JEE Chemistry - MCQ Practice Questions

Setting Up the ICE Table

For the reaction $2A(g)\rightleftharpoons B(g)$, we track pressures (in atm) using an ICE table:

**Step 1: Write the Expression for $K_p$

The equilibrium constant in terms of partial pressures is:

**Step 2: Substitute the Known Value of $K_p$

Given $K_p=0.5{\text{atm}}^{-1}$:

**Step 3: Solve for $x$

This gives $x=2$ or $x=\frac{1}{2}$.

Step 4: Apply Physical Constraint

If $x=2$, then $P_A=2-2(2)=-2\text{atm}$, which is physically impossible. Therefore, we must take $x=\frac{1}{2}=1\text{atm}$... wait — $x=0.5$.

Re-checking directly: $x=0.5$ gives $P_A=2-2(0.5)=1\text{atm}$, and

Result

The equilibrium pressure of $B$ is:

After 3 half-lives (30 min): ($\frac{1}{2}$)³ = $\frac{1}{8}$ = 12.5% remains

Using Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.$\frac{15}{0}$.1) = 4.74 + 0.48 = 5.22

In presence of 0.1 M Cl⁻: Ksp = [Ag⁺][Cl⁻]; 1.8 × 10⁻¹⁰ = s × 0.1; s = 1.8 × 10⁻⁹ M

ΔG° = -nFE°; For bivalent ion transfer: ΔG° = -2 × 96485 × 0.92 ≈ -177.6 kJ/mol

Benzene and toluene are both aromatic hydrocarbons with similar molecular structures, forming nearly ideal solutions

New rate = k(2[A])²(0.5[B]) = k × 4[A]² × 0.5[B] = 2k[A]²[B] = 2 × original rate

ΔG = ΔH - TΔS; For spontaneity ΔG < 0. Exothermic (ΔH < 0) with ΔS > 0 always gives ΔG < 0

At higher dilutions, weak electrolytes ionize more completely, increasing degree of ionization and thus molar conductivity

Initial: 2 atm of A. At equilibrium with α = 0.5: A dissociates to (1-0.5)×2 = 1 atm, producing B and C each at 0.5 atm. Total = 1 + 0.5 + 0.5 = 2 atm × (1 + α) = 2 × 1.5 = 3 atm

For weak acid: α = √(Ka/C) = √(1×10⁻⁵/0.1) = √(1×10⁻⁴) = 0.01 or 1%. Using exact formula: α ≈ 0.0316 (3.16%)

Williamson ether synthesis follows SN2 mechanism, which is favored by primary alkyl halides due to minimal steric hindrance. Reactivity: Primary > Secondary > Tertiary.

Aldol condensation requires an α-hydrogen. Acetaldehyde has α-hydrogens and undergoes aldol condensation. Benzaldehyde and formaldehyde lack α-hydrogens.

PCC oxidizes primary alcohols to aldehydes without further oxidation to carboxylic acids. KMnO4 and Na2Cr2O7 further oxidize aldehydes to carboxylic acids.

Basic hydrolysis of alkyl halides follows SN2 mechanism (bimolecular nucleophilic substitution) with OH- as nucleophile, especially for primary halides.

Nitration involves formation of NO2+ (nitronium ion) from HNO3 in presence of H2SO4. NO2+ acts as the electrophile in electrophilic aromatic substitution.

Dehydration follows Zaitsev's rule, producing the most substituted (stable) alkene. 2-methylbut-2-ene is the major product with the double bond at the tertiary position.

Cannizzaro reaction occurs with aldehydes lacking α-hydrogens. Acetaldehyde has α-hydrogens, so it undergoes aldol condensation instead of Cannizzaro reaction.

SN1 is a unimolecular reaction with rate = k[R-X]. It is first-order with respect to alkyl halide only, independent of nucleophile concentration.

Formic acid is most acidic (Ka = 1.8×10-4) due to favorable +I effect of hydrogen. Acidity decreases with increasing alkyl chain length due to +I effect: Formic > Acetic > Propionic.