Order, Molecularity and Half-Life Period order and molecularity of reactions, rate law, rate constant and its units, differential and integral forms of zero and first-orderreactions, their characteristics and half-lives, the effect of temperature on the rate of reactions,: NEET MCQs with Answers

15 free questions on Order, Molecularity and Half-Life Period order and molecularity of reactions, rate law, rate constant and its units, differential and integral forms of zero and first-orderreactions, their characteristics and half-lives, the effect of temperature on the rate of reactions, 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.

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Question 1

Previous-year question
The rate constant for a first order reaction is 4.606 × 10⁻³ s⁻¹. The time required to reduce 2.0 g of the reactant to 0.2 g is [NEET 2020]
  1. 200 s
  2. 500 s
  3. 1000 s
  4. 100 s
Show answer
Answer: (B) 500 s
Explanation: t = (2.303/k)log(2.0/0.2) = 2.303/(4.606×10⁻³) = 500 s.

Question 2

Previous-year question
If the rate constant for a first order reaction is k, the time t required for the completion of 99% of the reaction is given by [NEET 2019]
  1. t = 0.693/k
  2. t = 6.909/k
  3. t = 4.606/k
  4. t = 2.303/k
Show answer
Answer: (C) t = 4.606/k
Explanation: t = (2.303/k)log(100/1) = 2×2.303/k = 4.606/k.

Question 3

Previous-year question
The correct difference between first and second order reactions is that [NEET 2018]
  1. the rate of a first-order reaction does not depend on reactant concentrations while that of a second-order reaction does
  2. the half-life of a first-order reaction does not depend on [A]₀, while the half-life of a second-order reaction does depend on [A]₀
  3. the rate of a first-order reaction does depend on reactant concentrations, while that of a second-order reaction does not
  4. a first-order reaction can be catalyzed, a second-order reaction cannot be catalyzed
Show answer
Answer: (B) the half-life of a first-order reaction does not depend on [A]₀, while the half-life of a second-order reaction does depend on [A]₀
Explanation: For first order, t½ = 0.693/k is independent of the initial concentration; for second order, t½ = 1/k[A]₀ varies inversely with it.

Question 4

Previous-year question
A first order reaction has a specific reaction rate of 10⁻² sec⁻¹. How much time will it take for 20 g of the reactant to reduce to 5 g? [NEET 2017]
  1. 138.6 sec
  2. 346.5 sec
  3. 693.0 sec
  4. 238.6 sec
Show answer
Answer: (A) 138.6 sec
Explanation: 20 g → 5 g is two half-lives; t½ = 0.693/10⁻² = 69.3 s, so t = 138.6 s.

Question 5

Previous-year question
The rate of a first-order reaction is 0.04 mol L⁻¹ s⁻¹ at 10 seconds and 0.03 mol L⁻¹ s⁻¹ at 20 seconds after initiation of the reaction. The half-life period of the reaction is [NEET 2016]
  1. 24.1 s
  2. 34.1 s
  3. 44.1 s
  4. 54.1 s
Show answer
Answer: (A) 24.1 s
Explanation: k = (2.303/10)log(0.04/0.03) = 0.02876 s⁻¹; t½ = 0.693/0.02876 ≈ 24.1 s.

Question 6

Previous-year question
When the initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is [AIPMT 2015]
  1. first
  2. second
  3. more than zero but less than first
  4. zero
Show answer
Answer: (A) first
Explanation: Only for a first order reaction is t½ = 0.693/k, entirely independent of the starting concentration.

Question 7

Previous-year question
The rate constant of the reaction A → B is 0.6 × 10⁻³ mole per second. If the concentration of A is 5 M then the concentration of B after 20 minutes is [AIPMT 2015]
  1. 1.08 M
  2. 3.60 M
  3. 0.36 M
  4. 0.72 M
Show answer
Answer: (D) 0.72 M
Explanation: The unit shows a zero order reaction, so [B] = kt = 0.6×10⁻³×1200 = 0.72 M.
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Question 8

For a reaction A → products, a plot of half-life against initial concentration [A]₀ is a straight line through the origin with slope 50 s L mol⁻¹. The order of the reaction and its rate constant are:
  1. Second order; k = 2.0 × 10⁻² L mol⁻¹ s⁻¹
  2. First order; k = 1.4 × 10⁻² s⁻¹
  3. Zero order; k = 2.0 × 10⁻² mol L⁻¹ s⁻¹
  4. Zero order; k = 1.0 × 10⁻² mol L⁻¹ s⁻¹
Show answer
Answer: (D) Zero order; k = 1.0 × 10⁻² mol L⁻¹ s⁻¹
Explanation: A half-life directly proportional to [A]₀ is the mark of a zero-order reaction, for which t½ = [A]₀/(2k). A first-order half-life would be a horizontal line, and a second-order half-life would fall as [A]₀ rises. The slope equals 1/(2k), so k = 1/(2 × 50) = 1.0 × 10⁻² mol L⁻¹ s⁻¹.

