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 :

Options

A. `0*52` atm
B. `0*21` atm
C. ` 0*41`atm
D. `0*82` atm

Answer

C

Explanation

` {:(,2 SO_(2)(g),+,O_(2)(g),hArr,2 SO_(3)(g)),("Intial",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)`
`=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) " equilibrium = 3 moles"`
`= 2 + 3*5 + 3 = 8*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 "`