The most stable conformation of ethane is:

The correct answer is B: Staggered. In staggered conformation, bonds are maximally separated reducing electron-electron repulsion and torsional strain. Staggered is most stable with ~12 kJ/mol lower energy than eclipsed.

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.

In the extraction of copper using electrorefining, what is the anode material?

The correct answer is B: Blister copper. Blister copper (impure copper from smelting) acts as the anode. Pure copper is the cathode. Impurities fall as anode mud.

In a Daniel cell, the mass of zinc electrode decreases and copper electrode increases. This indicates:

The correct answer is A: Zn acts as anode and Cu as cathode. Anode undergoes oxidation (Zn → Zn²⁺ + 2e⁻), losing mass. Cathode undergoes reduction (Cu²⁺ + 2e⁻ → Cu), gaining mass.

The pKa of phenol (C6H5OH) is approximately 10, while that of aliphatic alcohol is around 16. This difference is due to:

The correct answer is B: Resonance stabilization of phenoxide ion through benzene ring. Phenoxide ion is stabilized by resonance with the benzene ring, making phenol more acidic than aliphatic alcohols.

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JEE Chemistry
Chemical Kinetics

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

12 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–12 of 12

Topics in JEE Chemistry

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

Q.11 Hard Chemical Kinetics

The rate constant for a reaction increases 4 times when temperature increases from 27°C to 47°C. What is the activation energy? (R = 8.314 J mol⁻¹ K⁻¹)

A 25.6 kJ/mol

B 51.2 kJ/mol

C 102.4 kJ/mol

D 12.8 kJ/mol

Correct Answer: B. 51.2 kJ/mol

EXPLANATION

Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ - 1/T₂): ln(4) = (Eₐ/8.314)(1/300 - 1/320), solving gives Eₐ ≈ 51.2 kJ/mol.

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Q.12 Hard Chemical Kinetics

For a reaction with mechanism: A ⇌ B (fast equilibrium), B + C → D (slow), the rate law is:

A rate = k[B][C]

B rate = k[A][C]

C rate = k[A]^(1/2)[C]

D rate = k[A][B][C]

Correct Answer: C. rate = k[A]^(1/2)[C]

EXPLANATION

From fast equilibrium: K = [B]/[A], so [B] = K[A]. The slow step rate law is rate = k'[B][C] = k'K[A][C] = k[A]^(1/2)[C] where k combines constants.

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JEE Chemistry
Chemical Kinetics