Chemical Kinetics
The half-life of a second-order reaction is 100 s when initial concentration is 0.5 M. What is the rate constant?
A
0.04 M⁻¹s⁻¹
B
0.02 M⁻¹s⁻¹
C
0.2 M⁻¹s⁻¹
D
0.01 M⁻¹s⁻¹
Correct Answer:
A. 0.04 M⁻¹s⁻¹
For second-order: t₁/₂ = 1/(k[A]₀); 100 = 1/(k × 0.5); k = 0.04 M⁻¹s⁻¹
The integrated rate law for second-order reaction is 1/[A] = 1/[A]₀ + kt. If [A]₀ = 0.5 M, k = 0.4 M⁻¹s⁻¹, find [A] after 5 seconds
A
0.2 M
B
0.25 M
C
0.1 M
D
0.15 M
Correct Answer:
B. 0.25 M
1/[A] = 2 + 0.4(5) = 2 + 2 = 4; [A] = 1/4 = 0.25 M
Which statement about potential energy diagrams is CORRECT?
A
Higher activation energy means faster reaction
B
An exothermic reaction has ΔH > 0
C
Activation energy is always greater than ΔH
D
Catalyst changes activation energy but not ΔH
Correct Answer:
D. Catalyst changes activation energy but not ΔH
Catalyst lowers Eₐ (forward and reverse) without changing ΔH, reaction enthalpy change
For an elementary reaction: 2A + B → Products, the rate law is:
A
Rate = k[A][B]
B
Rate = k[A]²[B]
C
Rate = k[A][B]²
D
Rate = k[A]³[B]
Correct Answer:
B. Rate = k[A]²[B]
For elementary reactions, rate law exponents = stoichiometric coefficients. Rate = k[A]²[B]
In the reaction A → Products, doubling [A] increases rate by 4 times. The order of reaction is:
A
Zero
B
First
C
Second
D
Third
Correct Answer:
C. Second
If doubling concentration increases rate by 4 times (2²), reaction is second order
The decomposition of N₂O₅ is a first-order reaction. If 50% decomposes in 30 minutes, what is the rate constant?
A
0.023 min⁻¹
B
0.035 min⁻¹
C
0.052 min⁻¹
D
0.069 min⁻¹
Correct Answer:
A. 0.023 min⁻¹
For first-order: k = 0.693/t₁/₂ = 0.693/30 = 0.0231 min⁻¹ ≈ 0.023 min⁻¹
At 25°C, the rate constant is 3 × 10⁻² s⁻¹ and at 35°C it is 6 × 10⁻² s⁻¹. What is the temperature coefficient (Q₁₀) for this reaction?
Q₁₀ = k(T+10)/k(T) = (6 × 10⁻²)/(3 × 10⁻²) = 2. For typical reactions, Q₁₀ = 2-3
The rate law for a reaction is Rate = k[A]¹[B]⁰. If [B] is doubled and [A] is halved, how does the rate change?
A
Increases by factor of 2
B
Decreases by factor of 2
C
Remains same
D
Increases by factor of 4
Correct Answer:
B. Decreases by factor of 2
Rate depends only on [A] (order = 1 w.r.t. A). Halving [A] decreases rate by factor of 2. [B] has no effect
The rate constant of a reaction increases from 4 × 10⁻³ s⁻¹ to 8 × 10⁻³ s⁻¹ when temperature increases from 300K to 310K. Calculate activation energy (R = 8.314 J/mol·K)
A
50.4 kJ/mol
B
60.8 kJ/mol
C
75.2 kJ/mol
D
85.6 kJ/mol
Correct Answer:
A. 50.4 kJ/mol
Using ln(k₂/k₁) = (Eₐ/R)(T₂-T₁)/(T₁T₂); ln(2) = (Eₐ/8.314)(10/93000); Eₐ ≈ 50.4 kJ/mol
The pre-exponential factor (A) in the Arrhenius equation is related to:
A
Only the number of collisions
B
Only the proper orientation of reactant molecules
C
The number of collisions with proper orientation and energy distribution
D
The activation energy only
Correct Answer:
C. The number of collisions with proper orientation and energy distribution
The pre-exponential factor accounts for collision frequency, proper orientation (steric factor), and the Maxwell-Boltzmann energy distribution of molecules
