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?

Explanation

`2SO_2 + O_2 hArr 2SO_3`
`K_c=([SO_3]^2)/([SO_2]^3[O_2])`
(i)Let `[SO_2]=[SO_3]=x`
`K_c=(x)^2/((x)^2[O_2])`=100
`therefore [O_2]=1/100 "mol L"^(-1)`
Now `[O_2]=("Moles of" O_2)/10=1/100` (volume =10L)
`therefore` Moles of `O_2=10/100`=0.1 mol
(ii) Let moles of `SO_2=y` , then moles of `SO_3 =2y`
`[SO_2]=y/10, [SO_3]=(2y)/10, [O_2]`= ?
`K_C=((2y)/10)^2/((y/10)^2[O_2])=100`
`4/([O_2])=100` or `[O_2]=4/100 = "0.04 mol L"^(-1)`
Now `[O_2]=("Moles of "O_2)/10 =0.04`
`therefore` Moles of `O_2`= 0.04 x 10 = 0.4 mol