What is #\DeltaS# for the below phase change if #\DeltaH# is 6010 #"J"/"mol"# at STD conditions?

#H_2O"(s)"\rightleftharpoonsH_2O"(l)"#


(Note: please use basic chemistry terms/concepts, I just started learning this)

1 Answer

Answer:

#DeltaS_"fus" = "22.0 J"/{"mol"cdot"K"}#

Explanation:

The change in entropy #\Delta S# due to heat exchange at constant pressure, #\Delta H#, at a constant temperature #T# in #"K"#, is given as

#\Delta S_"trans"=\frac{\Delta H_"trans"}{T}#

because a normal phase change at constant temperature and pressure is an equilibrium condition, for which

#cancel(DeltaG_"trans")^(0) = DeltaH_"trans" - TDeltaS_"trans"#.

As per given data, and knowing that melting requires energy input (i.e. #DeltaH > 0#),

#\Delta H_"fus"= "6010 J"/"mol"# & #T=0^\circ C=273.15\ K#
(freezing point of water)

#\Delta S_"fus"= \frac{"6010 J/mol"}{"273.15 K"}#

#\ \ \ ="22.0 J"/{"mol"cdot"K"}#