Question

The reversible chemical reaction

A+B?C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=3.4

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?

Part B

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

Answer 1

Part A:

            A   +   B<--->C   +   D

Initial:                          2         2         0         0

At equilibrium:         2-x        2-x       x          x

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

3.4 = x*x / {(2-x)(2-x)}

3.4 = {x / (2-x)}²

(x)/(2-x) = 1.844

x=3.69 - 1.844x

x= 1.3

Concentartion of A at equilibrium = 2-x =2 - 1.3 = 0.7 M

Part B:

            A   +   B<--->C   +   D

Initial:                          1        2         0         0

At equilibrium:         1-x        2-x       x          x

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

3.4 = x*x / {(1-x)(2-x)}

3.4 = x^2 / (2-3x+x^2)

6.8-10.2x+3.4x^2 =x^2

2.4x^2-10.2x+ 6.8=0

solving above quadratic equation we get, x = 3.4 and 0.83

since x can't be greater than 1 so, x = 0.83

Concentartion of D at equilibrium = x = 0.83 M