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At `K, K_(p) = 2*0 xx 10 ^(10)//"bar"` for the given reaction at equilibrium
` 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) `
What is `K_(c)` at this temperature ?

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For the given reaction , ` Delta n_(g) = 2 -3 = -1 `
` K_(p) = K_(c) (RT)^(Delta n) or K_(c) = K_(p) (RT)^(-Delta n) = K_(p) (RT) = ( 2*0 xx 10^(10)"bar"^(-1))(0*0831 L "bar" K^(-1) "mol" ^(-1) ) ( 450 K)`
` = 7.48 xx 10^(10 )L"mol"^(-1) = 7*48 xx 10^(11) L "mol"^(-1)`

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