Ncert Solutions Chemistry Class 11th

(a) The expression for K_p for the reaction is:

Kp = (pCO) x (pH₂) / (pCH₄) x (pH₂O)

(b) (i) By increasing the pressure, the number of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of K_p will decrease.

(ii) If the temperature is increased, according to Le Chatelier's principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of K_p will increase.

(iii) The addition of catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.

For the equilibrium reaction

                            2HBr (g) ? H₂(g) + Br₂(g) , the equilibrium constant is K = 1/1.6*10⁵.

Initial pressure     10               0           0

At equilibrium      10-p          p/2        p/2

The equilibrium constant expression is
            K_p=P_H2P_Br2 / (P_HBr)²=(p/2)*(p/2) / (10−p)²=1 / 1.6*10⁵
            p² / 4(10−p)²=1 / 1.6*10⁵
                        400p= 20−2p

                              p= 0.0498 bar

The equilibrium pressures are
P_H2=P_Br2=0.0498 / 2=0.0249 bar
P_HBr=10−0.0498=10 bar

Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different (i.e. when n_p ≠ n_r) (gaseous). In general,

• The reaction will go to the left if n_p> n_r.
• The reaction will go to the right if n_r> n_p.
  Keeping this in mind,

(i) Increase in pressure will favour backward reaction because n_p  (2) > n_r (1)

(ii) Increase in pressure will not affect equilibrium because n_p = n_r = 3.

(iii) Increase in pressure will favour backward reaction because n_p  (2) > n_r  (1)

(iv) Increase in pressure will favour forward reaction because n_p  (1) < n_r  (2)

(v) Increase in pressure will favour backward reaction because n_p  (1) > n_r (0).

(vi) Increase in pressure will favour backward reaction because n_p  (10) > n_r  (9).

Let the mixture has 100g as total mass.
So, The masses of CO and CO₂? are 90.55g and 100 − 90.55 = 9.45 g respectively.
Therefore, te number of moles of CO

                        n= 90.55 / 28 ?= 3.234.
The number of moles of CO₂? 

                        n= 9.45 / 44 ?= 0.215.
The mole fraction of CO

                         = 3.234? / (3.234+0.215) =0.938.
The mole fraction of CO₂?  

                         = 1−0.938=0.062.
The partial pressure of CO is the product of the mole fraction of CO and the total pressure.
It is 0.938*1=0.938 atm.
The partial pressure of carbon dioxide is 0.062*1=0.042 atm.
The expression for the equilibrium constant is: 
K_p?=? (P_CO)² / (P_CO2) ?=(0.938)² / (0.062) ?=14.19
Δn_g?= 2−1=1
K_c?= K_p?(RT)^−Δn=14.19*(0.0821*1127)⁻¹= 0.153.

 Let x moles of BrCl decompose in order to attain the equilibrium. The initial molar concentration and the molar concentration at equilibrium point of different species may be represented as follows:
2BrCl₂?(g) ?→?Ar_2?(g)+Cl₂?(g)

3.3*10⁻³m        0             0

3.3*10⁻³−2αα?             α

∴ Kc?= (α*α)? /(3.3*10−3−2α)²

⇒ α² / (3.3*10−3−2α)²?=32

⇒ α / (3.3*10−3−2α)^2?=4

⇒ α=18.67*10−3−8 α

⇒ (1+ 8 α =18.67*10⁻³

⇒ α= 1.5162*10⁻³

∴ Molar concentration of BrCl at equation 

[BrCl] =3.3*10⁻³−2α

=3.3*10⁻³−2*1.562*10⁻³

=0.267*10⁻³M