Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

Which of the following is a Lewis acid?

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ka of HF is 6.8 × 10⁻⁴ . What is the Kb of F- at 298 K?

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

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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

A weak acid HX ( Ka = 2.0 × 10⁻⁵ ) is mixed with 0.06 M NaX in a 1:2 volume ratio (acid:salt). If [HX] = 0.04 M after mi

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ksp of AgBr is 5.0 × 10⁻¹³ . What is [Ag+] in a saturated solution with 0.05 M NaBr ?

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

In the equilibrium A(g) 2B(g) , what happens if B is removed at constant volume?

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

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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 2NO(g) N₂(g) + O₂(g) , Kc = 0.01 at 300 K. If 0.4 mol NO and 0.1 mol N₂ are in a 2 L vessel, what is [O₂] at equilib

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ka of HCN is 6.2 × 10⁻¹⁰ . What is the Kb of CN- at 298 K?

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ksp of Ag₂CO₃ is 8.1 × 10⁻¹² . What is [Ag+] in a saturated solution with 0.01 M Na₂CO₃ ?

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

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

In the equilibrium 2A(g) + B(g) 3C(g) , increasing the concentration of A shifts the equilibrium towards which direction

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

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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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