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.

Step-by-step explanation:

Moles of
`CO` = 0.35 mole

Moles of
`H_2O` = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of
`CO` =
`(0.35mol)/(1.00L)=0.35M`

Initial concentration of
`H_2O` =
`(0.40mol)/(1.00L)=0.40M`

Equilibrium concentration of
`CO` =
`(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)=([CO_2]* [H_2])/([CO]* [H_2O])`

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

`K_(eq)=(0.16* 0.16)/((0.35-0.16)* (0.40-0.16))`

`K_(eq)=(0.16* 0.16)/((0.19)* (0.24))=0.56`

Thus
`K_(eq)` at the temperature of the experiment is 0.56.