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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?

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The correct Answer is:
A, C

For the reaction `2SO_(3) hArr 2SO_(2)(g)+O_(2)(g)`
`Deltan=n_(P)-n_(R)=3-2=1 "mol"`
Given,
`K_(p)=1.8xx10^(-3) Kpa, R=8.314 J K "mol"^(-1), T=700 K`
Using the relation,
`K_(p)=K_(p)(RT)^(Deltan)`
`:. K_(c)=K_(p)/((RT)^(Deltan))=(1.8xx10^(-3))/((8.314xx700)^(1))=3.09xx10^(-7) "mol" L^(-1)`
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