Question

At a certain temperature, 0.700 mol SO3 is placed in a 3.50 L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.180 mol O2 is present. Calculate c.

Answer 1

Answer: The value of
`K_c` for the given chemical equation is 0.0457.

Step-by-step explanation:

Given values:

Initial moles of
`SO_3` = 0.700 moles

Volume of conatiner = 3.50 L

The given chemical equation follows:

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

I: 0.700

C: -2x +2x x

E: 0.700-2x 2x x

Equilibrium moles of
`O_2` = x = 0.180 moles

Equilibrium moles of
`SO_2` = 2x =
`(2* 0.180)=0.360moles`

Equilibrium moles of
`SO_3` = 0.700 - 2x =
`0.700-(2* 0.180)=0.340moles`

Molarity is calculated by using the equation:

`Molarity=(Moles)/(Volume)`

So,

`[SO_3]_(eq)=(0.340)/(3.50)=0.0971M`

`[SO_2]_(eq)=(0.360)/(3.50)=0.103M`

`[O_2]_(eq)=(0.180)/(3.50)=0.0514M`

The expression of
`K_c` for above equation follows:

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

Plugging values in above expression:

`K_c=((0.0971)^2* 0.0514)/((0.103)^2)\\\\K_c=0.0457`

Hence, the value of
`K_c` for the given chemical equation is 0.0457.