Unit 8: Chemical Kinetics — Online MCQ Test
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Practice Unit 8: Chemical Kinetics with a free chapter-wise online MCQ test.
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....
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Unit 8: Chemical Kinetics — Important Questions & Answers
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.
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.
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.
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.
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).
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.
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).
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.
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.
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.
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.