Calculate the entropy change for a reaction:
`XrarrY`
Given that `DeltaH^(Theta) = 28.40 kJ mol^(-1)` and equilibrium constant is `1.8 xx 10^(-7) at 298K`.

Answer

`-33.8 J "mol"^(-1) K^(-1)`

Explanation

We know `DeltaG^(@) = -2.303 RT "log K`
`K = 1.8 xx 10^(-7), R = 8.314 J K^(-1) "mol"^(-1), T = 298K`
`DeltaG^(@) = -2.303 xx 8.314 xx 298 xx log 1.8 xx 10^(-7)`
` - 38.484 kJ "mol"^(-1)`
Now, `DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
or `DeltaS^(@) = (DeltaH^(@) - DeltaG^(@))/(T)`
` = (28.40 - 38.484)/(298)`
`= -0.0338 kJ "mol"^(-1) K^(-1)`
` = -33.8 J "mol"^(-1) K^(-1)`.