Chapter 7: Chemical Kinetics — Online MCQ Test
CHEMISTRY · CLASS 12th · Tamil Nadu State Board
Practice Chapter 7: Chemical Kinetics with a free chapter-wise online MCQ test.
This chapter covers: Focusing on reaction rates this chapter details rate laws order and molecularity of chemical reactions. It covers integrated rate equations half-life determination Arrhenius equati....
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Chapter 7: Chemical Kinetics — Important Questions & Answers
What is the unit of rate constant for a first-order reaction?
- A. mol L^-1 s^-1
- B. s^-1
- C. L mol^-1 s^-1
- D. mol^2 L^-2 s^-1
Answer: B. s^-1
For a first-order reaction, the rate constant has the unit time^-1, which is s^-1.
For a first-order reaction, the rate constant has the unit time^-1, which is s^-1.
The molecularity of an elementary reaction can be defined as the number of:
- A. products formed in the reaction
- B. steps in the mechanism
- C. reactant species taking part in a single step
- D. collisions per second
Answer: C. reactant species taking part in a single step
Molecularity refers to the number of reacting species involved in one elementary step.
Molecularity refers to the number of reacting species involved in one elementary step.
For a second-order reaction involving a single reactant A, which integrated rate law is correct?
- A. ln[A] = ln[A]0 - kt
- B. 1/[A] = 1/[A]0 + kt
- C. [A] = [A]0 - kt
- D. t1/2 = 0.693/k
Answer: B. 1/[A] = 1/[A]0 + kt
For a second-order reaction in one reactant, the integrated rate law is 1/[A] = 1/[A]0 + kt.
For a second-order reaction in one reactant, the integrated rate law is 1/[A] = 1/[A]0 + kt.
A reaction is slow at room temperature but becomes much faster on heating. Which reason best explains this?
- A. Heating decreases activation energy directly
- B. Heating increases the fraction of molecules having energy greater than activation energy
- C. Heating always increases molecularity
- D. Heating changes the products formed
Answer: B. Heating increases the fraction of molecules having energy greater than activation energy
On heating, more molecules acquire energy equal to or greater than the activation energy, so the reaction rate increases.
On heating, more molecules acquire energy equal to or greater than the activation energy, so the reaction rate increases.
A zero-order reaction has a rate constant k = 0.02 mol L^-1 s^-1. If the initial concentration is 0.10 mol L^-1, the time required for complete consumption is:
- A. 2 s
- B. 5 s
- C. 10 s
- D. 20 s
Answer: B. 5 s
For a zero-order reaction, t = [A]0/k. So t = 0.10/0.02 = 5 s.
For a zero-order reaction, t = [A]0/k. So t = 0.10/0.02 = 5 s.