#Kp calculation

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 2A(g) + B(g) 3C(g) , Kp = 2 at 700 K. If PA = 1 atm , PB = 0.5 atm initially, what is PC at equilibrium?

Let PC = 3x , PA = 1 - 2x , PB = 0.5 - x . Kp = ((PC)³/(PA)² PB) = ((3x)³/(1 - 2x)² (0.5 - x)) = 2 , 27x³ = 2 (1 - 2x)² (0.5 - x) , x ≈ 0.2 , PC = 0.6 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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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For N₂O₄(g) 2NO₂(g) , Kp = 0.16 at 298 K. If the initial pressure of N₂O₄ is 2 atm in a closed vessel, what is the total

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

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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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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For 2A(g) + B(g) 3C(g) , Kp = 9 at 500 K. If PA = 0.5 atm , PB = 0.5 atm , and PC = 1.5 atm at equilibrium, what happens

Initial: Q = ((1.5)³/(0.5)² (0.5)) = 27 > Kp = 9 , doubling pressure increases concentrations, Q increases further, shifts left.

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 A(g) + B(g) 2C(g) , Kp = 1 at 300 K. If the total pressure at equilibrium is 3 atm and PA = 1 atm , what is PC ?

Total pressure = PA + PB + PC = 3 , PB + PC = 2 , Kp = ((PC)²/PA PB) = ((PC)²/1 · (2 - PC)) = 1 , PC² = 2 - PC , PC ≈ 1.414 atm .

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

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Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For A(g) + 2B(g) 2C(g) , Kp = 1 at 400 K. If PA = 0.5 atm , PB = 1 atm initially, what is PC at equilibrium?

Let PC = 2x , PA = 0.5 - x , PB = 1 - 2x . Kp = ((PC)²/PA PB²) = ((2x)²/(0.5 - x)(1 - 2x)²) = 1 , 4x² = (0.5 - x)(1 - 2x)² , x ≈ 0.25 , PC = 0.5 atm .

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

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Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis

For 2A(g) + B(g) 2C(g) , Kp = 16 at 400 K. If the total pressure at equilibrium is 5 atm and PA = 2 atm , what is PC ?

Total pressure = PA + PB + PC = 5 , PB + PC = 3 , Kp = ((PC)²/PA² PB) = ((PC)²/4 (3 - PC)) = 16 , PC² = 192 - 64 PC , PC ≈ 2.85 atm .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis

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