CALCULALTIONS INVOLVING EQUILIBRIUM CONSTANT AND EQUILIBRIUM CONCENTRATIONS:

 

CALCULALTIONS INVOLVING EQUILIBRIUM CONSTANT AND EQUILIBRIUM CONCENTRATIONS:

 

            QUESTION:

 

            The numerical value for the equilibrium constant K_c for the following reaction.

 

            2A + B  C is 4

 

(a)       What are the units for K_c?

(b)       Calculate the equilibrium constants for the following reactions.

            (i)        C  2 A + B

            (ii)       2 C  4 A + 2B

 

 

SOLUTION:

 

(a)       2A + B  C

 

            K_c =

            Units of K_e,

                        =

 

(b)       (i)        C 2A + B                 

                         =

 

                         = ¹/_Kc 

                         =    

 

            (ii)       2C 4A + 2B 

                         =

                         = ²

                         = ( )²

                         = (¹/₄)² or ¹/₁₆

                         =   mol⁴. Dm^-12

 

QUESTION:

 

            It is formed that, if one mole of ethanoic acid and 0.5 moles of ethanol react to equilibrium at a certain temperature, 0.423 moles of ethyl ethanoate are produced calculate K_c.

 

SOLUTION:

 

            CH₃COOH+C₂H₅OH CH₃COOC₂H₅+H₂O

            t = 0                1                      0.5                   0                      0

            t = t                 (1-0.423)        0.5-0.423        0.423              0.423

            t = Eq              0.577              0.077              0.423              0.423

 

             =

           

            Let, v be the volume of mixture in dm^-3 and concentration is represented as mol. Dm^-3.

 

            K_c =

 

            K_c = 4.0

 

QUESTOION:

 

            One mole of the anhydrous ethanedioic acid and two moles of ethanol are mixed together in a sealed glass contains and allowed to reach equilibrium at a certain temperature. At equilibrium 36.8 grams of ethanol remains. Calculate the value of the equilibrium constant.

 

 

SOLUTION:

 

            COOH                        COOC₂H₅

                         + 2C₂H₅OH                         + 2H₂O

            COOH                          COOHC₂H₅

t=0      1                   2                 0                  0

 

At equilibrium 36.8 g of ethanol remains; no of moles of ethanol =  

                                                                                                               = 0.8 moles                                    

 

            Since, 0.8 moles of ethanol is left at equilibrium from initially present two moles of ethanol therefore 1.2 moles of ethanol has reacted. Reacting mole ratio between acid and ethanol in the balanced equation is 1:2  therefore, the mole ratio is, 0.6; 1.2 and mole ratio between the formed easter and water is 1:2.

 

            MOLE RATIO              1 : 2  1 : 2

            Amount                     

            Reacted/mole            0.6 1.2 - -

 

            Amount

            Formed/mole - - 0.6 1.2

            T = Eq . (1-0.6)          (2-1.2)            0.6       1.2

                   0.4                   0.8                  0.6        1.2

 

Let the volume of equilibrium mixture by V dm³.

 

K_c =

 

=

 

=

 

=

= K_c = 3.375