Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=4.7

Part A

Initially, only A and B are present, each at 2.00 M . What is the final concentration of A once equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

Express your answer to two significant figures and include the appropriate units

Answer 1

Part A :

                               A   +    B   ⇌ C +   D

initial conc            2.00      2.00       0        0

change                  -a         -a          +a      +a

Equb conc       2.00-a    2.00-a       a        a

Given equilibrium constant:, K_c = 4.7

K_c = ([C][D]) / ([A][B] )

4.7 = (a x a ) / [(2.00-a) (2.00-a) ]

a / ( 2.00-a ) = = 2.17

   a = 2.17 ( 2.00-a)

       = 4.34 - 2.17 a

3.17 a = 4.34

       a = 4.34 / 3.17

         = 1.37 M

So the Equilibrium concentration of A is 2.00-a = 2.00 - 1.37 = 0.63 M

Part B :

                               A   +    B   ⇌ C +   D

initial conc            1.00      2.00       0        0

change                  -a         -a          +a      +a

Equb conc       1.00-a    2.00-a       a        a

Given equilibrium constant:, K_c = 4.7

K_c = ([C][D]) / ([A][B] )

4.7 = (a x a ) / [(1.00-a) (2.00-a) ]

On solving we get a has two values one is 0.86 & other is 2.94 .The second value is not taken since it is more than the initia concentration 2.00 M .

Therefore the correct value is a = 0.86 M

So the Equilibrium concentration of D is = a = 0.86 M