#Kc calculation

CHEMISTRY

For the reaction 2COâ‚‚(g) 2CO(g) + Oâ‚‚(g), K_c = 0.01 at 1000 K. If initial [COâ‚‚] = 0.5 M, what is [CO] at equilibri

Given: For the reaction 2CO₂(g) 2CO(g) + O₂(g), K_c = 0.01 at 1000 K. If initial [CO₂] = 0.5 M, what is [CO] at equilibrium? These values define the system as per NCERT data. Formula: Let [CO] = 2x, [O₂] = x, [CO₂] = 0.5 - 2x. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: K_c = frac[CO]² [O₂][CO₂]² = (2x)² · x/(0.5 - 2x)² = 0.01 . 4x³/(0.5 - 2x)² = 0.01, solve: x approx 0.027, 2x approx 0.054 M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Thermodynamics, Equilibrium and Chemical Kinetics, Topic: Energetics, equilibrium constants and reaction rates.

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

For SO₂(g) + Cl₂(g) SO₂Cl₂(g) , Kc = 16 at 400 K. If 0.3 mol SO₂ and 0.2 mol Cl₂ are in a 1 L vessel, what is [SO₂Cl₂] a

Initial: [SO₂] = 0.3 M , [Cl₂] = 0.2 M , [SO₂Cl₂] = 0 . Let x = [SO₂Cl₂] , [SO₂] = 0.3 - x , [Cl₂] = 0.2 - x . Kc = ([SO₂Cl₂]/[SO₂][Cl₂]) = (x/(0.3 - x)(0.2 - x)) = 16 , x = 16 (0.06 - 0.5x + x²) , x ≈ 0.18 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 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

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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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For the reaction 2HI(g) H₂(g) + I₂(g) , if Kc = 0.04 at 700 K and [HI] = 0.2 M at equilibrium, what is [H₂] ?

Kc = ([H₂][I₂]/[HI]²) = 0.04 . Since [H₂] = [I₂] = x , 0.04 = (x²/(0.2)²) = (x²/0.04) , x² = 0.0016 , x = 0.04 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 2X(g) Y(g) , if Kp = 0.25 at 300 K ( R = 0.0831 bar L/mol K ), what is Kc ?

Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 300 = 24.93 . Thus, Kc = Kp × RT = 0.25 × 24.93 ≈ 6.23 .

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

In 2A(g) + B(g) 2C(g) , if Kc = 4 and the equilibrium mixture contains 0.2 mol A , 0.1 mol B , and 0.4 mol C in a 1 L ve

Initial equilibrium: [A] = 0.2 M , [B] = 0.1 M , [C] = 0.4 M , Kc = ((0.4)²/(0.2)²(0.1)) = 4 . After adding 0.1 mol A , [A] = 0.3 M , reaction shifts right to restore equilibrium.

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 2NO(g) + O₂(g) 2NO₂(g) , Kc = 100 at 300 K. If 0.2 mol NO and 0.1 mol O₂ are in a 1 L vessel, what is [NO₂] at equil

Initial: [NO] = 0.2 M , [O₂] = 0.1 M , [NO₂] = 0 . Let 2x = [NO₂] , [NO] = 0.2 - 2x , [O₂] = 0.1 - x . Kc = ([NO₂]²/[NO]²[O₂]) = ((2x)²/(0.2 - 2x)²(0.1 - x)) = 100 , (4x²/(0.2 - 2x)²(0.1 - x)) = 100 . Solving, x ≈ 0.09 , [NO₂] = 2 × 0.09 = 0.18 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 2A(g) B(g) , Kc = 0.25 at 500 K. If 0.4 mol A is in a 2 L vessel, what is the degree of dissociation?

Initial: [A] = 0.2 M , [B] = 0 . Let α be the degree of dissociation, [A] = 0.2 (1 - α) , [B] = 0.1α . Kc = ([B]/[A]²) = (0.1α/(0.2 - 0.2α)²) = 0.25 , α ≈ 0.36 .

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 A(g) 2B(g) , Kc = 0.16 at 400 K. If 0.5 mol A is in a 1 L vessel, what is [B] at equilibrium?

Initial: [A] = 0.5 M , [B] = 0 . Let 2x = [B] , [A] = 0.5 - x . Kc = ([B]²/[A]) = ((2x)²/0.5 - x) = 0.16 , 4x² = 0.08 - 0.16x , x ≈ 0.1 , [B] = 0.2 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 2NO₂(g) N₂O₄(g) , Kc = 200 at 298 K. If 0.1 mol NO₂ is placed in a 1 L vessel, what is [N₂O₄] at equilibrium?

Initial: [NO₂] = 0.1 M , [N₂O₄] = 0 . Let x = [N₂O₄] , [NO₂] = 0.1 - 2x . Kc = ([N₂O₄]/[NO₂]²) = (x/(0.1 - 2x)²) = 200 , x = 200 (0.1 - 2x)² , sqrtx = 14.14 (0.1 - 2x) , x ≈ 0.045 M (solving iteratively).

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 reaction H₂(g) + I₂(g) 2HI(g) , the equilibrium concentrations are [H₂] = 0.02 M , [I₂] = 0.02 M , and [HI] = 0.

Kc = ([HI]²/[H₂][I₂]) = ((0.16)²/(0.02)(0.02)) = (0.0256/0.0004) = 64 .

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

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