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The equilibrium constant for the reactio...

The equilibrium constant for the reaction given below is `2.0xx10^(-7)` at `300K.` Calculate the standard free energy change for the reaction,
`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`.
Also, calculate the standard entropy change if `DeltaH^@=28.40kJmol^(-1)`.

Text Solution

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`DeltaG^(@)=-2.303xx8.314xx300xxlog[2.0xx10^(-7)]`
`=+38479.8 J mol^(-1)`
`+38.48 k J mol^(-1)`
Also, `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`:. DeltaS^(@)=(DeltaH^(@)-DeltaG^(@))/T`
`=(28.40-38.48)/300`
`=-0.03356 k J=-33.6 J K mol^(-1)`
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