Question

initially, only a and b are present, each at 2.00 m. what is the final concentration of a once equilibrium is reached?

Answer 1

 The "REVERSIBLE" chemical reaction- 
A + B <--> C + D 
Kc = [C][D] / [A][B] = 6.8 
Concentration at the start and after equilibrium- 
[A] = 2.00 M --> (2.00 - X) M 
[B] = 2.00 M --> (2.00 - X) M 
[C] = 0.00 M --> X M 
[D] = 0.00 M --> X M 
6.8 = X^2 / (2.00 - X)^2 
take the square root of both sides- 
2.6077 = X / (2.00 - X) 
5.2154 - 2.6077 X = X 
5.2154 = 3.6077 X 
X = 1.4456 
at equilibrium- 
[A] = 0.55 M 
[B] = 0.55 M 
[C] = 1.45 M 
[D] = 1.45 M