Equilibrium constant for a reaction 1.6x...

Equilibrium constant for a reaction `1.6xx10^(-6)` at 298K. Calculate standard free energy change `(DeltaG^(0))` and standard entropy change `(DeltaS^(0))` of the reaction wat that pressure. Given, at 298K, `DeltaH^(0)=25.34kJ`.

Text Solution

Verified by Experts

We know, `DeltaG^(0)=-RTlnK`.
`therefore DeltaG^(0)=-8.314xx298ln(1.6xx10^(-6))=33064J=33.064kJ`
`therefore`The standard free energy change at 298K=+33.064 kJ
Again we know, `DeltaG^(0)=DeltaG^(0)-TDeltaS^(0)`. Here, `DeltaH^(0)=25.34kJ`
`therefore+33.064=25.34-298xxDeltaS^(0)`
`therefore DeltaS^(0)=0.0259kJ*K^(-1)=25.9J*K^(-1)`
so, standard entropy change for the reaction =25.9`J*K^(-1)`.

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