Equilibrium MCQs for NEET — Chemistry Questions with Answers

(c)      K_a = [H₃O⁺][F^-]/[HF][H₂O]

   and    K_b = [HF][OH^-]/[F^-][H₂O]

                .^.. K_a x K_b = [H₃O⁺][OH^-] = K_w

  (b) $K_{{\mathit{P}}_{\mathit{1}}}\mathit{=}\mathit{ }\frac{{\mathit{\eta }}_{\mathit{Y}\mathit{.}}{\mathit{\eta }}_{\mathit{Z}}}{{\mathit{\eta }}_{\mathit{X}}}\left[\frac{{\mathit{P}}_{\mathit{1}}}{\mathit{\Sigma }{\mathit{\eta }}_{\mathit{1}}}\right]\mathit{ }\mathit{ }\mathit{ }{K}_{{\mathit{P}}_{\mathit{2}}}\mathit{=}\mathit{ }\frac{\mathit{(}{\mathit{\eta }}_{\mathit{B}}{\mathit{)}}^{\mathit{2}}}{{\mathit{\eta }}_{\mathit{A}}}{\left[\frac{{\mathit{P}}_{\mathit{2}}}{\mathit{\Sigma }{\mathit{\eta }}_{\mathit{2}}}\right]}^{\mathit{1}}\mathit{ }\mathit{ }$

   For   X$\rightleftharpoons$ Y + Z                 for A $\rightleftharpoons$2B

           1     0    0                       1      0

         1-$\mathrm{\alpha }$   $\mathrm{\alpha }$     $\mathrm{\alpha }$                   1-$\mathrm{\alpha }$      $\mathrm{\alpha }$

     .^.. $\frac{K_{P_{\mathit{1}}}}{K_{P_{\mathit{2}}}}\mathit{ }\mathit{=}\mathit{ }\frac{P_{\mathit{1}}}{P_{\mathit{2}}}\mathit{ }\mathrm{x}\mathit{ }\frac{{\eta }_Y\mathit{.}{\eta }_Z}{{\eta }_X}\mathrm{x}\frac{{\eta }_A}{\mathit{(}{\eta }_B{\mathit{)}}^{\mathit{2}}}\mathrm{x}\left[\frac{\Sigma n_{\mathit{2}}}{\Sigma n_{\mathit{1}}}\right]$

          9 = $\frac{{\mathit{P}}_{\mathit{1}}}{{\mathit{P}}_{\mathit{2}}}\mathrm{x} \frac{\mathrm{\alpha }.\mathrm{\alpha }}{1-\mathrm{\alpha }}\mathrm{x}\frac{(1-\mathrm{\alpha })}{(2\mathrm{\alpha }{)}^2}\mathrm{x}\frac{(1+\mathrm{\alpha })}{(1-\mathrm{\alpha })}$

       .^.. $\frac{P_{\mathit{1}}}{P_{\mathit{2}}}$ = 36

(a) This is Handerson equation for acidic buffer mixtures.

(b) For the reaction,

     A₂ $\rightleftharpoons$2A  

      K=[A]²/[A₂]

The value of equilibrium constant is very less and hence, the product concentration is also very less. So, the reaction is slow. and endo thermic both,

2.303log$\frac{K_{\mathit{p}\mathit{1}}}{K_{\mathit{p}\mathit{2}}}$ = $\frac{∆\mathrm{H}}{\mathrm{R}}\frac{[{\mathrm{T}}_2-{\mathrm{T}}_1]}{{\mathrm{T}}_1{\mathrm{T}}_2}$

       Thus, if K_p2 > K_p1; T₂>T₁, then $∆\mathrm{H}$=+ ve

(a) The reaction shows a change in mole during the course of reaction and thus, increase in pressure will favour forward reaction. Also K_p changes with temperature. Catalyst has no effect on K_p. Thus, P and T influencce the equilibrium concentrations.

(a) In qualitative analysis of cations of second group H₂S gas is passed in presence of HCl, therefore due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in the form of their sulphides due to lower value of their solubility product (K_sp). Here, fourth group cations are not precipitated because it require  more sulphide ions for exceeding their ionic product to their solubility products which is not obtained here due to common ion effect.

(b) Gas + Liquid$\rightleftharpoons$Solution. An increase in P will favour forward reaction.

(c) For a saturated solution product of ionic concentrations $\ge$K_sp.

(a) ${\mathrm{K}}_{\mathrm{P}} = {\mathrm{K}}_{\mathrm{C}}(\mathrm{RT}{)}^{∆\mathrm{n}};$

     $\therefore {\mathrm{K}}_{\mathrm{P}} = 26 \times (0.821 \times 523{)}^{-1} = 0.605$

(c) [H⁺] in monobasic acid = molarity x degree of ionisation

                                        =0.1x 2/100 = 2x10^-3

       ionisation constant of water

          K_w  = [H⁺][OH^-]

           [OH^-] = K_w/[H⁺] = (1x10^-14)/(2x10^-3) = 5x10^-12