`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.

Explanation

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