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#reaction equilibrium

6 public questions tagged with this topic.

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

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Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For the equilibrium P(g) + Q(g) R(g) , if Kc = 2.0 and initial moles of P and Q are 1 each in a 1 L vessel, what is [R]

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

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Chemical Equilibrium - Law of Mass Action and Equilibrium Constant

For the reaction A(g) + 3B(g) 2C(g) , Kc = 125 at 400 K. If 1 mole of A and 4 moles of B are placed in a 2 L vessel, wha

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

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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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