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JEE Chemistry

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

225 Q 3 Subjects 12th (PCM)

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All Mathematics 30 JEE Physics 0 JEE Chemistry 457

Q.1 Medium JEE Chemistry 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

EXPLANATION

For second-order: t₁/₂ = 1/(k[A]₀); 100 = 1/(k × 0.5); k = 0.04 M⁻¹s⁻¹

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Q.2 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

1/[A] = 2 + 0.4(5) = 2 + 2 = 4; [A] = 1/4 = 0.25 M

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Q.3 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

Catalyst lowers Eₐ (forward and reverse) without changing ΔH, reaction enthalpy change

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Q.4 Medium JEE Chemistry Chemical Kinetics

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]

EXPLANATION

For elementary reactions, rate law exponents = stoichiometric coefficients. Rate = k[A]²[B]

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Q.5 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

If doubling concentration increases rate by 4 times (2²), reaction is second order

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Q.6 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

For first-order: k = 0.693/t₁/₂ = 0.693/30 = 0.0231 min⁻¹ ≈ 0.023 min⁻¹

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Q.7 Medium JEE Chemistry Chemical Kinetics

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?

A 1.5

B 2

C 3

D 4

Correct Answer: B. 2

EXPLANATION

Q₁₀ = k(T+10)/k(T) = (6 × 10⁻²)/(3 × 10⁻²) = 2. For typical reactions, Q₁₀ = 2-3

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Q.8 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

Rate depends only on [A] (order = 1 w.r.t. A). Halving [A] decreases rate by factor of 2. [B] has no effect

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Q.9 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

Using ln(k₂/k₁) = (Eₐ/R)(T₂-T₁)/(T₁T₂); ln(2) = (Eₐ/8.314)(10/93000); Eₐ ≈ 50.4 kJ/mol

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Q.10 Medium JEE Chemistry Chemical Kinetics

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

EXPLANATION

The pre-exponential factor accounts for collision frequency, proper orientation (steric factor), and the Maxwell-Boltzmann energy distribution of molecules

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