# Question cf988

Jan 13, 2016

$\Delta H = + \text{12.2 kJ}$

#### Explanation:

Yes, you are correct.

The molar enthalpy of vaporization, or simply the enthalpy of vaporization, tells you the enthalpy change that occurs when one mole of a substance goes from liquid at its boiling point to vapor at its boiling point.

Since heat is needed in order for a substance to undergo a liquid $\to$ vapor phase change, the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, will carry a positive sign.

$\Delta {H}_{\text{vap" = + "39.23 kJ/mol}}$

tells you that $\text{39.23 kJ}$ of heat are being given off when one mole of methanol undergoes a liquid $\to$ vapor phase change.

Convert the mass of methanol to moles by using the compound's molar mass

10.0 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.04color(red)(cancel(color(black)("g")))) = "0.3121 moles CH"_3"OH"

So, if one mole gives off $\text{39.23 kJ}$ of heat, it follows that $0.3121$ moles will give off

0.3121 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "39.23 kJ"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = "12.24 kJ"#

The enthalpy change will thus be

$\Delta H = \textcolor{g r e e n}{+ \text{12.2 kJ}}$

The answer is rounded to three sig figs.