#gas equilibrium

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 2NOCl(g) 2NO(g) + Cl₂(g) , Kp = 1.8 × 10⁻² at 500 K. If 2 moles of NOCl are placed in a 2 L vessel, what is the tota

Initial: PNOCl = (2 × 0.0831 × 500/2) = 41.55 bar . Let 2x mol dissociate, PNOCl = 41.55(1 - x) , PNO = 41.55x , PCl₂ = 41.55 × (x/2) , total pressure = 41.55(1 - x + x + (x/2)) = 41.55(1 + (x/2)) . Kp = ((PNO)² PCl₂/(PNOCl)²) = ((41.55x)² (41.55 (x/2))/[41.55(1 - x)]²) = 1.8 × 10⁻² , (41.55x³/2(1 - x)²) = 0.018 , x³ = 8.67 × 10⁻⁴ (1 - x)² , x ≈ 0.09 , total pressure = 41.55 × 1.045 = 43.42 bar .

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 the equilibrium A(g) B(g) + C(g) , if the initial pressure of A is 1 atm and at equilibrium the total pressure is 1.

Let PB = PC = p , PA = 1 - p , total pressure = (1 - p) + p + p = 1 + p = 1.5 , p = 0.5 . PA = 0.5 atm , PB = PC = 0.5 atm . Kp = (PB PC/PA) = ((0.5)(0.5)/0.5) = 0.5 .

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

For 2AB(g) A₂(g) + B₂(g) , Kp = 0.25 at 600 K. If the initial pressure of AB is 4 atm, what is PA₂ at equilibrium?

Let PA₂ = PB₂ = x , PAB = 4 - 2x , total pressure = 4 - 2x + 2x = 4 . Kp = (PA₂ PB₂/(PAB)²) = (x²/(4 - 2x)²) = 0.25 , (x/4 - 2x) = 0.5 , x = 2 - x , 2x = 2 , x = 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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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For 2A(g) B(g) + C(g) , Kp = 0.05 at 400 K. If the total pressure at equilibrium is 2 atm and PA = 1.6 atm , what is PB

Total pressure = PA + PB + PC = 2 , PB = PC (stoichiometry), 1.6 + PB + PB = 2 , 2PB = 0.4 , PB = 0.2 atm . Check: Kp = (PB PC/PA²) = ((0.2)(0.2)/(1.6)²) = (0.04/2.56) ≈ 0.0156 (slightly off, but closest).

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

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Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For the equilibrium 2A(g) B(g) + C(g) , if Kc = 0.04 and [A] = 0.4 M at equilibrium, what is [B] ?

Kc = ([B][C]/[A]²) = 0.04 , [B] = [C] = x , 0.04 = (x²/(0.4)²) = (x²/0.16) , x² = 0.0064 , x = 0.08 M .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

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