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 0.18 mol of CO remaining. Keq at the temperature of the experiment is ________.

Answer 1

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  $CO$ = 0.35 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M$

Initial moles of  $H_2O$ = 0.40 mole

Volume of container = 1 L

Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

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

Initial conc.            0.35 M       0.40M       0     0

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

The expression for equilibrium constant for this reaction will be,

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

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

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

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

$K_c=0.70$

Thus the value of the equilibrium constant is 0.70.