Chapter Exam

Unit 8: Chemical Kinetics — Online MCQ Test

Q: The rate constant for a first order reaction is 4.606 x 10^-3 s^-1. The time required to reduce 2.0 g of the reactant to 0.2 g is:

Answer: B. 500 s — For first order, t = (2.303 / k) * log(A0/At). t = (2.303 / 4.606e-3) * log(2.0/0.2) = 500 * log(10) = 500 s.

Q: If the half-life (t₁/₂) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

Answer: D. 10 minutes — For a first order reaction, t = (2.303 / k) * log(100 / (100-99.9)) = (2.303 / k) * log(1000). Since k = 0.693 / t₁/₂ = 0.693, t = 2.303 * 3 / 0.693 ≈ 10 minutes.

Q: If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

Answer: A. 69.3 s — First-order kinetics formula: t = (2.303/k) * log(A0/At). t = (2.303/0.03) * log(7.2/0.9) = (2.303/0.03) * log(8). log(8) = 3 * log(2) = 3 * 0.301 = 0.903. t = (2.303 * 0.903) / 0.03 ≈ 69.3 s.

Q: The addition of a catalyst during a chemical reaction alters which of the following quantities?

Answer: C. Activation energy — A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction without changing the internal energy, enthalpy, or entropy of the system.

Q: Which one of the following characteristics is associated with adsorption?

Answer: A. ΔG, ΔH and ΔS all are negative — Adsorption is a spontaneous process (ΔG < 0) that releases heat (ΔH < 0), and since gas molecules are fixed onto a surface, entropy decreases (ΔS < 0).

Q: 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:

Answer: D. 24.1 s — Using k = (2.303/t2-t1) * log(Rate1/Rate2), k = (2.303/10) * log(0.04/0.03) ≈ 0.0287 s^-1. Half-life t(1/2) = 0.693/k ≈ 24.1 s.

Q: 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 :

Answer: D. 24.1 s — For a first-order reaction, rate = k[A]. Using the ratio of rates at different times: k = (2.303 / (t2 - t1)) * log(rate1 / rate2). k = (2.303 / 10) * log(0.04 / 0.03) ≈ 0.0285 s⁻¹. t1/2 = 0.693 / k ≈ 24.3 s (approximated to 24.1 s).

Q: Activation energy of any chemical reaction can be calculated if one knows the value of

Answer: B. rate constant at two different temperatures. — Using the Arrhenius equation, activation energy can be calculated from rate constants at two temperatures.

Q: The addition of a catalyst during a chemical reaction alters which of the following quantities ?

Answer: D. Activation energy — A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction without altering thermodynamic properties like enthalpy or entropy.

Q: The correct difference between first- and second-order reactions is that

Answer: A. the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀ — For a first-order reaction, t₁/₂ = ln2/k, which is independent of initial concentration [A]₀. For a second-order reaction, t₁/₂ = 1/(k[A]₀), which depends on [A]₀. Both first- and second-order reactions can be catalyzed, and both have rates that depend on reactant concentrations.

CHEMISTRY · NEET (All) · NEET

Practice Unit 8: Chemical Kinetics with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: Rate of a chemical reaction factors affecting the rate of reactions: concentration temperature pressure and catalyst; elementary and complex reactions order and molecularity of rea.... AI-generated questions from basic to board-exam level, with instant results and explanations.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Unit 8: Chemical Kinetics — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 8: Chemical Kinetics, with answers and explanations. These are sample questions; the exam has its own separate question set.

The rate constant for a first order reaction is 4.606 x 10^-3 s^-1. The time required to reduce 2.0 g of the reactant to 0.2 g is:

A. 200 s
B. 500 s ✓
C. 1000 s
D. 100 s

Answer: B. 500 s
For first order, t = (2.303 / k) * log(A0/At). t = (2.303 / 4.606e-3) * log(2.0/0.2) = 500 * log(10) = 500 s.

If the half-life (t₁/₂) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

A. 2 minutes
B. 4 minutes
C. 5 minutes
D. 10 minutes ✓

Answer: D. 10 minutes
For a first order reaction, t = (2.303 / k) * log(100 / (100-99.9)) = (2.303 / k) * log(1000). Since k = 0.693 / t₁/₂ = 0.693, t = 2.303 * 3 / 0.693 ≈ 10 minutes.

If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

A. 69.3 s ✓
B. 23.1 s
C. 210 s
D. 21.0 s

Answer: A. 69.3 s
First-order kinetics formula: t = (2.303/k) * log(A0/At). t = (2.303/0.03) * log(7.2/0.9) = (2.303/0.03) * log(8). log(8) = 3 * log(2) = 3 * 0.301 = 0.903. t = (2.303 * 0.903) / 0.03 ≈ 69.3 s.

The addition of a catalyst during a chemical reaction alters which of the following quantities?

A. Internal energy
B. Enthalpy
C. Activation energy ✓
D. Entropy

Answer: C. Activation energy
A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction without changing the internal energy, enthalpy, or entropy of the system.

Which one of the following characteristics is associated with adsorption?

A. ΔG, ΔH and ΔS all are negative ✓
B. ΔG and ΔH are negative but ΔS is positive
C. ΔG and ΔS are negative but ΔH is positive
D. ΔG is negative but ΔH and ΔS are positive

Answer: A. ΔG, ΔH and ΔS all are negative
Adsorption is a spontaneous process (ΔG < 0) that releases heat (ΔH < 0), and since gas molecules are fixed onto a surface, entropy decreases (ΔS < 0).

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:

A. 34.1 s
B. 44.1 s
C. 54.1 s
D. 24.1 s ✓

Answer: D. 24.1 s
Using k = (2.303/t2-t1) * log(Rate1/Rate2), k = (2.303/10) * log(0.04/0.03) ≈ 0.0287 s^-1. Half-life t(1/2) = 0.693/k ≈ 24.1 s.

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 :

A. 34.1 s
B. 44.1 s
C. 54.1 s
D. 24.1 s ✓

Answer: D. 24.1 s
For a first-order reaction, rate = k[A]. Using the ratio of rates at different times: k = (2.303 / (t2 - t1)) * log(rate1 / rate2). k = (2.303 / 10) * log(0.04 / 0.03) ≈ 0.0285 s⁻¹. t1/2 = 0.693 / k ≈ 24.3 s (approximated to 24.1 s).

Activation energy of any chemical reaction can be calculated if one knows the value of

A. orientation of reactant molecules during collision.
B. rate constant at two different temperatures. ✓
C. rate constant at standard temperature.
D. probability of collision.

Answer: B. rate constant at two different temperatures.
Using the Arrhenius equation, activation energy can be calculated from rate constants at two temperatures.

The addition of a catalyst during a chemical reaction alters which of the following quantities ?

A. Entropy
B. Internal energy
C. Enthalpy
D. Activation energy ✓

Answer: D. Activation energy
A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction without altering thermodynamic properties like enthalpy or entropy.

The correct difference between first- and second-order reactions is that

A. the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀ ✓
B. a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed
C. the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
D. the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

Answer: A. the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀
For a first-order reaction, t₁/₂ = ln2/k, which is independent of initial concentration [A]₀. For a second-order reaction, t₁/₂ = 1/(k[A]₀), which depends on [A]₀. Both first- and second-order reactions can be catalyzed, and both have rates that depend on reactant concentrations.

← Previous Chapter

Unit 7: Redox Reactions and Electrochemistry

Next Chapter →

Unit 9: Classification of Elements and Periodicity in Properties