Equilibrium MCQs for NEET — Chemistry Questions with Answers

(c) Addition of inert gas at constant volume condition to equilibrium has no effect.

• Addition of an inert gas at constant volume:
  When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

• Addition of an inert gas at constant pressure:
  When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

(a) Higher is the value K, more are the products formed.

(d) Any change in the concentration, pressure and temperature of the reaction results in change in the direction of equilibrium. This change in the direction of equilibrium is governed by Le-Chateliers Principle. According to LeChatelier's principle, equilibrium shifts in the opposite direction to undo the change.

N₂ (g) + 3H₂(g) $\rightleftharpoons$2NH₃ (g) + heat

(a) Increasing the concentration of NH₃ (g): On Increasing the concentration of NH₃ (g), the equilibrium shifts in the backward direction, where the concentration of NH₃ (g) decreases.

(b) Decreasing the pressure: Since p $\propto$n (number of moles), therefore, equilibrium shifts in the backward direction, where the number of moles are increasing.

(c) Decreasing the concentration of N₂ (g)and H₂(g): Equilibrium shifts in the backward direction when the concentraion of N₂ (g)and H₂(g) decrease.

(d) Increasing pressure and decreasing temperature: On increasing pressure, equilibrium shifts in the forward direction, where number of moles decreases. it is an example of exothermic reaction therfore decreasing temperature favours the forward direction.

 (b)    K₁ = [H⁺][H₂P${\mathrm{O}}_4^{-}$]/[H₃PO₄];       K₂ = [H⁺][ HP${\mathrm{O}}_4^{2-}$]/[H₂P${\mathrm{O}}_4^{-}$] ;

   K₃ = [H⁺][P${\mathrm{O}}_4^{3-}$]/[H₂P${\mathrm{O}}_4^{-}$] Multiplying these three

   K₁ x K₂ x K₃ = [H⁺]³[P${\mathrm{O}}_4^{3-}$]/[H₃PO₄]

(a) K_c1 = 4.9x10-2 = $\frac{\left[{\mathrm{SO}}_2\right][{\mathrm{O}}_2{]}^{1/2}}{\left[{\mathrm{SO}}_3\right]}$

     K_c2 = $\frac{[{\mathrm{SO}}_3{]}^2}{\left[{\mathrm{SO}}_2{]}^2\right[{\mathrm{O}}_2]}$=[1/K_c1]² = [1/4.9x10-²]² = 416.5

(a) [Ag⁺]²[${\mathrm{CrO}}_4^{2-}$]     = K_sp or K_sp = (2s)² x s = 4s³

(b) Given, pH of Ba(OH)₂ = 12.

.^.. pOH = 14-pH = 14-12 = 2

We know that,  pOH = -log[OH]^-

                           2= -log[OH]^-

                       [OH]^- = antilog(-2)

                       [OH]^- = 1 x 10^-2 

Ba(OH)² dissolves in water as:

 Ba(OH)² (s) $\rightleftharpoons$ Ba²⁺ + 2OH^-

 S mol L-1           S        2S

.^.. [OH]^- = 2S = 1 x 10^-2

S=[OH]^-/2                             [Ba²⁺ =S]

Ba²⁺ = [OH]^-/2 = (1 x 10^-2)/2

K_sp = [Ba²⁺][OH^-]²

      =[(1 x10^-2)/2](1 x 10^-2)2

      = 0.5 x 10^-6 = 5 x 10^-7

(c) $K_{{\mathit{c}}_{\mathit{1}}}$ = 1/$K_{c_{\mathit{2}}}$ = 1/(2.4 x 10^-3) = 4.2 x 10²

(c) 2HI $\rightleftharpoons$H₂ + I₂;

           Kc = $\frac{{\alpha }^2}{4(1-\alpha {)}^2}$

where $\alpha$ is degree of dissociation,

   Also,         $\alpha$ =22/100

  .^.. K_c = 0.0199