Chemistry carries the highest scoring potential in JEE for anyone who keeps the three branches separate in revision. This set covers physical chemistry numericals, organic reaction mechanisms and named reactions, and inorganic chemistry including periodic trends, chemical bonding and coordination compounds. Organic questions show the mechanism arrow by arrow, so the reasoning transfers to reactions you have not seen before.
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.
Q.202Medium
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.203Medium
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.204Medium
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.205Medium
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.206Medium
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.207Medium
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.208Medium
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.212Medium
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.213Medium
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.214Medium
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)
