`K_(p)` for the reaction `SO_(2)+1/2 O_(2) hArr SO_(3)` at `600^(@)C` is `61.7`. Claculate `K_(p)`. What is the unit of `K_(p)` for the above equilibrium? `(R=0.0821 L-"atm deg"^(-1) mol^(-1))`

Answer

A, B

Explanation

`Deltan= "moles of product" - "moles of reactant"`
`=1-(1+1/2)=-1/2`
`K_(p)=K_(c )(RT)^(Deltan)`
`K_(p)=61.7{0.0821xx(600xx273)}^(-1//2)=7.29`
`K_(p)=p_(SO_(3))/(p_(SO_(2)). P^(1//2)o_(2))=("atm")/("atm. atm"^(1//2))`
Unit of `K_(p)` is `"atm"^(-1//2)`