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)
(Note: please use basic chemistry terms/concepts, I just started learning this)
1 Answer
Explanation:
The change in entropy
#\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.
#\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"}#