Rate of Reaction and Rate Expression Rate of a chemical reaction, factors affecting the rate of reactions – concentration: NEET MCQs with Answers

15 free questions on Rate of Reaction and Rate Expression Rate of a chemical reaction, factors affecting the rate of reactions – concentration for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: concentration, temperature, pressure, and catalyst; elementary and complex reactions,

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Question 1

Previous-year question
An increase in the concentration of the reactants of a reaction leads to a change in [NEET 2020]
  1. heat of reaction
  2. threshold energy
  3. collision frequency
  4. activation energy
Show answer
Answer: (C) collision frequency
Explanation: Higher concentration means more molecules per unit volume and hence more collisions per second; the energy terms (ΔH, threshold energy, Eₐ) are unaffected.

Question 2

Previous-year question
For the chemical reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the correct option is [NEET 2019]
  1. –(1/3)d[H₂]/dt = –(1/2)d[NH₃]/dt
  2. –d[N₂]/dt = 2d[NH₃]/dt
  3. –d[N₂]/dt = (1/2)d[NH₃]/dt
  4. 3d[H₂]/dt = 2d[NH₃]/dt
Show answer
Answer: (C) –d[N₂]/dt = (1/2)d[NH₃]/dt
Explanation: Rate = –d[N₂]/dt = –(1/3)d[H₂]/dt = +(1/2)d[NH₃]/dt, hence –d[N₂]/dt = (1/2)d[NH₃]/dt.

Question 3

On a curve of reactant concentration [R] against time t, how are the average rate between two times and the instantaneous rate at one time obtained?
  1. Average rate is the area under the curve between the two times; instantaneous rate is the height of the curve at that time
  2. Average rate is the height of the curve at the middle of the interval; instantaneous rate is the magnitude of the slope of the chord from that time to the end of the reaction
  3. Average rate is the magnitude of the slope of the tangent at the start of the interval; instantaneous rate is the magnitude of the slope of the chord from t = 0 to that time
  4. Average rate is the magnitude of the slope of the chord joining the two points; instantaneous rate is the magnitude of the slope of the tangent at that time
Show answer
Answer: (D) Average rate is the magnitude of the slope of the chord joining the two points; instantaneous rate is the magnitude of the slope of the tangent at that time
Explanation: Average rate is −Δ[R]/Δt, which is the slope of the straight line joining the two points on the curve, taken without its sign. As Δt is made very small, this chord turns into the tangent at the point. The instantaneous rate, −d[R]/dt, is therefore the magnitude of the tangent's slope.

Question 4

For 2N₂O₅(g) → 4NO₂(g) + O₂(g), the tangent to the curve of [NO₂] against time at a certain instant has a slope of +4.0 × 10⁻⁴ M s⁻¹. What is the slope of the tangent to the curve of [N₂O₅] against time at the same instant?
  1. −2.0 × 10⁻⁴ M s⁻¹
  2. −1.0 × 10⁻⁴ M s⁻¹
  3. +2.0 × 10⁻⁴ M s⁻¹
  4. −8.0 × 10⁻⁴ M s⁻¹
Show answer
Answer: (A) −2.0 × 10⁻⁴ M s⁻¹
Explanation: The tangent slopes are d[NO₂]/dt and d[N₂O₅]/dt, and the rate of reaction is −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt. The rate is therefore 4.0 × 10⁻⁴/4 = 1.0 × 10⁻⁴ M s⁻¹, which gives d[N₂O₅]/dt = −2 × 1.0 × 10⁻⁴ = −2.0 × 10⁻⁴ M s⁻¹. The slope is negative because the reactant concentration falls with time.

Question 5

For R → P, a curve of [R] against time is plotted. The tangent drawn to the curve at t = 40 s cuts the concentration axis at 0.80 M and the time axis at 160 s. What is the instantaneous rate of the reaction at t = 40 s?
  1. 4.0 × 10⁻³ M s⁻¹
  2. 6.7 × 10⁻³ M s⁻¹
  3. 5.0 × 10⁻³ M s⁻¹
  4. 2.0 × 10⁻² M s⁻¹
Show answer
Answer: (C) 5.0 × 10⁻³ M s⁻¹
Explanation: The instantaneous rate at a given time is the magnitude of the slope of the tangent to the concentration–time curve at that point. The tangent falls from 0.80 M at t = 0 to zero at t = 160 s, so its slope is −0.80/160 = −5.0 × 10⁻³ M s⁻¹. The intercepts belong to the tangent line, so the 40 s is used only to locate the point of contact.

