Practice questions and explanations on converting between Kp and Kc, and applying Le Chatelier's principle to predict how changes in conditions affect equilibrium.
A Lewis acid accepts an electron pair. Fe³⁺ has vacant orbitals and can accept an electron pair.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/6.8 × 10⁻⁴) ≈ 1.47 × 10⁻¹¹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let PB = PC = x , PA = 2 - 2x , total pressure = 2 - 2x + 2x = 2 . Kp = (PB PC/(PA)²) = (x²/(2 - 2x)²) = 0.5 , (x/2 - 2x) = sqrt0.5 ≈ 0.707 , x ≈ 0.828 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Total volume = 3V, [X-] = (0.06 × 2V/3V) = 0.04 M , pKa = 4.7 , pH = 4.7 + log (0.04/0.04) = 4.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For AgBr Ag+ + Br- , Ksp = [Ag+][Br-] = 5.0 × 10⁻¹³ , [Br-] ≈ 0.05 , [Ag+] = (5.0 × 10⁻¹³/0.05) = 1.0 × 10⁻¹¹ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
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 PNOCl = 2x , PNO = 1 - 2x , PCl₂ = 0.5 - x . Kp = ((PNOCl)²/(PNO)² PCl₂) = ((2x)²/(1 - 2x)² (0.5 - x)) = 9 , (4x²/(1 - 2x)² (0.5 - x)) = 9 , x ≈ 0.45 , PNOCl = 0.9 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Initial: [NO] = (0.4/2) = 0.2 M , [N₂] = (0.1/2) = 0.05 M , [O₂] = 0 . Let x = [O₂] , [NO] = 0.2 - 2x , [N₂] = 0.05 + x . Kc = ([N₂][O₂]/[NO]²) = ((0.05 + x)x/(0.2 - 2x)²) = 0.01 , x ≈ 0.004 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰) ≈ 1.61 × 10⁻⁵ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For Ag₂CO₃ 2Ag+ + CO₃²⁻ , Ksp = [Ag+]²[CO₃²⁻] = 8.1 × 10⁻¹² , [CO₃²⁻] ≈ 0.01 , [Ag+]² = (8.1 × 10⁻¹²/0.01) = 8.1 × 10⁻¹⁰ , [Ag+] = 2.85 × 10⁻⁵ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Per Le Chatelier’s principle, increasing [A] shifts the equilibrium right to form more C and consume A .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let the pressure of X at equilibrium be 2p , X₂ = 2 - p , total pressure = (2 - p) + 2p = 2 + p = 3 , p = 1 . PX₂ = 1 atm , PX = 2 atm . Kp = ((PX)²/PX₂) = ((2)²/1) = 4 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
