ and I⁻, it reacts extremely rapidly with S₂O₃²⁻: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Only after all the thiosulphate is completely consumed does free iodine accumulate and form the blue complex with starch.

medium

11. A student performs the iodine clock experiment to determine the order of the reaction with respect to H₂O₂. The volume of 0.1 M KI is kept constant at 10 mL, while the volume of 0.05 M H₂O₂ is varied. If the time taken for the blue colour to appear is 60 s with 5 mL of H₂O₂, what will be the approximate time taken when 15 mL of H₂O₂ is used (total volume kept constant with distilled water)?

  1. 180 s
  2. 60 s
  3. 15 s
  4. 20 s
Show answer and explanation

Answer: (D) 20 s

The rate is directly proportional to 1/t. Since the reaction is first order with respect to H₂O₂, tripling the concentration/volume of H₂O₂ (from 5 mL to 15 mL) triples the rate of the reaction. Because Rate ∝ 1/t, the time required will be reduced to one-third of 60 s, which is 20 s.

medium

12. Which chemical species is oxidized during the fast reaction that delays the appearance of the blue colour in the clock experiment?

  1. I⁻
  2. S₂O₃²⁻
  3. I₂
  4. H₂O₂
Show answer and explanation

Answer: (B) S₂O₃²⁻

In the reaction 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻, the oxidation state of sulfur increases from +2 in S₂O₃²⁻ to +2.5 in tetrathionate (S₄O₆²⁻). Thus, thiosulphate (S₂O₃²⁻) is the species undergoing oxidation.

Keep going with Principles Related to Practical Chemistry. All 487 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Principles Related to Practical Chemistry