Question

2.00 moles of CO2 and 1.50 moles of H2 are placed into a rigid 5.00-L container and they react according to the equation CO2(g) + H2(g) ⇄ CO(g) + H2O(g) K = 2.50 What will be the concentration of carbon monoxide when equilibrium is reached? 0.191 M 0.091 M 0.209 M (Your correct answer) 0.913 M 1.05 M

Answer 1

Answer: 0.209 M

Explanation:

Moles of  $CO_2$ = 2.00 mole

Moles of  $H_2$ = 1.50 mole

Volume of container = 5.00 L

Initial concentration of $CO_2$ = $\frac{moles}{Volume}=\frac{2.00mol}{5.00L}=0.400M$

Initial concentration of $H_2$ = $\frac{moles}{Volume}=\frac{1.50mol}{5.00L}=0.300M$

The given balanced equilibrium reaction is,

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

Initial conc.          0.400 M          0.300M             0 M        0 M

   At eqm. conc.     (0.400-x) M   (0.300-x) M   (x) M      (x)  M

The expression for equilibrium constant for this reaction will be,

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

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

$2.50=\frac{[x]\times [x]}{[0.300-x]\times [0.400-x]}$

By solving the term 'x', we get :

x = 0.209 M

Concentration of $CO$ at equilibrium = x M  =  0.209 M