Latest questions in this category.
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
Let [A₂] = [B₂] = x , [AB] = 0.5 . Kc = ([A₂][B₂]/[AB]²) = (x²/(0.5)²) = 0.0625 . Solving, x² = 0.0625 × 0.25 = 0.015625 , x = 0.125 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Initial: [A] = 0.6 M , [B] = [C] = 0 . Let α be the degree of dissociation, [A] = 0.6 (1 - α) , [B] = [C] = 0.3α . Kc = ([B][C]/[A]²) = ((0.3α)²/(0.6 - 0.6α)²) = 0.09 , α ≈ 0.25 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
The solubility of a gas in a liquid is proportional to its pressure above the liquid, as stated by Henry's Law.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
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
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
Adding C increases Q , shifts left to reduce C .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
The conjugate acid of NH₃ is formed by accepting a proton ( H+ ), resulting in NH₄+ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Total pressure = PA + PB + PC = 3 , PB + PC = 2 , Kp = ((PC)²/PA PB) = ((PC)²/1 · (2 - PC)) = 1 , PC² = 2 - PC , PC ≈ 1.414 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Let [R] = x , [P] = 1 - x , [Q] = 1 - x . Kc = ([R]/[P][Q]) = (x/(1 - x)²) = 2.0 . Solving, x = 2(1 - x)² , let y = 1 - x , 1 - y = 2y² , 2y² + y - 1 = 0 , y = 0.5 , x = 1 - 0.5 = 0.5 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Initial: [A] = (1/2) = 0.5 M , [B] = (4/2) = 2 M , [C] = 0 . Let 2x be moles of C formed, so A decreases by x , B by 3x . At equilibrium: [A] = 0.5 - x , [B] = 2 - 3x , [C] = x . Kc = ([C]²/[A][B]³) = ((x)²/(0.5 - x)(2 - 3x)³) = 125 . Solving, test x = 0.4 : ((0.4)²/(0.1)(0.2)³) = (0.16/0.0008) = 200 (too high), x = 0.35 , ((0.35)²/(0.15)(0.35)³) = (0.1225/0.0064) ≈ 19 (too low), x ≈ 0.38 , [C] = 0.38 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
The constant vapor pressure indicates that the rate of evaporation equals the rate of condensation, establishing a dynamic equilibrium between the liquid and its vapor.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
