53 public questions tagged with this topic.
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
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
According to Le Chatelier’s principle, increasing pressure favors the side with fewer moles of gas. Here, 4 moles (reactants) become 2 moles (products), so it shifts right.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Initial: PN₂O₄ = (0.5 × 0.0831 × 400/1) = 16.62 bar . Let 2x = PNO₂ , PN₂O₄ = 16.62 - x , total pressure = 16.62 + x . Kp = ((PNO₂)²/PN₂O₄) = ((2x)²/16.62 - x) = 0.16 , 4x² = 0.16 (16.62 - x) , x ≈ 0.8 , PNO₂ = 1.6 bar .
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 , Ka = (Kw/Kb) = (1.0 × 10⁻¹⁴/1.8 × 10⁻⁵) ≈ 5.56 × 10⁻¹⁰ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let PC = 2x , PA = 1 - x , PB = 2 - 3x . Kp = ((PC)²/PA (PB)³) = ((2x)²/(1 - x)(2 - 3x)³) = 0.125 , 4x² = 0.125 (1 - x)(2 - 3x)³ , x ≈ 0.25 , PC = 0.5 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For CuS Cu²⁺ + S²⁻ , Ksp = [Cu²⁺][S²⁻] = 6.3 × 10⁻³⁶ , [S²⁻] = 6.3 × 10⁻²⁴ . For H₂S 2H+ + S²⁻ , K = Ka₁ × Ka₂ = 9.5 × 10⁻²⁷ , [S²⁻] = (K [H₂S]/[H+]²) , assume [H₂S] = 0.1 M , 6.3 × 10⁻²⁴ = (9.5 × 10⁻²⁷ × 0.1/[H+]²) , [H+]² = 1.51 × 10⁻⁴ , [H+] = 1.23 × 10⁻² , pH ≈ 1.91 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Let PC = 2x , PA = 2 - 2x , PB = 1 - x , total pressure = 3 - x . Kp = ((PC)²/PA² PB) = ((2x)²/(2 - 2x)² (1 - x)) = 16 , (4x²/4(1 - x)² (1 - x)) = 16 , (x²/(1 - x)³) = 4 , (x/1 - x) = 2 , x = 2 - 2x , 3x = 2 , x = (2/3) , PC = 2 × (2/3) = 1.33 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Initial: [A] = 1 M , [B] = 3 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by x , B by 2x . At equilibrium: [A] = 1 - x , [B] = 3 - 2x , [C] = 3x . Kc = ([C]³/[A][B]²) = ((3x)³/(1 - x)(3 - 2x)²) = 27 , (27x³/(1 - x)(3 - 2x)²) = 27 , (x³/(1 - x)(3 - 2x)²) = 1 . Solving, x³ = (1 - x)(3 - 2x)² , test x = 0.5 : (0.5)³ = 0.125 , (1 - 0.5)(3 - 1)² = 0.5 × 4 = 2 (not equal). Solving numerically, x ≈ 0.75 , [C] = 3 × 0.75 = 2.25 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Kc = ([C]²/[A][B]²) = 4 , 4 = ((0.8)²/[A](0.4)²) = (0.64/[A] × 0.16) , 4 × 0.16 [A] = 0.64 , 0.64 [A] = 0.64 , [A] = 1.0 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
