36 public questions tagged with this topic.
Removing B decreases its concentration, shifting the equilibrium right to produce more B , per Le Chatelier’s principle.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let PNO₂ = 2x , PN₂O₄ = 2 - x , total pressure = 2 - x + 2x = 2 + x . Kp = ((PNO₂)²/PN₂O₄) = ((2x)²/2 - x) = 0.16 , (4x²/2 - x) = 0.16 , 4x² = 0.32 - 0.16x , 4x² + 0.16x - 0.32 = 0 , x² + 0.04x - 0.08 = 0 , x = (-0.04 pm sqrt0.0016 + 0.32/2) , x ≈ 0.27 . Total pressure = 2 + 0.27 = 2.27 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Initial: [A] = 1.5 M , [B] = 0.5 M , [C] = 0 . Let 2x be moles of C formed, so A decreases by 3x , B by x . At equilibrium: [A] = 1.5 - 3x , [B] = 0.5 - x , [C] = 2x . Kc = ([C]²/[A]³[B]) = ((2x)²/(1.5 - 3x)³ (0.5 - x)) = 8 , (4x²/(1.5 - 3x)³ (0.5 - x)) = 8 . Solving iteratively, x ≈ 0.25 , (0.5)² / [(0.75)³ × 0.25] = 0.25 / 0.1055 ≈ 2.37 (adjust), x ≈ 0.4 , [C] = 2 × 0.4 = 0.8 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Halving volume doubles concentrations: Q = ((1.8)³/(0.6)² (0.2)) = 81 > 9 , shifts left.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Initial: [X] = 1 M . Let α be the degree of dissociation, [X] = 1 - α , [Y] = [Z] = (α/2) . Kc = ([Y][Z]/[X]²) = (((α/2))²/(1 - α)²) = 0.01 , (α²/4(1 - α)²) = 0.01 , (α/1 - α) = 0.2 , α = 0.2 - 0.2α , 1.2α = 0.2 , α = 0.167 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Total pressure = PA + PB + PC = 4 , PB + PC = 3 . Kp = ((PC)²/PA (PB)²) = ((PC)²/1 · (3 - PC)²) = 0.25 , PC = 0.5 (3 - PC) , PC = 1.5 - 0.5 PC , 1.5 PC = 1.5 , PC = 1 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Initial: [A] = 0.4 M , [B] = 0 . Let 2x = [B] , [A] = 0.4 - x . Kc = ([B]²/[A]) = ((2x)²/0.4 - x) = 0.25 , 4x² = 0.1 - 0.25x , x ≈ 0.125 , [B] = 0.25 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Let PC = 2x , PA = 2 - 2x , PB = 1 - x . Kp = ((PC)²/PA² PB) = ((2x)²/(2 - 2x)² (1 - x)) = 4 , (4x²/(4 - 4x + 4x²)(1 - x)) = 4 , x ≈ 0.8 , PC = 1.6 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Adding an inert gas at constant volume does not change the partial pressures or concentrations of the reactants and products, so the equilibrium position remains unaffected.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Initial equilibrium: [A] = 0.2 M , [B] = 0.1 M , [C] = 0.4 M , Kc = ((0.4)²/(0.2)²(0.1)) = 4 . After adding 0.1 mol A , [A] = 0.3 M , reaction shifts right to restore equilibrium.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Initial: [NO] = 0.2 M , [O₂] = 0.1 M , [NO₂] = 0 . Let 2x = [NO₂] , [NO] = 0.2 - 2x , [O₂] = 0.1 - x . Kc = ([NO₂]²/[NO]²[O₂]) = ((2x)²/(0.2 - 2x)²(0.1 - x)) = 100 , (4x²/(0.2 - 2x)²(0.1 - x)) = 100 . Solving, x ≈ 0.09 , [NO₂] = 2 × 0.09 = 0.18 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Increasing pressure (decreasing volume) favors the side with fewer moles of gas. Here, 2 mol → 2 mol , so Δ n = 0 , and there is no shift.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
