Question

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2 (g) H2 (g) In an experiment, 0.35 mol of CO and 0.40 mol of H2O were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.19 mol of CO remaining. Keq at the temperature of the experiment is __________. A) 5.47 B) 1.0 C) 1.78 D) 0.75 E) 0.56

Answer 1

Answer: $K_{eq}$ at  the temperature of the experiment is 0.56.

Explanation:

Moles of  $CO$ = 0.35 mole

Moles of  $H_2O$ = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ = $\frac{0.35mol}{1.00L}=0.35M$

Initial concentration of $H_2O$ = $\frac{0.40mol}{1.00L}=0.40M$

Equilibrium concentration of $CO$ = $\frac{0.19mol}{1.00L}=0.19M$

The given balanced equilibrium reaction is,

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

Initial conc.          0.35 M         0.40 M                  0 M        0M

    At eqm. conc.     (0.35-x) M   (0.40-x) M   (x) M      (x) M

Given:  (0.35-x) = 0.19

x= 0.16 M

The expression for equilibrium constant for this reaction will be,

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

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

$K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}$

$K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56$

Thus $K_{eq}$ at  the temperature of the experiment is 0.56.