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At 450 K, K(p) = 2.0 xx 10^(10)//"bar" f...

At `450 K`, `K_(p) = 2.0 xx 10^(10)//"bar"` for the given reaction at equilibrium
`2SO_(2)(g) + O_(2)(g) hArr 2SO_(2)(g)`
What is `K_(c)` at this temperature?

Text Solution

Verified by Experts

For the given reaction, `Deltan_(g)=2-3=-1`
`K_(p)=K_(c )(RT)^(Deltan) or K_(c )=K_(p)(RT)^(-Deltan)=K_(p)(RT)`
`=(2.0xx10^(10) "bar"^(-1))(0.0831 L "bar" K^(-1) mol^(-1))(450 K)`
`=74.8xx10^(10)L mol^(-1)=7.48xx10^(11) L mol^(-1)`
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