Equilibrium MCQs for NEET — Chemistry Questions with Answers

(a) A^pB^q $\rightleftharpoons$pA⁺ + qB^- ; Let solubility be s mol/litre, Thus

                  p.s       q.s

                    K_sp = [A⁺]^p[B^-]^q = (ps)^p.(qs)^q 

                          = p^pq^q(s)^p+q

(b) 1 mole of N₂ reacts with 3 moles of H₂ thus, for 

    N₂ + 3H₂ $\rightleftharpoons$ 2NH₃ ; (a-x) > (a-3x)

    a         a

(a-x)    (a-3x)     2x

(d) Acid $\stackrel{-{\mathrm{H}}^{+}}{\to }$conjugate base;

     Base$\stackrel{+{\mathrm{H}}^{+}}{\to }$conjugate acid.

(b)                NH₄HS(s)     $\rightleftharpoons$NH₃ (g) + H₂S(g)

            Pressure at equlibrium                  P             P

             .^.. Total pressure at equilibrium = 2P =1.12 atm

                                                       P =1.12/2 atm

                ^..^. K_p = $P{\mathit{\text{'}}}_{N{H}_{\mathit{3}}}\mathit{ }\mathrm{x}\mathit{ }P{\mathit{\text{'}}}_{H_{\mathit{2}}S}$ 

                         K_p = (1.12/2) x (1.12/2) = 0.3136 atm²

(c) Hg exists as ${\mathrm{Hg}}_2^{2+}$ and not as Hg⁺. Thus ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Hg}}_2^{2+}\right][{\mathrm{I}}^{-}{]}^2$

(b) In aqueous solution of 10^-8 M HCl, [H⁺] is based upon the concentration of H+ ion of 10^-8 M HCl and concentration of H+ ion of water.

             K_w of H₂O = 10^-14 = [H⁺][OH^-]

       or [H⁺] = 10^-7 M (due to its neutral behaviour)

  So , in aqueous solution of 10^-8 M HCl,

          [H⁺] = [H⁺] of HCl + [H⁺] of water = 10^-8 + 10^-7 = 11x10^-8 M$\approx$1.10 x 10^-7

(a) Lower is the value of K, lesser will be the tendency to show forward reaction.

(d) Given, CH₃COOH $\rightleftharpoons$CH₃COO^- + H⁺

     K_a = 1.5 x 10^-5                             .............(i)

    HCN$\rightleftharpoons$H⁺ + CN^-,         K_a1 = 4.5 x 10^-10.....(ii)

For CN^- + CH₃COOH $\rightleftharpoons$HCN + CH₃COO^-

K=?

On subtracting Eq (ii) from Eq. (i), we get

CH₃COOH + CN^- $\rightleftharpoons$ HCN+CH₃COO^-

           K=K_a/K_a1 = (1.5 x10^-5)/(4.5 x 10^-10) =10⁵/3 = 3.33 x 10⁴ $\approx$ 3 x 10⁴

(a) K_c = [A][B]/[AB];

         If [A] =2x[A]

        To have K_c constant [B] should be [B] x 1/2