Question

At `700 K`, the equilibrium constant `K_(p)` for the reaction
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
is `1.80xx10^(-3) kPa`. What is the numerical value of `K_(c )` in moles per litre for this reaction at the same temperature?

Answer 1

For the reaction
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
Here `n_(p)=3 "mol", n_(r)=2 "mol"`
`:.Deltan=n_(p)-n_(r)=3-2=1 "mole"`
`K_(p)=1.80xx10^(-3) kPa`
`R=8.314 J K^(-1) "mol"^(-1)`
`T=700 K`
Using the relation, `K_(p)=K_(c )(RT)^(Deltan)`
`K_(c )=K_(p)/(RT)=(1.80xx10^(-3))/(8.314xx700)=3.09xx10^(-7) "mol" L^(-1)`