NEETprep Bank NEET Chemistry Equilibrium Questions Solved


The conjugate base of H3BO3 is:

(a) B(OH)4-                                      (b) H2BO3-

(c) HBO32-                                       (d) H4BO3+

(a)  H3BO3 accepts OH- to form its conjugate base B(OH)4-,

    H3BO3 + H2B(OH)4- + H+

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The degree of dissociation of PCl5 (α) obeying the equilibrium,

PCl5 (g PCl3 (g) + Cl2 (g), is approximately related to the pressure at equilibrium by:

(a) αP                                         (b)  α1/P

(c)  α1/P2                                   (d)  α1/P4

(b)  PCl5 (g PCl3 (g) + Cl2 (g)

         1               0              0

        1-α             α              α 

...   Kpα2(1-α)P1+α=α2P1-α2

         or α=KpP  if 1-α2=1

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The equilibrium constants KP1 and KP2 for the reactions X2Y and Z  P + Q respectively are in the ratio of 1:9. If the degree of dissociation ox X and Z be equal then the ratio of total pressure at these equilibria is:

(a) 1:9                 (b) 1:36                 (c) 1:1                     (d) 1:3

(b)    Given :  KP1KP2 = 1/9     and αx=αz

        X  2Y1        01-α    2α      ...     KP1 = (2αx)2(1-αx) x P11+ αx1        

         Z   P + Q1        01-α   2α    ...      KP2 = (αz)2(1-αz) x P21+ αz1                                                                            

                          ...           P1/P2KP1KP2x 1/4 = 1/9 x 1/4 = 1/36

 

 

 

 

 

 

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In a system: A(s)      2B(g) + 3C(g). If the concentration of C at eqilibrium is increased by a factor 2, it will cause the eqilibrium concentration of B to change to:

(a) two times of its original value

(b) one half of its original value

(c) 22 times of its original value

(d) 122 times of its original value

For A(s)      2B(g) + 3C(g).

 Kc=C3B2; if C becomes twice,Then let conc. of B becomes B' then                    Kc=2C3B'2  or    C3.B2=2C3B'2   B'B=18=122

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For the reaction:

 CO(g) + 12O2(g) CO2(g), Kp/Kis:

(a) RT              (b) (RT)-1               (c) (RT)-1/2                (d) (RT)1/2

(c) n=1-1-1/2 = -1/2

      ...    Kp =Kc(RT)-1/2

Difficulty Level:

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 A weak acid, HA has a Ka of 1.00 x 10-5. If 0.100 mole of this acid is dissolved in one litre of water, the percentage of acid dissociated at equibrium is closest to

(a) 99.9%                             (b) 1.00%                      

(c) 99.9%                             (d) 0.100%

(b) HA H+ + A-

     At equlibrium, [H+]=[A-]

         Ka[H+][A-][HA] =  [H+]2[HA]

             [H+] =Ka[HA] = 1x10-5 x0.1=1x10-6 =  10-3x 1

             α=Actual ionisationMolar ionisation = 10-3/0.1 = 10-2

        % of acid dissociated = 10-2 x 100 = 1.00%

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If little heat is added to ice  liquid equilibrium in a sealed container, then:

(a) pressure will rise                            (b) temperature will rise

(c) temperature will fall                        (d) no change in P and T

(d) Heat will be used to melt ice.

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In the equilibrium,

2SO2(g) + O2 (g) 2SO3(g), the partial pressure of SO2, O2 and SO3 are 0.662, 0.101 and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO2 and SO3 are equal.

(a) 0.4 atm                    (b) 1.0 atm

(c) 0.8 atm                    (d) 0.25 atm

(a) Kp =(pSO3)2(pSO2)2(pO2) =(0.331)2(0.662)2(0.101)=2.5

       Now,     Kp =(pSO3)2(pSO2)2(pO2)

                pSO3=pSO2 

           Then, pO2 = 1/Kp  =1/0.25 = 4 atm

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Ionisation constant of CH3COOH is 1.7 X 10-5 and concentration of H+ ions is 3.4 X 10-4.Then, find out initial concentration of CH3COOH molecules.             [2001]

(a) 3.4 X 10-4 

(b) 3.4 X 10-3

(c) 6.8 X 10-4

(d) 6.8 X 10-3

 

(d)

CH3COOH       CH3COO- + H+Ka=CH3COO-H+CH3COOHGiven that,       CH3COO-=H+=3.4 X 10-4 MKa for CH3COOH = 1.7 X 10-5CH3COOH is week acid, so in it CH3COOH is equal to initial concentration. Hence,1.7 × 10-5=3.4 ×10-43.4 ×10-4CH3COOHCH3COOH=3.4 ×10-43.4 ×10-41.7 ×10-5=6.8 × 10-3 M

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The formation of phosgene is represented as,

            CO + Cl2 COCl2

The reaction is carried out in 500 mL flask. At equilibrium o.3 mole of phosgene, 0.1 mole of CO and 0.1 mole of Cl2 are present. The equilibrium constant of the reaction is:

(a) 30                     (b) 15

(c) 5                       (d) 3

(b) [CO] =0.1/0.5

     [Cl2] =0.1/0.5

     [COCl2] = 0.3/0.5

      ..Kc0.3/0.50.10.5x0.10.5 = 15

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