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
Truong-Son N. ยท Harish Chandra Rajpoot
Jul 25, 2018

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"}

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