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19 public questions tagged with this topic.

Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For 2A(g) B(g) + C(g) , Kp = 0.5 at 600 K. If the initial pressure of A is 2 atm, what is PB at equilibrium?

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For 2NO(g) + Cl₂(g) 2NOCl(g) , Kp = 9 at 500 K. If initial pressures are PNO = 1 atm , PCl₂ = 0.5 atm , what is PNOCl at

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For the equilibrium X₂(g) 2X(g) , if the initial pressure of X₂ is 2 atm and at equilibrium the total pressure is 3 atm,

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For N₂O₄(g) 2NO₂(g) , Kp = 0.16 at 400 K. If 0.5 moles of N₂O₄ are placed in a 1 L vessel, what is PNO₂ at equilibrium (

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For A(g) + 3B(g) 2C(g) , Kp = 0.125 at 600 K. If PA = 1 atm , PB = 2 atm initially, what is PC at equilibrium?

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

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

For 2A(g) + B(g) 2C(g) , Kp = 16 at 500 K. If initial pressures are PA = 2 atm , PB = 1 atm , what is PC at equilibrium?

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

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

For 2NO₂(g) 2NO(g) + O₂(g) , Kp = 0.05 at 600 K. If 1 mole of NO₂ is placed in a 2 L vessel, what is PO₂ at equilibrium

Initial: PNO₂ = (1 × 0.0831 × 600/2) = 24.93 bar . Let 2x dissociate, PNO₂ = 24.93 - 2x , PNO = 2x , PO₂ = x , total pressure = 24.93 + x . Kp = ((PNO)² PO₂/(PNO₂)²) = ((2x)² x/(24.93 - 2x)²) = 0.05 , 4x³ = 0.05 (24.93 - 2x)² , x ≈ 0.71 , PO₂ = 0.71 bar .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

For 2SO₂(g) + O₂(g) 2SO₃(g) , Kp = 100 at 800 K. If initial pressures are PSO₂ = 1 atm , PO₂ = 0.5 atm , what is PSO₃ at

Let PSO₃ = 2x , PSO₂ = 1 - 2x , PO₂ = 0.5 - x . Kp = ((PSO₃)²/PSO₂² PO₂) = ((2x)²/(1 - 2x)² (0.5 - x)) = 100 , (4x²/(1 - 2x)² (0.5 - x)) = 100 , (2x/1 - 2x) · (1/sqrt0.5 - x) = 10 , x ≈ 0.45 , PSO₃ = 0.9 atm .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

For 2SO₃(g) 2SO₂(g) + O₂(g) , Kp = 0.01 at 900 K. If 1 mole of SO₃ is placed in a 1 L vessel, what is the partial pressu

Initial: PSO₃ = (1 × 0.0831 × 900/1) = 74.79 bar . Let 2x mol of SO₃ dissociate, so PSO₃ = 74.79(1 - 2x) , PSO₂ = 74.79 × 2x , PO₂ = 74.79x , total pressure = 74.79. Kp = ((PSO₂)² PO₂/(PSO₃)²) = ((74.79 × 2x)² (74.79x)/[74.79(1 - 2x)]²) = 0.01 . Simplifying, (4x² · 74.79x/(1 - 2x)²) = 0.01 , 299.16x³ = 0.01 (1 - 2x)² . Solving, x ≈ 0.013 , PO₂ = 74.79 × 0.013 ≈ 0.972 bar .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

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Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

For N₂(g) + 3H₂(g) 2NH₃(g) , Kp = 4.0 × 10⁻³ at 600 K. If 1 mole of N₂ and 3 moles of H₂ are in a 2 L vessel, what is PN

Initial: PN₂ = (1 × 0.0831 × 600/2) = 24.93 bar , PH₂ = 74.79 bar . Let 2x be PNH₃ , PN₂ = 24.93 - x , PH₂ = 74.79 - 3x . Kp = ((PNH₃)²/PN₂ (PH₂)³) = ((2x)²/(24.93 - x)(74.79 - 3x)³) = 4.0 × 10⁻³ . Solving, x ≈ 0.8 , PNH₃ = 2 × 0.8 = 1.6 bar .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

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Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

For A(g) + 2B(g) 2C(g) , Kp = 0.25 at 400 K. If the total pressure at equilibrium is 4 atm and PA = 1 atm , what is PC ?

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

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Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

For A₂(g) + B₂(g) 2AB(g) , Kp = 9 at 600 K. If initial pressures are PA₂ = 1 atm , PB₂ = 1 atm , what is PAB at equilibr

Let 2x be PAB , PA₂ = 1 - x , PB₂ = 1 - x , total pressure = (1 - x) + (1 - x) + 2x = 2 . Kp = ((PAB)²/PA₂ PB₂) = ((2x)²/(1 - x)²) = 9 , (4x²/(1 - x)²) = 9 , (2x/1 - x) = 3 , 2x = 3 - 3x , 5x = 3 , x = 0.6 , PAB = 2 × 0.6 = 1.2 atm .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

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