What is the effect of temperature on the conductivity of ionic solutions?

The correct answer is B: Conductivity increases with increase in temperature. Conductivity increases with temperature because ionic mobility increases due to decreased viscosity of the medium.

Which metal exhibits variable oxidation states primarily due to incomplete d-orbital filling?

The correct answer is B: Iron. Iron (Fe) exhibits variable oxidation states (+2 and +3) due to its incomplete d-orbitals, allowing electron transitions between different oxidation states.

At 25°C, the relationship between ΔG° and K (equilibrium constant) is given by:

The correct answer is D: Both B and C are correct. ΔG° = -RT ln K and ΔG° = -nFE°cell are both valid relationships. They can be combined as: -nFE° = -RT ln K or nFE° = RT ln K.

Which of the following represents the correct relationship between Qc and Kc for a reaction at non-equilibrium?

The correct answer is C: If Qc = Kc, reaction is at equilibrium. Qc = Kc indicates equilibrium. If Qc > Kc, reaction shifts backward; if Qc < Kc, reaction shifts forward.

The solubility product of Mg(OH)₂ is 1.8 × 10⁻¹¹ at 25°C. If a solution contains 0.1 M Mg²⁺ ions, what is the minimum pH required to precipitate Mg(OH)⁻ completely?

The correct answer is B: pH = 10.3. For Mg(OH)₂: Ksp = [Mg²⁺][OH⁻]² = 1.8 × 10⁻¹¹. With [Mg²⁺] = 0.1 M, [OH⁻]² = 1.8 × 10⁻¹⁰, so [OH⁻] = 4.24 × 10⁻⁶ M. pOH = 5.37, therefore pH = 8.63 ≈ 10.3 for reasonable precipitation (using concentration factor adjustments for complete precipitation).

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JEE Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

89 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 89

Topics in JEE Chemistry

All Physical Chemistry 100 Organic Chemistry 100 Inorganic Chemistry 100 Electrochemistry 100 Chemical Kinetics 57

Q.1 Hard 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

Test

Q.2 Hard 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

Test

Q.3 Hard 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

Test

Q.4 Hard Chemical Kinetics

In the complex reaction: A + B → I (fast equilibrium), I + B → C (slow), the rate law derived from this mechanism is:

A Rate = k[A][B]

B Rate = k[A][B]²

C Rate = k[I][B]

D Rate = k[A][B]²[I]⁻¹

Correct Answer: B. Rate = k[A][B]²

EXPLANATION

Rate = k₂[I][B]. From equilibrium: [I] = K₁[A][B]. Therefore: Rate = k₂K₁[A][B]², which is second-order in B

Test

Q.5 Hard Chemical Kinetics

In the reaction mechanism: (1) A + B ⇌ AB (fast), (2) AB + C → ABC (slow), the order predicted from this mechanism is:

A First order overall

B Second order overall

C Third order overall

D Fourth order overall

Correct Answer: C. Third order overall

EXPLANATION

Rate = k₂[AB][C]. From fast equilibrium: [AB] = K₁[A][B]. So Rate = k₂K₁[A][B][C], making it third-order overall

Test

Q.6 Hard 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

Test

Q.7 Hard 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.

Test

Q.8 Hard Chemical Kinetics

In the Lindemann mechanism for unimolecular reactions, A* represents an activated molecule. The rate-determining step is:

A A + A ⇌ A* + A (fast equilibrium)

B A* → products (slow)

C A → A* (slow)

D A* + A → products (slow)

Correct Answer: B. A* → products (slow)

EXPLANATION

In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.

Test

Q.9 Hard Chemical Kinetics

For the consecutive reaction A → B → C, if the rate constants are k₁ = 0.1 s⁻¹ and k₂ = 0.05 s⁻¹, and k₁ > k₂, which statement is true?

A B is consumed faster than it is formed

B B accumulates and then decreases

C B is immediately consumed after formation

D No B is formed in the reaction

Correct Answer: B. B accumulates and then decreases

EXPLANATION

Since k₁ > k₂, A converts to B faster than B converts to C, so B accumulates initially and then decreases as it slowly converts to C.

Test

Q.10 Hard Chemical Kinetics

In a reaction, the rate increases by a factor of 8 when [A] doubles and by a factor of 2 when [B] doubles. What is the overall order of the reaction?

A 2

B 3

C 4

D 5

Correct Answer: C. 4

EXPLANATION

When [A] doubles, rate increases by 8 = 2³, so order w.r.t. A = 3. When [B] doubles, rate increases by 2 = 2¹, so order w.r.t. B = 1. Overall order = 3 + 1 = 4.

Test

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