Question

At a certain temperature, 0.900 mol SO 3 is placed in a 2.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.110 mol O 2 is present. Calculate K c .

Answer 1

Answer: The value of
`K_c` is 0.0057

Step-by-step explanation:

Initial moles of
`SO_3` = 0.900 mole

Volume of container = 2.00 L

Initial concentration of
`SO_3=(moles)/(volume)=(0.900moles)/(2.00L)=0.450M`

equilibrium concentration of
`O_2=(moles)/(volume)=(0.110mole)/(2.00L)=0.055M` [/tex]

The given balanced equilibrium reaction is,

`2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)`

Initial conc. 0.450 M 0 0

At eqm. conc. (0.450 -2x) M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

`K_c=([O_2][SO_2]^2)/([SO_3]^2)`

`K_c=(x* (2x)^2)/(0.450-2x)^2)`

we are given : x = 0.055

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

`K_c=(0.055* (2* 0.055)^2)/(0.450-2* 0.055)^2)`

`K_c=0.0057`

Thus the value of the equilibrium constant is 0.0057