Question 9

Phosphine decomposes on a hot tungsten surface. As the pressure of PH₃ is raised from very low to high values at constant temperature, the observed order of the reaction changes:
  1. from about 1 to about 0, because the surface becomes saturated with gas molecules
  2. from about 1 to about 2, because surface collisions become twice as frequent
  3. not at all, because the rate constant is fixed at a given temperature
  4. from about 0 to about 1, because the surface becomes less covered
Show answer
Answer: (A) from about 1 to about 0, because the surface becomes saturated with gas molecules
Explanation: At low pressure only a small part of the surface is covered, so the rate is proportional to the pressure and the order is about 1. At high pressure the surface is fully covered, and adding more gas does not change the amount adsorbed. The rate then becomes constant and the order falls to about 0.

Question 10

In a first-order reaction, the concentration of the reactant is 0.64 M at t = 10 min and 0.16 M at t = 30 min. The concentration at the start of the reaction was:
  1. 1.28 M
  2. 2.56 M
  3. 0.96 M
  4. 0.80 M
Show answer
Answer: (A) 1.28 M
Explanation: A fall from 0.64 M to 0.16 M is a fall to one-fourth, which takes two half-lives. These two half-lives span 20 min, so the half-life is 10 min. Going back 10 min from t = 10 min doubles the concentration, giving 1.28 M.

Question 11

A reaction follows the rate law rate = k[A][B], with k = 2.0 × 10⁻³ L mol⁻¹ s⁻¹. It is run with [A]₀ = 1.0 × 10⁻³ M and B in large excess at a constant 0.50 M. The half-life of A under these conditions is about:
  1. 693 s
  2. 347 s
  3. 1386 s
  4. 3.47 × 10⁵ s
Show answer
Answer: (A) 693 s
Explanation: Since [B] stays practically constant, the reaction is pseudo first order with k′ = k[B] = 2.0 × 10⁻³ × 0.50 = 1.0 × 10⁻³ s⁻¹. The half-life is 0.693/k′ = 693 s. It does not depend on [A]₀, but it does depend on the concentration of the excess reactant.

Question 12

A gas X decomposes by first-order kinetics in a rigid vessel at constant temperature: X(g) → Y(g) + Z(g), with k = 2.0 × 10⁻³ s⁻¹. Pure X is introduced at 0.50 atm. At the instant the total pressure reads 0.70 atm, the rate at which the partial pressure of X is falling is:
  1. 1.4 × 10⁻³ atm s⁻¹
  2. 1.0 × 10⁻³ atm s⁻¹
  3. 6.0 × 10⁻⁴ atm s⁻¹
  4. 4.0 × 10⁻⁴ atm s⁻¹
Show answer
Answer: (C) 6.0 × 10⁻⁴ atm s⁻¹
Explanation: If p atm of X has reacted, the total pressure is (0.50 − p) + p + p = 0.50 + p, so p = 0.20 atm. The partial pressure of X left is 0.50 − 0.20 = 0.30 atm. Only the reactant pressure enters the first-order rate law, so the rate is k × 0.30 = 2.0 × 10⁻³ × 0.30 = 6.0 × 10⁻⁴ atm s⁻¹.

Question 13

An ester is hydrolysed in two separate experiments: once in dilute HCl with water in large excess, and once with aqueous NaOH at a concentration comparable to that of the ester. Which statement correctly describes the kinetics?
  1. Both are pseudo first order because water is in large excess in each experiment
  2. With HCl it is second order because H⁺ is consumed; with NaOH it is pseudo first order because OH⁻ is regenerated
  3. With HCl it is pseudo first order because water is in excess and H⁺ is regenerated; with NaOH it is second order because OH⁻ is consumed as a reactant
  4. With HCl it is zero order because the rate depends only on H⁺; with NaOH it is first order in OH⁻ alone
Show answer
Answer: (C) With HCl it is pseudo first order because water is in excess and H⁺ is regenerated; with NaOH it is second order because OH⁻ is consumed as a reactant
Explanation: In acid, H⁺ acts only as a catalyst and is regenerated, and water is in such excess that its concentration stays almost constant, so the rate depends on the ester alone. In base, OH⁻ is used up as a reactant in forming the carboxylate salt, so its concentration falls during the reaction. The rate then depends on both ester and OH⁻, giving second order kinetics.

Question 14

The rate of a gas phase decomposition is followed by pressure measurements and obeys −dp/dt = k p^(3/2), with p in atm and t in seconds. The unit of k is
  1. s⁻¹
  2. atm^(−1/2) s⁻¹
  3. atm^(−3/2) s⁻¹
  4. atm^(1/2) s⁻¹
Show answer
Answer: (B) atm^(−1/2) s⁻¹
Explanation: The rate −dp/dt has the unit atm s⁻¹. So k = (atm s⁻¹)/(atm^(3/2)) = atm^(−1/2) s⁻¹. In general, for order n in pressure terms, the unit of k is atm^(1−n) s⁻¹.

Question 15

For a first order reaction, a plot of log₁₀[A] against time is a straight line of slope −3.01 × 10⁻³ min⁻¹. The time taken for [A] to fall to one-eighth of its initial value is (log 2 = 0.301)
  1. 100 min
  2. 300 min
  3. 690 min
  4. 800 min
Show answer
Answer: (B) 300 min
Explanation: The line follows log[A] = log[A]₀ − (k/2.303)t, so log₁₀[A] must fall by log 8 = 0.903. With a fall of 3.01 × 10⁻³ per minute, the time is 0.903/(3.01 × 10⁻³) = 300 min. As a check, the half-life is 0.301/(3.01 × 10⁻³) = 100 min, and one-eighth is reached after three half-lives.

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