Govt. Exams
Entrance Exams
Using ICE table: Let x be change. At equilibrium: Kc = (2x)²/((1-x)(1-x)) = 50. Solving: 4x² = 50(1-x)², gives x ≈ 0.82. [HI] = 2x ≈ 1.64 M.
Rate law depends on the rate-determining step (slowest step). Step 1 is slow: A + B → C, so Rate = k[A][B]. The fast step doesn't appear in the rate law unless intermediates need elimination.
For first-order reaction: t₁/₂ = 0.693/k = 0.693/(1.4 × 10⁻³) ≈ 495 min. Half-life is independent of initial concentration for first-order reactions.
Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). With k₂/k₁ = 2, T₁ = 300K, T₂ = 310K: ln(2) = (Ea/8.314)(1/300 - 1/310). Solving gives Ea ≈ 52.85 kJ/mol.
The rate-determining step (slowest step) controls the overall reaction rate. The RDS is the bottleneck that limits how fast reactants can be converted to products.
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Cu²⁺/Cu is cathode (reduction), Zn²⁺/Zn is anode (oxidation).
The integrated rate law for a second-order reaction is 1/[A] = kt + 1/[A]₀. Option B shows rate law (differential form), option D shows first-order integrated rate law.
Moles of acid = 0.1 × 0.1 = 0.01 mol. Moles of base = 0.05 × 0.1 = 0.005 mol. After reaction: acid = 0.005 mol, conjugate base = 0.005 mol. Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74. (Slight adjustment needed: pH ≈ 4.07)
For AgCl: Ksp = [Ag⁺][Cl⁻] = s². Therefore, s = √Ksp = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L
For isothermal expansion: ΔS = nR ln(V₂/V₁) = 1 × 8.314 × ln(8/2) = 8.314 × ln(4) = 8.314 × 1.386 = 11.53 J/(mol·K)