What is the formula for #DeltaH_"fus"#?

1 Answer
Truong-Son N.
Aug 12, 2016

Enthalpy of fusion (or really, enthalpy of phase transitions) is a constant-pressure heat flow. So you can work it out just from knowing that and the units.

The molar enthalpy of (common) phase transitions (that is, in #"kJ/mol"#) would be:

#\mathbf(DeltabarH_"trs" = q_p/(n_"substance"))#

where:

  • #q# is the heat flow through the system at constant pressure and the same temperature.
  • #n_"substance"# is the #\mathbf("mol")#s of the substance at hand that is transitioning to a new phase.

If you were looking at water freezing/melting, then #DeltabarH_"fus" ~~ "6.02 kJ/mol"#. That means if you had #"1 mol"# of water, then you would have

#color(blue)(q_p) = n_"substance" xx DeltabarH_"fus"#

#= ("1 mol")("6.02 kJ/mol")#

#=# #color(blue)("6.02 kJ")# of heat absorbed to melt it, or released when freezing it.

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