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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 1–10 of 457
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Q.1 Hard JEE Chemistry Chemical Kinetics
A reaction has Eₐ = 60 kJ/mol. How many times faster will it be at 327°C compared to 27°C? (R = 8.314 J/mol·K, assume A constant)
A 100 times
B 256 times
C 500 times
D 1000 times
Correct Answer:  C. 500 times
EXPLANATION

Using ln(k₂/k₁) = (Eₐ/R)[(T₂-T₁)/(T₁T₂)]; ln(k₂/k₁) = (60000/8.314)[(300)/(600×300)] ≈ 6.2; k₂/k₁ ≈ 500

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Q.2 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.3 Hard JEE Chemistry Chemical Kinetics
For the reaction: (1) Cl₂ ⇌ 2Cl (fast), (2) Cl + H₂ → HCl + H (slow), (3) H + Cl₂ → HCl + Cl (fast). The overall reaction is:
A Cl₂ + H₂ → 2HCl
B 2Cl₂ + H₂ → 2HCl + Cl
C H₂ + Cl₂ → 2HCl
D Cl + H₂ → HCl + H
Correct Answer:  A. Cl₂ + H₂ → 2HCl
EXPLANATION

Adding all steps and canceling intermediates (Cl, H): Cl₂ + H₂ → 2HCl

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Q.4 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.5 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.6 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.7 Hard JEE Chemistry Chemical Kinetics
For a reaction with Eₐ = 50 kJ/mol and A = 2 × 10¹³ s⁻¹ (Arrhenius pre-exponential factor), calculate k at 300K (R = 8.314 J/mol·K)
A 2.3 × 10⁻² s⁻¹
B 1.8 × 10⁻¹ s⁻¹
C 3.4 × 10⁻³ s⁻¹
D 5.1 × 10⁻² s⁻¹
Correct Answer:  C. 3.4 × 10⁻³ s⁻¹
EXPLANATION

k = A·exp(-Eₐ/RT) = 2 × 10¹³ × exp(-50000/8.314×300) = 2 × 10¹³ × exp(-20.03) ≈ 3.4 × 10⁻³ s⁻¹

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Q.8 Easy JEE Chemistry Chemical Kinetics
Which factor does NOT affect the rate constant k of a reaction?
A Temperature
B Concentration of reactants
C Nature of reactants
D Presence of catalyst
Correct Answer:  B. Concentration of reactants
EXPLANATION

Rate constant k is independent of reactant concentration; it depends on T, nature of reactants, and catalyst

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Q.9 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.10 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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