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)`.

The equilibrium constant for the reaction given below is `2.0xx10^(-7)

1.	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)`.
Answer» `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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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)`.

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