# How many joules of heat energy are required to change 64 grams of methane from a liquid to a gas at constant temperature and pressure?

## The heat of vaporization of methane is 9.2 kJ/mol?

Mar 11, 2016

$\text{37 kJ}$

#### Explanation:

The key to this problem is the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, for methane.

In essence, the enthalpy change of vaporization tells you how much heat is needed in order to convert one mole of a substance from liquid at its boiling point to vapor at its boiling point.

The enthalpy change of vaporization expresses the amount of heat needed for one mole of a substance to undergo a liquid $\to$ vapor phase change.

In this case, you know that

$\Delta {H}_{\text{vap" = "9.2 kJ mol}}^{- 1}$

This tells you that one mole of liquid methane at its boiling point must absorb $\text{9.2 kJ}$ of heat to go to vapor at the same temperature.

Use methane's molar mass to determine how many moles of methane you have in that $\text{64-g}$ sample

64 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "3.99 moles CH"_4

So, if one mole requires $\text{9.2 kJ}$ of heat, it follows that this many moles would require

3.99color(red)(cancel(color(black)("moles CH"_4))) * overbrace("9.2 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))))^(color(purple)(DeltaH_text(vap)color(white)(a) "for methane")) = "36.708 kJ"

Rounded to two sig figs, the answer will be

"heat required" = color(green)(|bar(ul(color(white)(a/a)"37 kJ"color(white)(a/a)|)))