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24 public questions tagged with this topic.

CHEMISTRY

For PCl₅(g) PCl₃(g) + Cl₂(g), K_c = 0.04 at 250°C. If 1 mole of PCl₅ is placed in a 10 L vessel, what is the equilibrium

Given: For PCl₅(g) PCl₃(g) + Cl₂(g), K_c = 0.04 at 250°C. If 1 mole of PCl₅ is placed in a 10 L vessel, what is the equilibrium concentration of Cl₂ ? These values define the system as per NCERT data. Formula: Initial [PCl₅] = 1 / 10 = 0.1 M. This is standard NCERT relation. Substitution & Calculation: Let x be the concentration of Cl₂ at equilibrium. Then, [PCl₅] = 0.1 - x, [PCl₃] = x, [Cl₂] = x . K_c = frac[PCl₃][Cl₂][PCl₅] = x · x/0.1 - x = 0.04 . x² = 0.04 (0.1 - x), x² + 0.04x - 0.004 = 0 . Solving: x = frac-0.04 pm sqrt(0.04)² + 4 · 0.0042 = 0.0336 M (positive root). Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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

For the reaction A(g) + 2B(g) 3C(g) , Kc = 27 at 400 K. If 1 mole of A and 3 moles of B are placed in a 1 L vessel, what

Initial: [A] = 1 M , [B] = 3 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by x , B by 2x . At equilibrium: [A] = 1 - x , [B] = 3 - 2x , [C] = 3x . Kc = ([C]³/[A][B]²) = ((3x)³/(1 - x)(3 - 2x)²) = 27 , (27x³/(1 - x)(3 - 2x)²) = 27 , (x³/(1 - x)(3 - 2x)²) = 1 . Solving, x³ = (1 - x)(3 - 2x)² , test x = 0.5 : (0.5)³ = 0.125 , (1 - 0.5)(3 - 1)² = 0.5 × 4 = 2 (not equal). Solving numerically, x ≈ 0.75 , [C] = 3 × 0.75 = 2.25 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

For CH₄(g) + H₂O(g) CO(g) + 3H₂(g) , Kc = 0.1 at 800 K. If 0.5 mol CH₄ and 0.5 mol H₂O are in a 2 L vessel, what is [H₂]

Initial: [CH₄] = [H₂O] = 0.25 M , [CO] = [H₂] = 0 . Let 3x = [H₂] , [CO] = x , [CH₄] = [H₂O] = 0.25 - x . Kc = ([CO][H₂]³/[CH₄][H₂O]) = (x (3x)³/(0.25 - x)²) = 0.1 , 27x⁴ = 0.1 (0.25 - x)² , x ≈ 0.06 , [H₂] = 0.18 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

For 2X(g) Y(g) + Z(g) , Kc = 0.01 at 400 K. If 1 mol X is in a 1 L vessel, what is the degree of dissociation at equilib

Initial: [X] = 1 M . Let α be the degree of dissociation, [X] = 1 - α , [Y] = [Z] = (α/2) . Kc = ([Y][Z]/[X]²) = (((α/2))²/(1 - α)²) = 0.01 , (α²/4(1 - α)²) = 0.01 , (α/1 - α) = 0.2 , α = 0.2 - 0.2α , 1.2α = 0.2 , α = 0.167 .

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) , Kc = 0.02 at 700 K. If 0.4 mol SO₃ is in a 2 L vessel, what is [O₂] at equilibrium?

Initial: [SO₃] = 0.2 M , [SO₂] = [O₂] = 0 . Let x = [O₂] , [SO₂] = 2x , [SO₃] = 0.2 - 2x . Kc = ([SO₂]²[O₂]/[SO₃]²) = ((2x)² x/(0.2 - 2x)²) = 0.02 , 4x³ = 0.02 (0.04 - 0.8x + 4x²) , x ≈ 0.016 M .

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 H₂(g) + I₂(g) 2HI(g) , Kc = 50 at 700 K. If 0.2 mol H₂ and 0.3 mol I₂ are in a 1 L vessel, what is [HI] at equilibri

Initial: [H₂] = 0.2 M , [I₂] = 0.3 M , [HI] = 0 . Let 2x = [HI] , [H₂] = 0.2 - x , [I₂] = 0.3 - x . Kc = ([HI]²/[H₂][I₂]) = ((2x)²/(0.2 - x)(0.3 - x)) = 50 , 4x² = 50 (0.06 - 0.5x + x²) , 46x² - 25x + 3 = 0 , x ≈ 0.15 , [HI] = 0.3 M .

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 COCl₂(g) CO(g) + Cl₂(g) , Kc = 0.16 at 700 K. If 0.2 mol COCl₂ and 0.1 mol CO are in a 1 L vessel, what is [Cl₂] at

Initial: [COCl₂] = 0.2 M , [CO] = 0.1 M , [Cl₂] = 0 . Let x = [Cl₂] , [COCl₂] = 0.2 - x , [CO] = 0.1 + x . Kc = ([CO][Cl₂]/[COCl₂]) = ((0.1 + x)x/0.2 - x) = 0.16 , 0.1x + x² = 0.032 - 0.16x , x² + 0.26x - 0.032 = 0 , x ≈ 0.11 M .

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

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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For the reaction 2A(g) + 2B(g) 3C(g) , Kc = 64 at 500 K. If 2 moles of A and 2 moles of B are placed in a 1 L vessel, wh

Initial: [A] = 2 M , [B] = 2 M , [C] = 0 . Let 3x be moles of C formed, so A and B decrease by 2x . At equilibrium: [A] = 2 - 2x , [B] = 2 - 2x , [C] = 3x . Kc = ([C]³/[A]²[B]²) = ((3x)³/(2 - 2x)² (2 - 2x)²) = (27x³/(2 - 2x)⁴) = 64 , (27x³/(2 - 2x)⁴) = 64 , (3x/2 - 2x) = 4 , 3x = 8 - 8x , 11x = 8 , x ≈ 0.727 , [C] = 3 × 0.727 ≈ 2.18 M .

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

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