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A mixture of SO(3), SO(2) and O(2) gases...

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?

Text Solution

Verified by Experts

`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`
At equilibrium, let the number of moles of each of `SO_(2)` and `SO_(3)` be `n_(1)` and of oxygen `n_(2)`, i.e.
`[SO_(2)]=n_(1)/10`
`[O_(2)]=n_(2)/10`
`[SO_(3)]=n_(1)/10`
Applying the law of mass action,
`K_(c )=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=(n_(1)/10)^(2)/((n_(1)/10)^(2)(n_(2)/10))=100`
`n_(2)=0.1 "mol"`
Oxygen `=0.1 "mol"`
b. Let the number of moles of `SO_(2)` be `=n_(1)`
So number of moles of `SO_(3)=2n_(1)`
Let the number of moles of oxygen be `=n_(2)`
`K_(c )=((2n_(1))/100)^(2)/((n_(1)/10)^(2)(n_(2)/10))` or `100=40/n_(2)`
`n_(2)=0.4 "mol"`
Oxygen `=0.4 "mol"`
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