Question

At a certain temperature, 0.740 mol SO 3 0.740 mol SO3 is placed in a 2.50 L 2.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.180 mol O 2 0.180 mol O2 is present. Calculate K c

Answer 1

Answer:
`K_c` = 0.0046

Step-by-step explanation:

Moles of
`SO_3` = 0.740 mole

Volume of solution = 2.50 L

Initial concentration of
`SO_3` =
`(moles)/(volume)=(0.740)/(2.50)=0.296M`

The given balanced equilibrium reaction is,

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

Initial conc. 0.296 M 0M 0M

At eqm. conc. (0.296-2x)M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

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

Given : moles of
`O_2` at equilibrium = 0.180M

Concentration of
`O_2` at equilibrium=
`(moles)/(volume)=(0.180)/(2.50)=0.072M`

x = 0.072 M

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

`K=((2x)^2* (x)^2)/((0.296-2x)^2)`

`K=((2* 0.072)^2* (0.072)^2)/((0.296-2* 0.072)^2)`

`K=0.0046`

Thus the value of
`K_c` is 0.0046