The equilibrium constant `(K_(p))` for the reaction, `2SO_(2)+O_(2)hArr2SO_(3)` at `1000 K` is `3.5`. The partial pressure of oxygen gas to give equal mole of `SO_(2)` and `SO_(3)` is :

Options

A. `0.29 atm`
B. `35 atm`
C. `0.53 atm`
D. `1.87 atm`

Explanation

`2SO_(2)+O_(2)hArr2SO_(3)`,
Let at equilibrium, mole of `SO_(2)` and `SO_(3)` be same, then
`P_(SO_(3))=P_(SO_(2))`
`K_(p)=((P_(SO_(3)))^(2))/((P_(SO_(2)))^(2)*P_(O_(2)))`
or `P_(O_(2))=(1)/(K_(P))=(1)/(3.5)=0.29 atm`