JEE Main Chemistry Test- 2 - JEE MCQ

• Catalytic hydrogenation involves the addition of hydrogen (H₂) across the double bonds in alkenes, facilitated by a catalyst such as palladium, platinum, or nickel.
• The rate of hydrogenation is influenced by the stability of the alkene. Less substituted alkenes (fewer alkyl groups attached) react faster because they are less stable.
• Alkene A, being less substituted compared to the others, will react the fastest under catalytic hydrogenation conditions because it is less stable and thus more reactive.

• To find the molar entropy change (ΔS) during the melting of ice, use the formula:

• ΔS = ΔH/T

• Where:
  • ΔH is the enthalpy of fusion = 1.435 kcal/mol = 1435 cal/mol (since 1 kcal = 1000 cal)
  • T is the temperature in Kelvin = 0°C = 273.15 K

• Now, calculate ΔS:

• ΔS = 1435 cal/mol / 273.15 K ≈ 5.26 cal/(mol K)

• Thus, the correct answer is option A: 5.260 cal/(mol K).

Answer is D: -3273.26 kJ.

• In the given gaseous equilibrium, the reaction's favorability is influenced by Le Chatelier's Principle.
• High pressure is favorable when the reaction results in a decrease in gas moles. This principle asserts that the system will shift to reduce pressure.
• Low temperature favors exothermic reactions by shifting equilibrium to the product side.
• Thus, the correct conditions are high pressure for reducing gas volume and low temperature for exothermic reactions, making option A the correct choice.

1. Reaction: SO₂Cl₂ ⇌ SO₂ + Cl₂

2. Degree of dissociation (α): 90% = 0.9

3. Moles at equilibrium:

   • Moles of SO₂Cl₂ = 1 - 0.9 = 0.1
   • Moles of SO₂ = 0.9
   • Moles of Cl₂ = 0.9
   • Total moles = 0.1 + 0.9 + 0.9 = 1.9

4. Partial pressures:

   • Partial pressure of SO₂= 0.9 ÷ 1.9
   • Partial pressure of Cl₂ = 0.9 ÷ 1.9
   • Partial pressure of SO₂Cl₂ = 0.1 ÷ 1.9

5. Expression for Kp:
   K_p = (Partial pressure of SO₂ × Partial pressure of Cl₂) ÷ Partial pressure of SO₂Cl₂

6. Substitute values:
   Kp = (0.9 ÷ 1.9) × (0.9 ÷ 1.9) ÷ (0.1 ÷ 1.9)
   Kp = (0.81 ÷ 0.1) ÷ 1.9
   Kp = 8.1 ÷ 1.9 = 4.26

Answer: Kp = 4.26

• Hydrochloric acid (HCl) is a strong acid that completely dissociates in water.
• The pH of a solution is determined by the concentration of hydrogen ions H⁺.
• For strong acids like HCl, H⁺ is equal to the concentration of the acid.
• A pH of 7 is neutral, indicating neither acidic nor alkaline.
• A pH of 6.6990 suggests a slightly acidic solution, which is unusual for strong HCl.
• Therefore, the pH of HCl is typically around zero, reflecting its strong acidic nature.

• In a reversible chemical reaction at equilibrium, the equilibrium constant (K) is determined by the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients.
• The equilibrium constant is dependent only on temperature and not on the concentrations of reactants or products.
• Therefore, doubling the concentration of a reactant does not affect the value of the equilibrium constant.
• Hence, the equilibrium constant will remain the same.

The reactivity of hydrogen atoms in alkanes varies based on the type of carbon atom they are attached to.

• Tertiary hydrogen atoms are most reactive.
• Next in reactivity are Secondary hydrogen atoms.
• Primary hydrogen atoms are the least reactive.

This order of reactivity is due to the stability of the resulting radicals formed when the hydrogen is removed. Tertiary radicals are the most stable, followed by secondary, and then primary.

The reaction of ethene with Br₂ in CCl₄ produces 1,2-dibromoethane. This process involves the addition of bromine across the double bond of ethene.

