Question

`DeltaH` and `DeltaS` for the system `H_(2)O(l) hArr H_(2)O(g)` at `1atm` are `40.63 k J mol^(-1)`and `108.8 J K^(-1) mol^(-1)`, respectively. Calculate the temperature at which the rates of forward and backward reactions will be the same. Predict the sign of free enegy for this transformation above this temperature.

Answer 1

Given, `DeltaH=40.63kJ" "mol^(-1)` ltbr `DeltaS=108.8JK^(-1)" "mol^(-1)=0.1088kJ" "K^(-1)" "mol^(-1)`
`DeltaG=0` (when the system is in equilibrium)
Applying `DeltaG=DeltaH-TDeltaS`
`0=40.63-Txx0.1088`
`T=(40.63)/(0.1088)=373.4K`
The sign of `DeltaG` above 373 K, i.e., say 374 K, may be calculated as follows:
Again applying `DeltaG=DeltaH-TDeltaS`
`=40.63-374xx0.1088`
`=40.63-40.69=-0.06kJ`
`DeltaG` will be negative, hence, the reaction will be spontaneous.