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%)