• Ethene is a simple alkene with a double bond between two carbon atoms.
• When bromine (Br₂) is added, each bromine atom attaches to a different carbon atom.
• This reaction happens in a solvent like CCl₄, which does not participate in the reaction.
• The final product, 1,2-dibromoethane, is a saturated compound with bromine atoms on adjacent carbon atoms.

When isobutane is treated with chlorine in the presence of light, the major product formed is tert-butyl chloride. This reaction is an example of free radical halogenation, which tends to favour the formation of more stable carbocations.

• Isobutane is a branched alkane with a tertiary carbon.
• Chlorine reacts with the hydrogen attached to this tertiary carbon.
• This is because tertiary radicals are more stable than primary or secondary radicals.
• The major product is tert-butyl chloride due to this stability.

The compound involves elements with specific oxidation states:

• X has an oxidation number of +2.
• Y has an oxidation number of +5.
• Z has an oxidation number of -2.

To find the possible formula:

• The total oxidation number must equal zero for a neutral compound.
• Calculate the number of each atom needed to balance these charges.
• The correct formula balances the positive and negative charges from each element.

Hence, the appropriate formula is C.

• Sulfur dioxide (SO₂) can act as both an oxidizing and a reducing agent.
• As a reducing agent, it can donate electrons to other substances. For example, SO₂ is oxidized to sulfate (SO₄²⁻) in its reactions with stronger oxidizing agents.
• As an oxidizing agent, SO₂ can accept electrons, being reduced to elemental sulfur or sulfide (S²⁻) in reactions with strong reducing agents.
• This dual behavior makes SO₂ versatile in chemical reactions, hence option C is correct.

H₃PO₃ → 2H⁺ + HPO₃^–2

2H⁺ + 2OH^– → 2H₂O
ΔH = –55.84 × 2 = –116.68
Now
–106.68 = ΔH_ion – 55.84 × 2
ΔH_ion = 5KJ/mol

SF₄ has a see-saw geometry due to the presence of one lone pair on the sulfur atom.

• The most stable carbocation is the benzylcarbocation.
• Stability in carbocations is enhanced by resonance and hyperconjugation.
• Benzylcarbocation benefits from resonance with the aromaticring, allowing electrondelocalization.
• Delocalization spreads the positive charge over a larger area, reducing its intensity.
• In contrast, methyl and isopropylcarbocations lack significant resonance stabilization.
• The allylcarbocation is also resonance-stabilized but less than the benzylcarbocation.

• The strongest acid among the given options is HI (hydroiodic acid).
• Acid strength in hydrohalic acids (HF, HCl, HBr, HI) increases as the bond strength between hydrogen and the halogen decreases.
• HI has the weakest H-X bond due to the large size of iodine, making it easier to dissociate into H⁺ ions.
• As the bond strength decreases from HF to HI, the acidity increases: HF < HCl < HBr < HI.
• Therefore, HI is the strongest acid.

NH₄Cl  is a salt of strong acid and weak base for solutions of such salts.
pH = 1/2 [pK_W – log C – pK_b]
⇒ 10.26 = 14 – log C – 4.74
⇒ log C = 9.26 – 10.26 = –1.0
∴ C = 10^–1 M
[NH₄Cl] = 10^–1 M
= 10^–1 × 53.5 gL^–1
= 5.35 gL^–1

• To find the oxidation state of chromium in potassium dichromate K₂Cr₂O₇, start by knowing that the sum of all oxidation states in a compound is zero.
•  Potassium (K) has an oxidation state of +1, and there are two potassium atoms, contributing +2 in total.
•  Oxygen (O) has an oxidation state of -2, and there are seven oxygen atoms, contributing -14 in total.
•  Let the oxidation state of each chromium (Cr) atom be x. There are two chromium atoms, so 2x.
•  The equation is 2 + 2x - 14 = 0.
•  Solving gives 2x = +12, so x = +6.
•  Therefore, the oxidation state of chromium in K₂Cr₂O₇ is +6.