Daily Olympiad: Chemistry - Chemical Kinetics [20260511] – 6 Practice Problems & Printable PDF

1. For a reaction, the rate constant triples when the temperature is increased from 300 K to 310 K. The activation energy (Ea) for the reaction is close to (R = 8.314 J/mol·K):

A. 23000 J/mol B. 50000 J/mol C. 46000 J/mol D. 35000 J/mol

Solution

Correct: C

Using the Arrhenius equation in logarithmic form: ln(k2/k1) = -Ea/R (1/T2 - 1/T1). Given k2/k1 = 3, substituting values and solving for Ea gives approximately 46000 J/mol. The calculation involves solving the equation with the temperature difference and natural logarithm of 3.

2. Which of the following statements about the rate-determining step (RDS) is incorrect?

A. RDS directly depends on the highest energy transition state in the mechanism B. The rate constant of the RDS governs the overall reaction rate C. The RDS has the highest activation energy among all steps D. The RDS always corresponds to an elementary reaction

Solution

Correct: D

The rate-determining step does not always correspond to an elementary reaction; it is determined by the slowest step with the highest activation energy, which may involve intermediates and non-elementary steps. The rate law for the RDS must consider the mechanism.

3. A catalyst lowers the activation energy of the forward and reverse reactions by the same amount. How does this affect the equilibrium constant (K)?

A. K increases exponentially B. K decreases exponentially C. K remains unchanged D. K becomes independent of temperature

Solution

Correct: C

Catalysts lower the activation energies of both forward and reverse reactions equally, altering the rate constants equally. Since K = k_forward / k_reverse, and both k_forward and k_reverse are proportionally increased, the ratio (K) remains unchanged. Equilibrium is unchanged by catalysts, only the time to reach equilibrium is reduced.

4. A second-order reaction has a half-life of 20 minutes at an initial concentration of 0.1 M. What will be the half-life when the initial concentration is 0.05 M?

A. 40 minutes B. 10 minutes C. 5 minutes D. 20 minutes

Solution

Correct: A

For a second-order reaction, t₁/₂ is inversely proportional to the initial concentration. Since [A]₀ is halved, t₁/₂ doubles. Original t₁/₂ = 1/(k·0.1) = 20 min. New t₁/₂ = 1/(k·0.05) = 40 min.

5. A plot of ln[A] vs. time is linear for a reaction. What is the order of the reaction?

A. Zero-order B. First-order C. Second-order D. Pseudo-first order

Solution

Correct: B

For a first-order reaction, ln[A] vs time forms a linear plot with slope = -k. Zero-order shows [A] vs time as linear, while second-order shows 1/[A] vs time as linear. Linear ln[A] vs time confirms first-order kinetics.

6. The rate constant k for a reaction at 400 K is 2×10⁻³ s⁻¹. If the activation energy is 80 kJ/mol, what is the rate constant at 410 K? (R = 8.314 J/mol·K)

A. 3.15×10⁻³ s⁻¹ B. 4.0×10⁻³ s⁻¹ C. 1.5×10⁻³ s⁻¹ D. 6.3×10⁻³ s⁻¹

Solution

Correct: A

Use the Arrhenius equation: ln(k2/k1) = -Ea/R (1/T2 - 1/T1). Substitute values for T1=400K, T2=410K, Ea=80000 J/mol, k1=2×10⁻³. Solve for k2. The calculation yields ln(k2/2×10⁻³) ≈ 0.366, leading to k2 ≈ 3.15×10⁻³ s⁻¹.

Daily Olympiad: Chemistry - Chemical Kinetics [20260511]

1. For a reaction, the rate constant triples when the temperature is increased from 300 K to 310 K. The activation energy (Ea) for the reaction is close to (R = 8.314 J/mol·K):

2. Which of the following statements about the rate-determining step (RDS) is incorrect?

3. A catalyst lowers the activation energy of the forward and reverse reactions by the same amount. How does this affect the equilibrium constant (K)?

4. A second-order reaction has a half-life of 20 minutes at an initial concentration of 0.1 M. What will be the half-life when the initial concentration is 0.05 M?

5. A plot of ln[A] vs. time is linear for a reaction. What is the order of the reaction?

6. The rate constant k for a reaction at 400 K is 2×10⁻³ s⁻¹. If the activation energy is 80 kJ/mol, what is the rate constant at 410 K? (R = 8.314 J/mol·K)

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