What is the enthalpy required to vaporize #"63.5 g"# of water at #100^@ "C"#? #DeltabarH_"vap" = "40.66 kJ/mol"#.

Question

Mateo Aguilar · Accepted Answer

It gets easier if you look at your units.

Let us define #DeltabarH_"vap"# as the molar enthalpy of vaporization in #"kJ/mol"#, and #DeltaH_"vap"# as the enthalpy of vaporization in #"kJ"#.

An easy formula that connects them is:

#\mathbf(DeltaH_"vap" = n_("H"_2"O")DeltabarH_"vap")#

where #n_("H"_2"O")# is just the number of #"mol"#s of water.

You have a mass in #"g"#, and you have a molar enthalpy that is on a per-mol basis, so you should either convert the mass of the water to #"mol"#s of water or convert the molar enthalpy to be on a per-gram basis.

I'll be converting the mass of the water to #"mol"#s of water.

#n_("H"_2"O") = 63.5# #cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O"))#

#= color(green)("3.525 mol H"_2"O")#

Thus, we are left with:

#DeltaH_"vap" = n_("H"_2"O")DeltabarH_"vap"#

#= ("3.525" cancel("mol H"_2"O"))("40.66 kJ/"cancel("mol H"_2"O"))#

#=# #color(blue)"143.3 kJ"#

Henry Antrim · Answer

undefined $ \Delta H_{\text{vap}} = 40.66 \, \text{kJ/mol} $. The enthalpy required to vaporize "63.5 g" of water at 100°C is $ \frac{63.5 \, \text{g}}{18.01528 \, \text{g/mol}} \times 40.66 \, \text{kJ/mol} $.