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#Kc value

4 public questions tagged with this topic.

Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

For X₂(g) 2X(g) , Kc = 0.25 at 500 K. If 0.4 mol X₂ is in a 2 L vessel, what is the degree of dissociation at equilibriu

Initial: [X₂] = (0.4/2) = 0.2 M . Let α be the degree of dissociation, [X₂] = 0.2(1 - α) , [X] = 0.2 × 2α = 0.4α . Kc = ([X]²/[X₂]) = ((0.4α)²/0.2(1 - α)) = (0.16α²/0.2(1 - α)) = 0.25 , 0.8α² = 0.25(1 - α) , 0.8α² + 0.25α - 0.25 = 0 , α ≈ 0.5 .

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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Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For the reaction 2A(g) + B(g) 3C(g) , Kc = 64 at 300 K. If 2 moles of A and 1 mole of B are placed in a 2 L vessel, what

Initial: [A] = (2/2) = 1 M , [B] = (1/2) = 0.5 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by 2x , B by x . At equilibrium: [A] = 1 - x , [B] = 0.5 - x , [C] = 1.5x . Kc = ([C]³/[A]²[B]) = ((1.5x)³/(1 - x)²(0.5 - x)) = 64 . Simplifying, (3.375x³/(1 - x)²(0.5 - x)) = 64 . Trial: x = 0.4 , [A] = 0.6 , [B] = 0.1 , [C] = 0.6 , ((0.6)³/(0.6)²(0.1)) = (0.216/0.036) = 6 , too small. Solving approximately, x ≈ 0.48 , [C] = 1.5 × 0.48 = 0.72 M .

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

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Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis

For A₂(g) 2A(g) , Kc = 0.36 at 300 K. If 0.5 mol A₂ is in a 1 L vessel, what is the degree of dissociation?

Initial: [A₂] = 0.5 M , [A] = 0 . Let α be the degree of dissociation, [A₂] = 0.5 (1 - α) , [A] = α . Kc = ([A]²/[A₂]) = ((α)²/0.5 (1 - α)) = 0.36 , α² = 0.18 (1 - α) , α ≈ 0.36 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis

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