`K_(c)` 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))`

Explanation

`trianglen=` moles of product -moles of reactant
`=1-(1+1/2) =-1/2`
`K_p=K_c (RT)^(triangle n)`
`K_p=61.7 [0.0821 xx (600 +273)]^(-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 atmosphere`""^(-1/2)`.