Question 6

In the hydrolysis of an alkyl chloride at a fixed temperature, the reactant concentration is 0.200 M at 0 s, 0.170 M at 40 s, 0.146 M at 80 s and 0.128 M at 120 s. Which conclusion is correct?
  1. The average rate is 6.0 × 10⁻⁴ M/s over each of the three 40 s intervals
  2. The instantaneous rate at 0 s equals the average rate over 120 s, which is 6.0 × 10⁻⁴ M/s
  3. The instantaneous rate at 120 s is greater than 4.5 × 10⁻⁴ M/s
  4. The instantaneous rate at 0 s is greater than 7.5 × 10⁻⁴ M/s
Show answer
Answer: (D) The instantaneous rate at 0 s is greater than 7.5 × 10⁻⁴ M/s
Explanation: The average rates over the three intervals are 7.5 × 10⁻⁴, 6.0 × 10⁻⁴ and 4.5 × 10⁻⁴ M/s, and over the full 120 s the average is 6.0 × 10⁻⁴ M/s. The rate falls continuously because the reactant is being used up. So the instantaneous rate at the start exceeds the first interval's average, while the rate at 120 s is below the last interval's average. This is why a rate at a chosen moment is better given as an instantaneous rate.

Question 7

For the reaction A + 3B → 2C + 4D, B is consumed at 0.060 mol L⁻¹ s⁻¹ at a certain instant. At that instant, what are the rate of consumption of A and the rate of formation of D, respectively, in mol L⁻¹ s⁻¹?
  1. 0.020 and 0.080
  2. 0.020 and 0.040
  3. 0.060 and 0.240
  4. 0.180 and 0.045
Show answer
Answer: (A) 0.020 and 0.080
Explanation: The rate of reaction equals each species' rate of change divided by its stoichiometric coefficient. Here the rate is 0.060/3 = 0.020 mol L⁻¹ s⁻¹. A has coefficient 1, so it is consumed at 0.020 mol L⁻¹ s⁻¹. D has coefficient 4, so it forms at 4 × 0.020 = 0.080 mol L⁻¹ s⁻¹.
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Question 8

A reaction proceeds through three consecutive first-order steps A → B → C → D with k₁ = 1 × 10⁻⁴ s⁻¹, k₂ = 1 × 10⁻¹ s⁻¹ and k₃ = 10 s⁻¹. Once the reaction is running steadily, which change would nearly double the rate of formation of D?
  1. Doubling both k₂ and k₃, leaving k₁ unchanged
  2. Doubling k₁ only
  3. Doubling k₃ only
  4. Doubling k₂ only
Show answer
Answer: (B) Doubling k₁ only
Explanation: The step with the smallest rate constant is the slowest and is the rate-determining step here. B and C are removed almost as soon as they form, so the rate of formation of D is close to k₁[A]. Speeding up the later fast steps cannot push material through any faster than the first step supplies it.

Question 9

In the decomposition 2H₂O₂(aq) → 2H₂O(l) + O₂(g), the concentration of H₂O₂ falls from 0.80 M to 0.50 M in 25 min. The average rate of formation of O₂ over this interval is:
  1. 2.0 × 10⁻⁴ M s⁻¹
  2. 1.2 × 10⁻² M s⁻¹
  3. 1.0 × 10⁻⁴ M s⁻¹
  4. 6.0 × 10⁻³ M s⁻¹
Show answer
Answer: (C) 1.0 × 10⁻⁴ M s⁻¹
Explanation: The average rate of loss of H₂O₂ is 0.30 M/(25 × 60 s) = 2.0 × 10⁻⁴ M s⁻¹. One O₂ forms for every two H₂O₂ consumed, so the rate of formation of O₂ is half of this. That gives 1.0 × 10⁻⁴ M s⁻¹.

Question 10

For the first-order reaction 2A → 3B, the rate law is written in two ways: −d[A]/dt = k_A[A] and d[B]/dt = k_B[A]. The ratio k_B/k_A is:
  1. 3
  2. 3/2
  3. 1
  4. 2/3
Show answer
Answer: (B) 3/2
Explanation: The stoichiometric relation is −(1/2) d[A]/dt = (1/3) d[B]/dt. This gives d[B]/dt = (3/2)(−d[A]/dt), and so k_B[A] = (3/2) k_A[A]. The numerical value of a rate constant therefore depends on which species the rate is expressed in.

