229 public questions tagged with this topic.
log 2 → Ea ≈ 72.6 kJ mol⁻¹.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Rate of Reaction - Average and Instantaneous Rate and Rate Law
E°cell = (0.059/n) log Kc , 0.354 = (0.059/2) log Kc , log Kc = 12, Kc = 10¹² .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Δr G° = -n F E°cell = -2 × 96500 × 0.77 = -148610 J/mol = -148.61 kJ/mol .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Relative lowering = (p⁰ - p/p⁰) . (24 - 22.8/24) = (1.2/24) = 0.05 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Kw = [H+][OH-] = 1.0 × 10⁻¹² , [H+] = sqrt1.0 × 10⁻¹² = 1.0 × 10⁻⁶ , pH = -log(1.0 × 10⁻⁶) = 6 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Initial: [A] = 0.5 M , [B] = 2 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by x , B by 2x . At equilibrium: [A] = 0.5 - x , [B] = 2 - 2x , [C] = 3x . Kc = ([C]³/[A][B]²) = ((3x)³/(0.5 - x)(2 - 2x)²) = 27 , (27x³/(0.5 - x)(4 - 4x + 4x²)) = 27 , x³ = (0.5 - x)(1 - x)² , x ≈ 0.3 , [C] = 3 × 0.3 = 0.9 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Increasing pressure (halving volume) favors the side with fewer moles of gas. Here, 2 mol → 2 mol , Δ n = 0 , so there is no shift.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Kp = Kc (RT)Δ n , Δ n = 2 - 1 = 1 , RT = 0.0831 × 298 ≈ 24.76 . Thus, Kc = (Kp/RT) = (0.98/24.76) ≈ 0.04 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
At equilibrium: SO₂ = 2 - 1 = 1 M , O₂ = 1 - 0.5 = 0.5 M , SO₃ = 1 M . Kc = ([SO₃]²/[SO₂]²[O₂]) = ((1)²/(1)²(0.5)) = (1/0.5) = 2 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Let PC = 2x , PA = 0.5 - x , PB = 1 - 2x . Kp = ((PC)²/PA PB²) = ((2x)²/(0.5 - x)(1 - 2x)²) = 1 , 4x² = (0.5 - x)(1 - 2x)² , x ≈ 0.25 , PC = 0.5 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 500 = 41.55 . Thus, Kc = Kp × RT = 0.1 × 41.55 = 4.155 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Total pressure = PA + PB + PC = 5 , PB + PC = 3 , Kp = ((PC)²/PA² PB) = ((PC)²/4 (3 - PC)) = 16 , PC² = 192 - 64 PC , PC ≈ 2.85 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
