Question #04705

1 Answer
Dec 29, 2015

DeltaH_"fus" = -"6.005 kJ/mol"

Explanation:

As you know, the molar enthalpy change of fusion, DeltaH_"fus", is defined as the change in enthalpy when one mole of a substance undergoes a liquid to solid or a solid to liquid phase change at its melting / freezing point.

More specifically, the molar enthalpy change of fusion will tell you

  • how much heat must be added to one mole of a substance at its melting point in order for a solid -> liquid phase change to take place

  • how much heat must be given off by one mole of a substance at its freezing point in order for a liquid -> solid phase change to take place

In your case, you are interested in finding out the molar enthalpy change of fusion for the freezing of "30.00 g" of water at 0^@"C". This means that the molar enthalpy change of fusion must carry a negative sign, since it represents heat lost.

The equation you'll use here looks like this

color(blue)(q = n * DeltaH_"fus")" ", where

q - heat lost / absorbed
n - the number of moles of the substance
DeltaH_"fus" - the molar enthalpy change of fusion

Use water's molar mass to determine how many moles you have in that "30.00-g" sample

30.00 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.6653 moles H"_2"O"

No,w you know that "10.00 kJ" of heat are given off when this sample of water goes from liquid water at 0^@"C" to ice at 0^@"C".

This means that the value of q will be

q = -"10.00 kJ"

Again, the negative sign is used to symbolize heat lost.

This means that you have

DeltaH_"fus" = q/n

DeltaH_"fus" = (-"10.00 kJ")/"1.6653 moles" = color(green)(-"6.005 kJ/mol")

The answer is rounded to four sig figs.

So, you can say that the molar enthalpy change of fusion for water will be equal to

  • DeltaH_"fus" = +"6.005 kJ/mol" -> when ice at 0^@"C" melts to liquid water at 0^@"C"
  • DeltaH_"fus" = -"6.005 kJ/mol" -> when liquid water at 0^@"C" freezes to ice at 0^@"C"