Class 11th

(i) The concentration ratio (Concentration quotient), Qc for the reaction is:

Q_c = [CH₃COOC₂H₅] [ H₂O] / [CH₃COOH] [C₂H₅OH]

(ii)

Applying

K_c = [CH₃COOC₂H₅] [H₂O] / [CH₃COOH] [C₂H₅OH]

= (0.171 mol) x (0.171 mol) / (0.829 mol) (0.009 mol)

 = 3.92

(iii)

Q_c = [CH₃COOC₂H₅] [H₂O] / [CH₃COOH] [C₂H₅OH]

     = (0.214 mol) x (0.214 mol) / (0.786 mol) (0.286 mol)

     = 0.204

Since Q_c is less than K_c,  this means that the equilibrium has not been reached. The reactants are still taking part in the reaction to form the products.

The equilibrium reaction is

C_2?H₆?(g) ? C₂?H₄?(g)+H₂?(g).

The expression for the equilibrium constant is Kp?= (?P_C2?H4?) (P_H2) / P_C2?H6?.
Substituting the values in the above equation, we get
                        0.04=x² / (4−x)

? or                         x=0.38
Thus, the pressure of ethane is, P_C2?H6?=3.62atm.

Suppose at equilibrium, the molar concentration of both I₂ and Cl₂ is x mol L^-1.

 K_c = [I₂] [ Cl₂] / [ICl]²= x² / (0.78 – 2x)²

=>x/ (0.78 – 2x) = (0.14)^1/2 = 0.374

=> x= 0.167

[ICl] = (0.78 – 2 x 0.167) = (0.78 – 0.334) = 0.446 M

[I₂] = 0.167 M,

[Cl₂] = 0.167 M

H₂ (g) + I₂ (g)? 2HI (g); K=64, T=700K

2HI? H₂ + I₂ K=1/64700K

  a         0      0

a (1−α)   aα/2   aα/2

0.5         x        x

x² / (0.5)²  = 1 / 54.8

x² =   0.25 / 54.8

x =  = 0.068 M

At equilibrium, [H₂] = [I₂] = 0.068 M

Number of moles of water originally present = 1 mol
Percentage of water reacted= 40%
Number of moles of water reacted= 1 x 40/100 = 0.4 mol
Number of moles of water left= (1 – 0.4) = 0.6 mole

According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar conc., per litre of the reactants and products before the reaction and at the equilibrium point are as follows:

Equilibrium constant, K_c= [H₂] [CO₂] / [H₂O] [CO]

= [ (0.4/10) x (0.4/10)] / [ (0.6/10) x (0.6/10)]

= 0.16 / 0.36 = 0.44

Balanced chemical equation for the reaction is

4 NO (g) + 6 H₂O (g)? 4 NH₃ (g) + 5 O₂ (g)

According to the given equation, concentration quotient,

Qc = [NH₃]² / [N₂] [ H₂]³

= (8.13/20 mol L^-1)² / (1.57 / 20 mol L^-1) x (1.92 / 20 mol L^-1)³

= 2.38 x 10³

The equilibrium constant (K_c) for the reaction = 1.7 x 10^-2

As Q_c ≠ K_c, the reaction is not in a state of equilibrium. 

pHI = 0.04 atm, pH₂ = 0.08 atm, pI₂ = 0.08 atm

K_p= (pH₂ x pI₂) / (p_HI)² = (0.08 x 0.08) / (0.04 x 0.04)

= 4.0

Kp = Kc (RT)^{? ng}

=> Kc = Kp (RT)^{-? ng}

Putting the values of Kp= 2.0x10¹⁰ bar^-1, R= 0.083 L bar K^-1 mol^-1, T = 450 K, and? ng = 2-3= -1

=>  Kc = (2.0 x 10¹⁰ bar^-1) x [ (0.083 L bar K^-1 mol^-1) x (450 K)]^{- (-1)}

= 7.47 x 10¹¹ mol^-1 L

= 7.47 x 10¹¹ M^-1