NEETAceToday's NEET Daily

Zero- and first-order integrated/differential forms and half-lives: NEET MCQs with answers

84 questions on this topic of Chemical Kinetics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Chemical Kinetics freeAll Chemical Kinetics questions

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

easy

1. A reaction following first-order kinetics has k = 0.85 min⁻¹. The time taken for half the reactant to be consumed is:

  1. 1.18 min
  2. 0.815 min
  3. 1.23 min
  4. 0.408 min
Show answer and explanation

Answer: (B) 0.815 min

For a first-order reaction, t½ = 0.693/k = 0.693/0.85 = 0.815 min.

Quick tip: First-order half-life is independent of initial concentration.

medium

2. A first-order reaction has a half-life of 5 min. The time needed for 93.75% of the reactant to decompose is:

  1. 25 min
  2. 80 min
  3. 9.38 min
  4. 20 min
Show answer and explanation

Answer: (D) 20 min

93.75% decomposed means 1/16 remains, which takes 4 half-lives: 4 × 5 = 20 min.

hard

3. A reaction follows first-order kinetics. When the initial concentration of reactant is a, the fraction decomposed in time t is x. The plot of ln(1 - x) versus t gives:

  1. A straight line passing through the origin with slope -k
  2. A straight line with y-intercept ln a and slope -k
  3. A straight line passing through the origin with slope k
  4. A curve asymptotic to the time axis
Show answer and explanation

Answer: (A) A straight line passing through the origin with slope -k

Fraction decomposed is x = ([A]₀ - [A]) / [A]₀, so [A]/[A]₀ = 1 - x. From first-order integrated rate law, [A] = [A]₀ e^(-kt) ⇒ ln([A]/[A]₀) = -kt ⇒ ln(1 - x) = -kt. Plotting ln(1 - x) against t yields a straight line with slope = -k and intercept = 0 (passes through origin).

medium

4. A first-order reaction has a rate constant of 2.303 × 10⁻³ s⁻¹. How much time will it take for 40.0 g of this reactant to be reduced to 10.0 g? (Given: log 2 = 0.3010)

  1. 602 s
  2. 300 s
  3. 1000 s
  4. 1204 s
Show answer and explanation

Answer: (A) 602 s

The amount remaining is 10 g from an initial 40 g, which represents two half-lives (40 g → 20 g → 10 g). The rate constant k = 2.303 × 10⁻³ s⁻¹. Using t = (2.303 / k) log(40/10) = (2.303 / (2.303 × 10⁻³)) × log(4) = 1000 × 2 × 0.3010 = 602 s.

easy

5. A reaction following first-order kinetics has k = 0.6 min⁻¹. The time taken for half the reactant to be consumed is:

  1. 1.16 min
  2. 0.866 min
  3. 0.58 min
  4. 1.67 min
Show answer and explanation

Answer: (A) 1.16 min

For a first-order reaction, t½ = 0.693/k = 0.693/0.6 = 1.16 min.

Quick tip: First-order half-life is independent of initial concentration.

hard

6. In a certain first-order reaction A → Products, the rate of reaction is 0.04 mol L⁻¹ s⁻¹ when [A] = 0.4 mol L⁻¹. The time required for the reaction rate to fall from 0.04 mol L⁻¹ s⁻¹ to 0.01 mol L⁻¹ s⁻¹ is: (Given: ln 2 = 0.693)

  1. 6.93 s
  2. 3.46 s
  3. 27.72 s
  4. 13.86 s
Show answer and explanation

Answer: (D) 13.86 s

For a first-order reaction, rate = k[A]. Thus, rate is directly proportional to concentration: rate₁/rate₂ = [A]₁/[A]₂. Since rate decreases from 0.04 to 0.01 (a factor of 4 = 2²), two half-lives have elapsed. From the given data, k = rate/[A] = 0.04 / 0.4 = 0.1 s⁻¹. Half-life t₁/₂ = 0.693 / 0.1 = 6.93 s. Total time = 2 × 6.93 = 13.86 s.

medium

7. A first-order reaction has a half-life of 10 min. The time needed for 87.5% of the reactant to decompose is:

  1. 40 min
  2. 80 min
  3. 17.5 min
  4. 30 min
Show answer and explanation

Answer: (D) 30 min

87.5% decomposed means 1/8 remains, which takes 3 half-lives: 3 × 10 = 30 min.

easy

8. For a zero-order reaction A → Products, which of the following statements is INCORRECT?

  1. A plot of concentration of A versus time gives a straight line passing through the origin.
  2. The half-life of the reaction is directly proportional to the initial concentration of A.
  3. The rate of the reaction is constant throughout the course of the reaction.
  4. The rate constant has the units mol L⁻¹ s⁻¹.
Show answer and explanation

Answer: (A) A plot of concentration of A versus time gives a straight line passing through the origin.

The integrated rate law for a zero-order reaction is [A] = [A]₀ - kt. A plot of [A] vs t is a straight line with a y-intercept of [A]₀ and slope of -k; it does not pass through the origin.

medium

9. A first-order reaction has a half-life of 5 min. The time needed for 75% of the reactant to decompose is:

  1. 20 min
  2. 10 min
  3. 7.5 min
  4. 15 min
Show answer and explanation

Answer: (B) 10 min

75% decomposed means 1/4 remains, which takes 2 half-lives: 2 × 5 = 10 min.

easy

10. For a first-order process with rate constant k = 0.15 min⁻¹, the half-life is:

  1. 2.31 min
  2. 4.62 min
  3. 6.67 min
  4. 0.216 min
Show answer and explanation

Answer: (B) 4.62 min

For a first-order reaction, t½ = 0.693/k = 0.693/0.15 = 4.62 min.

Quick tip: First-order half-life is independent of initial concentration.

easy

11. For a zero-order reaction A → Products with rate constant k and initial concentration [A]₀, the time required for complete consumption of the reactant is given by:

  1. [A]₀ / (2k)
  2. [A]₀ / k
  3. 2[A]₀ / k
  4. ln 2 / k
Show answer and explanation

Answer: (B) [A]₀ / k

For a zero-order reaction, the integrated rate equation is [A] = [A]₀ - kt. When the reaction goes to completion, [A] = 0, which yields t_completion = [A]₀ / k. Note that [A]₀ / (2k) is the half-life.

easy

12. The unit of rate constant for a reaction is identical to the unit of the rate of reaction for:

  1. A first-order reaction
  2. A zero-order reaction
  3. A second-order reaction
  4. A pseudo-first-order reaction
Show answer and explanation

Answer: (B) A zero-order reaction

For a zero-order reaction, Rate = k[A]⁰ = k. Therefore, the unit of k is mol L⁻¹ s⁻¹, which is identical to the unit of reaction rate.

Keep going with Chemical Kinetics. All 514 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Chemical Kinetics