At 0^(@)C, ice and water in equilibrium and Delta H =6.00 kJ mol^(-1) for the processH_(2)O (s) rarr H_(2)O(l). What will be Delta S and Delta G for the conversino of ice to liquid water ?

At 0^(@)C, ice and water in equilibrium and Delta H =6.00 kJ mol^(-1)

1.	At 0^(@)C, ice and water in equilibrium and Delta H =6.00 kJ mol^(-1) for the processH_(2)O (s) rarr H_(2)O(l). What will be Delta S and Delta G for the conversino of ice to liquid water ?
Answer» Solution :Since the given PROCESS inequilibrium , `Delta G = 0` Putting this value in the relationship , `Delta G = DELTAH - T DELTAS ` weget `0= Delta H-T Delta S`or `T DeltaS = DeltaH ` or `Delta S = (DeltaH)/(T)` We are given `DeltaH = 6.0 kJ mol^(-1) = 6000 J mol^(-1) `and `T = 0^(@) C = 273 K` `:. Delta S = ( 6000 J mol^(-1))/( 273K ) = 21.98 J K^(-1) mol^(-1)`