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At 700 K, the equilibrium constant K(p) ...

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?

Text Solution

Verified by Experts

According to the available data,
`K_(p)=1.8xx10^(-3) kPa = ((1.8 xx 10^(-3) kPa))/((101.3k " Pa atm "^(-1)))=1.78 xx 10^(-5) atm`
`R=0.082 L " atm " K^(-1) mol^(-1), Delta^(ng) =3-2=2,T=700 K`
Using the relation `K_(p)=K_(c) (RT) Delta^(ng)`
`K_(c) =(K_(p))/((RT)^(Delta ng))=((1.78xx10^(-5)atm))/((0.082 L atm K^(-1) mol^(-1))xx(700K))=3.1 xx 10^(-7) mol L^(-1)`
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