For a first-order reaction, if the initial concentration is [A]₀ = 0.5 M and after 30 seconds it becomes 0.25 M, what is the rate constant?
Answer: A
Using ln([A]₀/[A]ₜ) = kt, ln(0.05.25) = k × 30, ln(2) = k × 30, k = 0.30693 = 0.0231 s⁻¹
Q.2Medium
The mechanism of a reaction is: Step 1: A + B → C (slow), Step 2: C + D → E + F (fast). What is the overall reaction and the rate law?
Answer: A
The overall reaction is obtained by adding all steps and canceling intermediates: A + B + D → E + F. Rate law is determined by the slow step: rate = k[A][B]
Q.3Medium
Which of the following graphs represents a first-order reaction?
Answer: B
For a first-order reaction, ln[A] = ln[A]₀ - kt, so a plot of ln[A] vs t gives a straight line with slope -k.
Q.4Medium
According to Arrhenius equation, k = Ae^(-Eₐ/RT), a catalyst increases reaction rate by:
Answer: B
A catalyst provides an alternative reaction pathway with lower activation energy, thus increasing the rate constant k without affecting A or T.
Q.5Medium
For the reaction A → B, the integrated rate law for zero-order kinetics is:
Answer: A
For zero-order reaction: d[A]/dt = -k, integrating gives [A] = [A]₀ - kt, which is a linear equation.
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Q.6Medium
At 300 K, a reaction has a half-life of 10 minutes. At 310 K, the half-life becomes 5 minutes. What is the approximate value of temperature coefficient (assuming RRT ≈ 2)?
Answer: B
For a first-order reaction, if half-life decreases from 10 to 5 minutes (becomes half) with a 10 K increase, this indicates the reaction rate doubles per 10 K, giving a temperature coefficient of 2.
Q.7Medium
If a reaction is first-order with rate constant k = 0.1 min⁻¹, what fraction of the reactant remains after 5 half-lives?
Answer: A
After n half-lives, fraction remaining = (21)ⁿ. After 5 half-lives: (21)⁵ = 321.
Q.8Medium
Which of the following is an example of a homogeneous catalyst?
Answer: B
A homogeneous catalyst is in the same phase as reactants. H₂SO₄ (liquid) catalyzes esterification of reactants (liquid), making it homogeneous. Others are heterogeneous catalysts.
Q.9Medium
A reaction has activation energy of 50 kJ/mol. If the temperature is increased from 300 K to 310 K, the rate constant increases by a factor of approximately (R = 8.314 J/mol·K):
Answer: B
Using Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(T₂-T₁)/(T₁T₂). With Ea = 50,000 J/mol, ΔT = 10 K, this gives log(k₂/k₁) ≈ 0.30, so k₂/k₁ ≈ 2.0
Q.10Medium
In the decomposition of N₂O₅, the rate constant at 320 K is 1.7 × 10⁻⁵ s⁻¹ and at 330 K is 5.0 × 10⁻⁵ s⁻¹. The activation energy is approximately:
Answer: A
Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(5.10.7) = (Ea/8.314)(3201 - 3301), solving gives Ea ≈ 50 kJ/mol
Q.11Medium
The half-life of a first-order reaction is independent of the initial concentration. If t₁/₂ = 30 minutes for a reaction, the time for the concentration to reduce to 41th of initial value is:
Answer: C
For first-order reaction, [A]ₜ = [A]₀(21)^(t/t₁/₂). For [A]ₜ = 41[A]₀, we need (21)^(t/30) = 41, so t/30 = 2, giving t = 60 minutes
Q.12Medium
In the reaction 2A + B → C, if the concentration of A is doubled and B is tripled, the rate increases by 12 times. The rate law is:
In enzyme catalysis, the Michaelis constant (Km) represents:
Answer: B
Km is a characteristic constant for an enzyme-substrate pair, representing substrate concentration when v = Vmax/2
Q.16Medium
The temperature coefficient (Q₁₀) for a reaction is 2.5. If the rate at 300 K is r, then the rate at 320 K is approximately:
Answer: B
Q₁₀ = rate at (T+10)/rate at T. For 300K to 320K (two 10K intervals), rate = r × 2.5² = 6.25r
Q.17Medium
The pre-exponential factor (A) in the Arrhenius equation is related to:
Answer: C
The pre-exponential factor accounts for collision frequency, proper orientation (steric factor), and the Maxwell-Boltzmann energy distribution of molecules
Q.18Medium
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)
