A mixture of `SO_(3), SO_(2)` and `O_(2)` gases is maintained in a `10 L` flask at a temperature at which the equilibrium constant for the reaction is `100`:
`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`
a. If the number of moles of `SO_(2)` and `SO_(3)` in the flask are equal. How many moles of `O_(2)` are present?
b. If the number of moles of `SO_(3)` in flask is twice the number of moles of `SO_(2)`, how many moles of oxygen are present?

Answer

(a) `0.1` (b) `0.4`

Explanation

(a) `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g) K_(C)=100` molelit
Initial mole `{:(a,b,0),(a-2x,b-x,2x),((a)/(2),b-(a)/(4),(a)/(2)):}`
But according to question.
No. of mole of `SO_(2)="No. of mole of" SO_(3)`
`=a-2x=2x`
`a=4x`.
`x=(a)/(4)`.
Now, `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`
But No. of mole of `SO_(3) and SO_(2)` are equal at eq. so.
`K_(C)=(1)/([O_(2)])`
`[O_(2)]=(1)/(K_(C)) , [O_(2)]= (1)/(100)`
But `[O_(2)]=("mole of"O_(2) "at" eq.)/(10)=(1)/(100)`
So No. of mole of `O_(2)=(1)/(10)=0.1`
(b) `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=`
`(((2n_(So_(2)))/(V))^(2))/((n_(SO_(2))/(v))^(2)xx(n_(O_(2))/(V))`
`K_(C)=(4)/((n_(O_2)/(V))) : n_(O_2)=(4xxV)/(K_(c))=(4xx10)/(100)=0.4`