Question

Carbon Monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K, the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00mol of each component is mixed in a 1.00L flask.

Answer 1

Answer: Concentration of $CO_2$ at equilibrium= 1.386 M

Concentration of $H_2$ at equilibrium = 1.386 M

Concentration of $CO$ at equilibrium = 0.614 M

Concentration of $H_2O$ at equilibrium= 0.614 M

Explanation:

Moles of $CO$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Moles of $CO_2$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ =$\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2O$ =$\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $CO_2$ =$\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2$ =$\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

The given balanced equilibrium reaction is,

                         $CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2$

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,

$K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}$

Now put all the given values in this expression, we get :

$5.10=\frac{(1.00+x)^2}{(1.00-x)^2}$

By solving the term 'x', we get :

x =  0.386

Concentration of $CO_2$ at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $H_2$ = (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $CO$ = (1.00-x) M = (1.00-0.386) M = 0.614 M

Concentration of $H_2O$ = (1.00-x) M = (1.00-0.386) M = 0.614 M