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

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

The Ksp of CuS is 6.3 × 10⁻³⁶ . What is the pH at which [Cu²⁺] = 1.0 × 10⁻¹² M in a saturated solution, given Ka of H₂S

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

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

In A(g) + 2B(g) 2C(g) , Kc = 4 and [A] = 0.1 M , [B] = 0.2 M , [C] = 0.4 M at equilibrium. What happens if temperature i

For an exothermic reaction, increasing temperature shifts equilibrium left, favoring reactants.

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

In a physical equilibrium H₂O(l) H₂O(g) , what happens if the temperature is increased?

Increasing temperature favors the endothermic process (evaporation), shifting the equilibrium towards the vapor phase.

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 CO(g) + 2H₂(g) CH₃OH(g) , Kc = 10 at 400 K. If 0.1 mol CO and 0.3 mol H₂ are in a 1 L vessel, what is [CH₃OH] at equ

Initial: [CO] = 0.1 M , [H₂] = 0.3 M , [CH₃OH] = 0 . Let x = [CH₃OH] , [CO] = 0.1 - x , [H₂] = 0.3 - 2x . Kc = ([CH₃OH]/[CO][H₂]²) = (x/(0.1 - x)(0.3 - 2x)²) = 10 . Solving iteratively, x ≈ 0.09 , 10 = (0.09/(0.01)(0.12)²) ≈ 625 (too high), adjust x ≈ 0.06 , (0.06/(0.04)(0.18)²) ≈ 46 (still high), x ≈ 0.03 , (0.03/(0.07)(0.24)²) ≈ 7.44 , close to 10.

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

A weak acid HZ has Ka = 2.0 × 10⁻⁴ . What is the pH of a 0.1 M solution of HZ ?

For HZ H+ + Z- , Ka = (x²/0.1 - x) ≈ (x²/0.1) = 2.0 × 10⁻⁴ . Solving, x² = 2.0 × 10⁻⁵ , x = 4.47 × 10⁻³ , pH = -log(4.47 × 10⁻³) ≈ 2.35 .

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 the reaction 3A(g) + B(g) 2C(g) , Kc = 8 at 500 K. If 1.5 moles of A and 0.5 moles of B are placed in a 1 L vessel,

Initial: [A] = 1.5 M , [B] = 0.5 M , [C] = 0 . Let 2x be moles of C formed, so A decreases by 3x , B by x . At equilibrium: [A] = 1.5 - 3x , [B] = 0.5 - x , [C] = 2x . Kc = ([C]²/[A]³[B]) = ((2x)²/(1.5 - 3x)³ (0.5 - x)) = 8 , (4x²/(1.5 - 3x)³ (0.5 - x)) = 8 . Solving iteratively, x ≈ 0.25 , (0.5)² / [(0.75)³ × 0.25] = 0.25 / 0.1055 ≈ 2.37 (adjust), x ≈ 0.4 , [C] = 2 × 0.4 = 0.8 M .

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

What is the conjugate base of H₂O according to the Bronsted-Lowry concept?

H₂O loses a proton ( H+ ) to form OH- , which is its conjugate base.

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

The Ka of HSO₄- is 1.2 × 10⁻² . What is the Kb of SO₄²⁻ at 298 K?

Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/1.2 × 10⁻²) = 8.33 × 10⁻¹³ .

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 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

A weak acid HX ( Ka = 4.0 × 10⁻⁶ ) is mixed with 0.03 M NaX in a 2:1 volume ratio (acid:salt). If [HX] = 0.08 M after mi

Total volume = 3V, [X-] = (0.03V/3V) = 0.01 M , pKa = 5.4 , pH = 5.4 + log (0.01/0.08) = 5.4 - 0.903 = 4.5 .

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

In 2A(g) + B(g) 3C(g) , Kc = 9 and [A] = 0.3 M , [B] = 0.1 M , [C] = 0.9 M at equilibrium. What happens if volume is hal

Halving volume doubles concentrations: Q = ((1.8)³/(0.6)² (0.2)) = 81 > 9 , shifts left.

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

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