Question 11

The decomposition of H₂O₂ in alkaline solution containing iodide ions occurs in two steps: H₂O₂ + I⁻ → H₂O + IO⁻ (slow) and H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast). Which statement is correct?
  1. I⁻ is a catalyst, IO⁻ is a reaction intermediate, and rate = k[H₂O₂][I⁻]
  2. I⁻ is a catalyst, IO⁻ is a reaction intermediate, and rate = k[H₂O₂]²
  3. I⁻ and IO⁻ are both reaction intermediates, and rate = k[H₂O₂]
  4. IO⁻ is a catalyst, I⁻ is a reaction intermediate, and rate = k[H₂O₂][IO⁻]
Show answer
Answer: (A) I⁻ is a catalyst, IO⁻ is a reaction intermediate, and rate = k[H₂O₂][I⁻]
Explanation: I⁻ is used in the first step and regenerated in the second, so it is a catalyst. IO⁻ is formed in the first step and consumed in the second, so it is an intermediate and does not appear in the overall equation 2H₂O₂ → 2H₂O + O₂. The slow first step fixes the rate, so rate = k[H₂O₂][I⁻], even though I⁻ is absent from the overall equation.

Question 12

Which of the following reactions proceeds at a conveniently measurable rate at room temperature, rather than almost instantly?
  1. Hydrolysis of ethyl acetate by aqueous NaOH
  2. Precipitation of BaSO₄ on mixing BaCl₂ and H₂SO₄ solutions
  3. Precipitation of AgCl on mixing AgNO₃ and NaCl solutions
  4. Neutralisation of aqueous HCl by aqueous NaOH
Show answer
Answer: (A) Hydrolysis of ethyl acetate by aqueous NaOH
Explanation: Precipitation and neutralisation in solution are reactions between ions, which need no covalent bonds to be broken, so they are practically instantaneous. Ester hydrolysis involves breaking and forming covalent bonds in molecules. It therefore occurs at a moderate rate that can be followed with time.

Question 13

The decomposition of H₂O₂ in alkaline medium in the presence of I⁻ is proposed to occur as: H₂O₂ + I⁻ → H₂O + IO⁻ (slow), followed by H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast). Which statement is correct?
  1. I⁻ is a catalyst, IO⁻ is an intermediate, and rate = k[H₂O₂]²
  2. I⁻ is an intermediate, IO⁻ is a product, and rate = k[H₂O₂]²[I⁻]
  3. IO⁻ is a catalyst, I⁻ is an intermediate, and rate = k[H₂O₂][IO⁻]
  4. I⁻ is a catalyst, IO⁻ is an intermediate, and rate = k[H₂O₂][I⁻]
Show answer
Answer: (D) I⁻ is a catalyst, IO⁻ is an intermediate, and rate = k[H₂O₂][I⁻]
Explanation: I⁻ is consumed in the first step and regenerated in the second, so it is a catalyst, and it appears in the rate law. IO⁻ is formed in the first step and used up in the second, so it is an intermediate and does not appear in the overall equation. The slow step decides the rate, giving rate = k[H₂O₂][I⁻].

Question 14

A complex reaction occurs in two first-order steps: A → B with k₁ = 2 × 10⁻⁴ s⁻¹, followed by B → C with k₂ = 5 s⁻¹. Which statement about this reaction is correct?
  1. The overall rate constant of the reaction is close to 5 s⁻¹
  2. The overall rate constant of the reaction is the sum k₁ + k₂
  3. The rate of formation of C is approximately k₁[A], since A → B is the rate-determining step
  4. B accumulates to a high concentration and B → C controls the overall rate
Show answer
Answer: (C) The rate of formation of C is approximately k₁[A], since A → B is the rate-determining step
Explanation: The overall rate of a stepwise reaction is governed by its slowest step, here A → B, which has the far smaller rate constant. B is consumed almost as soon as it forms, so it stays at a very low concentration. C therefore appears at about the rate at which A reacts, k₁[A].

Question 15

In a reaction involving only the species X, Y and Z, measurements over the same time interval show that [X] falls at 0.06 mol L⁻¹ min⁻¹, [Y] falls at 0.02 mol L⁻¹ min⁻¹ and [Z] rises at 0.04 mol L⁻¹ min⁻¹. Which balanced equation fits these data?
  1. 2X + Y → 3Z
  2. X + 3Y → 2Z
  3. 3X + 2Y → Z
  4. 3X + Y → 2Z
Show answer
Answer: (D) 3X + Y → 2Z
Explanation: Rates of change of concentration are proportional to stoichiometric coefficients. The ratio 0.06 : 0.02 : 0.04 simplifies to 3 : 1 : 2. X and Y fall, so they are reactants, and Z rises, so it is the product.

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