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JEE Chemistry

JEE Main MCQ questions — Mathematics, Physics, Chemistry for engineering entrance.

457 Q 3 Subjects 12th (PCM)

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Difficulty: All Easy Medium Hard 31–40 of 457

Q.31 Hard JEE Chemistry Chemical Kinetics

The rate constant for a reaction at 298 K is 2 × 10⁻⁵ s⁻¹ with Ea = 80 kJ/mol. What is the frequency factor (A) if rate = Ae^(-Ea/RT)?

A 1.5 × 10⁻² s⁻¹

B 8.0 × 10⁸ s⁻¹

C 3.2 × 10¹¹ s⁻¹

D 5.6 × 10¹³ s⁻¹

Correct Answer: D. 5.6 × 10¹³ s⁻¹

EXPLANATION

A = k/e^(-Ea/RT) = (2 × 10⁻⁵)/e^(-80000/2.478) ≈ 5.6 × 10¹³ s⁻¹

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Q.32 Easy JEE Chemistry Chemical Kinetics

For the reaction: A + B → Products with Rate = k[A][B]², what is the overall order of reaction?

A 1

B 2

C 3

D 4

Correct Answer: C. 3

EXPLANATION

Overall order = sum of exponents in rate law = 1 + 2 = 3 (third order reaction)

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Q.33 Easy JEE Chemistry Chemical Kinetics

In a reaction mechanism with fast pre-equilibrium followed by slow step, which statement is correct?

A The slow step determines the overall rate

B The fast step determines the overall rate

C Both steps equally contribute to rate

D The mechanism cannot be determined from rate law

Correct Answer: A. The slow step determines the overall rate

EXPLANATION

The rate-determining step (slowest step) controls the overall reaction rate, regardless of how many fast equilibrium steps precede it

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Q.34 Medium JEE Chemistry Chemical Kinetics

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 1/4th of initial value is:

A 30 minutes

B 45 minutes

C 60 minutes

D 90 minutes

Correct Answer: C. 60 minutes

EXPLANATION

For first-order reaction, [A]ₜ = [A]₀(1/2)^(t/t₁/₂). For [A]ₜ = 1/4[A]₀, we need (1/2)^(t/30) = 1/4, so t/30 = 2, giving t = 60 minutes

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Q.35 Easy JEE Chemistry Chemical Kinetics

A zero-order reaction has an initial concentration of 0.5 M and rate constant k = 0.02 M/s. The time taken for the concentration to reduce to 0.1 M is:

A 10 s

B 15 s

C 20 s

D 25 s

Correct Answer: C. 20 s

EXPLANATION

For zero-order reaction: [A]₀ - [A]ₜ = kt. So 0.5 - 0.1 = 0.02 × t, giving t = 20 s

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Q.36 Medium JEE Chemistry Chemical Kinetics

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:

A 50 kJ/mol

B 100 kJ/mol

C 150 kJ/mol

D 200 kJ/mol

Correct Answer: A. 50 kJ/mol

EXPLANATION

Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(5.0/1.7) = (Ea/8.314)(1/320 - 1/330), solving gives Ea ≈ 50 kJ/mol

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Q.37 Easy JEE Chemistry Chemical Kinetics

For the reaction 2NO + Cl₂ → 2NOCl, the rate law is found to be Rate = k[NO]²[Cl₂]. The order with respect to NO is:

A First order

B Second order

C Third order

D Negative order

Correct Answer: B. Second order

EXPLANATION

The exponent of [NO] in the rate law is 2, making the reaction second order with respect to NO

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Q.38 Medium JEE Chemistry Chemical Kinetics

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):

A 1.5

B 2.0

C 2.5

D 3.0

Correct Answer: B. 2.0

EXPLANATION

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

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Q.39 Medium JEE Chemistry Chemical Kinetics

Which of the following is an example of a homogeneous catalyst?

A Fe₃O₄ in ammonia synthesis

B H₂SO₄ in esterification reaction

C Pt surface in hydrogenation

D MnO₂ in decomposition of KMnO₄

Correct Answer: B. H₂SO₄ in esterification reaction

EXPLANATION

A homogeneous catalyst is in the same phase as reactants. H₂SO₄ (liquid) catalyzes esterification of reactants (liquid), making it homogeneous. Others are heterogeneous catalysts.

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Q.40 Hard JEE Chemistry Chemical Kinetics

For a pseudo-first-order reaction where [B]₀ >> [A]₀, the rate law simplifies to first-order even though the actual order is higher. This is because:

A [B] remains essentially constant throughout the reaction

B [A] is in large excess

C The reaction temperature is constant

D The rate constant becomes independent of [B]

Correct Answer: A. [B] remains essentially constant throughout the reaction

EXPLANATION

When [B]₀ >> [A]₀, the concentration of B doesn't change significantly during the reaction, so it can be incorporated into the rate constant, making the reaction appear first-order in A only.

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