Skip to content
Login Register

#equilibrium concentration

18 public questions tagged with this topic.

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

View Details
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

View Details
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

View Details
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

View Details
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

View Details
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

View Details
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

View Details
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

View Details
Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For the reaction CO(g) + Cl₂(g) COCl₂(g) , if Kc = 9 and initial concentrations are [CO] = 0.3 M , [Cl₂] = 0.3 M , what

Let [COCl₂] = x , [CO] = 0.3 - x , [Cl₂] = 0.3 - x . Kc = ([COCl₂]/[CO][Cl₂]) = (x/(0.3 - x)²) = 9 . Solving, sqrt(x/0.3 - x) = 3 , (x/0.3 - x) = 9 , x = 2.7 - 9x , 10x = 2.7 , x = 0.27 M .

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

View Details
Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For CO(g) + Cl₂(g) COCl₂(g) , Kc = 25 at 500 K. If 0.2 mol CO and 0.3 mol Cl₂ are in a 1 L vessel, what is [COCl₂] at eq

Initial: [CO] = 0.2 M , [Cl₂] = 0.3 M , [COCl₂] = 0 . Let x = [COCl₂] , [CO] = 0.2 - x , [Cl₂] = 0.3 - x . Kc = ([COCl₂]/[CO][Cl₂]) = (x/(0.2 - x)(0.3 - x)) = 25 , x ≈ 0.18 M .

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

View Details