5 moles of `SO_(2)`and 5 moles of `O_(2)` are allowed to react .At equilibrium , it was foumnd that `60%` of `SO_(2)` is used up .If the pressure of the equilibrium mixture is one aatmosphere, the parital pressure of `O_(2)` is :

Answer

`0.41`

Explanation

`{:(,2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g)),("Initial moles"," 5 5 0"):}`
As 60% `SO_(2)` is used up, no. of moles of `SO_(2)` used up `= (60)/(100) xx 5 = 3 `
`:.` No. of moles of `SO_(2)` at equilibrium = 5 - 3 = 2
As 2 moles of `SO_(2)` react with 1 mole of `O_(2)`
3 moles of `SO_(2)` will react with `O_(2)`
`= (1)/(2) xx 3 = 1.5` moles
i.e. No. of moles of `O_(2)` at equilibrium
= 5 - 1.5 = 3.5
As 2 moles of `SO_(2)` produce 2 moles of `SO_(3)`
`:.` No. of moles of `SO_(3)` at equilibrium = 3 moles
`:.` Total no. of moles at equilibrium
= 2 + 3.5 + 3 = 8.5
` :. P_(O_(2)) = (n_(O_(2)))/(n_("total")) xx P_("total") = (3.5)/(8.5) xx 1 atm ` = 0.41 atm