# Question a239e

Feb 4, 2016

$\Delta {H}_{\text{vap" = "24.3 kJ/mol}}$

#### Explanation:

Yes, you are right on the money.

In simple terms, the molar enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, tells you the amount of heat needed to convert one mole of a substance from liquid at its boiling point to vapor at its boiling point.

In your case, you know that the evaporation of $\text{4.00 g}$ of liquid butane requires an enthalpy change of $\text{1.67 kJ}$. In simple terms, you need to provide $\text{1.67 kJ}$ of heat in order to boil $\text{4.00 g}$ of liquid butane.

This means that in order to find the molar enthalpy change of vaporization, you need to convert the mass of butane to moles.

Use the compound's molar mass to determine how many moles of butane you have in that $\text{4.00-g}$ sample

4.00 color(red)(cancel(color(black)("g"))) * ("1 mole C"_4"H"_10)/(58.12color(red)(cancel(color(black)("g")))) = "0.06882 moles C"_4"H"_10

So, if that many moles of butane require $\text{1.67 kJ}$ of heat, it follows that one mole will require

1 color(red)(cancel(color(black)("mole C"_4"H"_10))) * "1.67 kJ"/(0.06882color(red)(cancel(color(black)("moles C"_4"H"_10)))) = "24.266 kJ"

This means that the molar enthalpy change of vaporization for butane will be equal to

DeltaH_"vap" = color(green)("24.3 kJ/mol")#

The answer is rounded to three sig figs.