Answer: Concentration of  at equilibrium= 1.386 M
 at equilibrium= 1.386 M 
Concentration of  at equilibrium = 1.386 M
 at equilibrium = 1.386 M 
Concentration of  at equilibrium = 0.614 M
 at equilibrium = 0.614 M
Concentration of  at equilibrium= 0.614 M
 at equilibrium= 0.614 M 
Explanation:
Moles of  = 1.00 mole
 = 1.00 mole
Moles of  = 1.00 mole
 = 1.00 mole
Moles of  = 1.00 mole
 = 1.00 mole
Moles of  = 1.00 mole
 = 1.00 mole
Volume of solution = 1.00 L
Initial concentration of  =
 =
Initial concentration of  =
 =
Initial concentration of  =
 =
Initial concentration of  =
 =
The given balanced equilibrium reaction is,
                          
Initial conc.          1.00M          1.00 M          1.00 M     1.00 M
At eqm. conc.     (1.00-x) M   (1.00-x) M   (1.00+x) M   (1.00+x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x =  0.386
Concentration of  at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M
 at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M 
Concentration of  = (1.00+x) M= (1.00+0.386)= 1.386 M
 = (1.00+x) M= (1.00+0.386)= 1.386 M 
Concentration of  = (1.00-x) M = (1.00-0.386) M = 0.614 M
 = (1.00-x) M = (1.00-0.386) M = 0.614 M
Concentration of  = (1.00-x) M = (1.00-0.386) M = 0.614 M
 = (1.00-x) M = (1.00-0.386) M = 0.614 M