Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=6.7

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

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

Please include all steps!

Answer 1

           A   +    B⇌C   +    D

I         2          2     0          0

C       -x      -x        +x        +x

E    2-x     2-x         +x      +x

    Kc    =   [C][D]/[A][B]

   6.7     = x*x/(2-x)(2-x)

   6.7     = (x/2-x)^2

6.7   = x/2-x

   2.53    = x/2-x

2.53*(2-x) = x

    x   = 1.43

[A]   = 2-x = 2-1.43 = 0.57M

B.

           A   +    B ⇌ C   +    D

I         1        2     0          0

C       -x      -x        +x        +x

E      1-x     2-x         +x      +x

    Kc    =   [C][D]/[A][B]

   6.7     = x*x/(1-x)(2-x)

   6.7     =    x*x/(1-x)(2-x)

6.7*(1-x)(2-x) = x^2

    x   = 0.89

[A]   = 1-x   = 1-0.89   = 0.11M

[B]    = 2-x   = 2-0.89    = 1.11M

[D]    = x      = 0.89M

[C]   = x     = 0.89M