# Question 98d79

Dec 31, 2015

$\text{8.40 kJ}$

#### Explanation:

Before doing any calculations, make sure that you understand what a substance's enthalpy of condensation, $\Delta {H}_{\text{cond}}$, tells you.

For a given substance, the enthalpy of condensation expresses the amount of heat that must be given off to convert a specific sample of that substance from vapor to liquid at constant temperature (and pressure).

Usually, that specific sample is a mole of that substance. Since heat is given off when a vapor $\to$ liquid phase change takes place (condensation), you can say that heat will be absorbed when a liquid $\to$ vapor phase change takes place (vaporization).

This is why the enthalpy of condensation, which tells you how much heat is given off when one mole of a substance undergoes condensation, carries a negative sign.

You can thus say that

$\textcolor{b l u e}{\Delta {H}_{\text{vap" = -DeltaH_"cond}}}$

This means that you can use $\Delta {H}_{\text{vap}}$ as a way to express both the enthalpy of vaporization and the enthalpy of condensation

$\Delta {H}_{\text{vap" = color(red)(+)"8.17 kJ/mol}} \to$ enthalpy of vaporization

$\Delta {H}_{\text{vap" = color(red)(-)"8.17 kJ/mol}} \to$ enthalpy of condensation

Now, phase changes always take place at constant temperature, so don't worry about the given value for the boiling point of methane.

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

DeltaH_"vap" = - (-"8.17 kJ/mol")

$\Delta {H}_{\text{vap" = +"8.17 kJ/mol}}$

All you have to do now is figure out how many moles are contained in that $\text{16.5-g}$ sample of methane. To do that, use the compound's molar mas

16.5 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_4)/(16.04color(red)(cancel(color(black)("g")))) = "1.0287 moles CH"_4

Well, if one mole must absorb $\text{8.17 kJ}$ of heat, it follows that $1.0287$ moles must absorb

1.0287 color(red)(cancel(color(black)("moles CH"_4))) * "8.17 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))) = color(green)("8.40 kJ")#

The answer is rounded to three sig figs.

Therefore, you can say that in order to convert $\text{16.5 g}$ of methane from liquid at $- {161.6}^{\circ} \text{C}$ to vapor at $- {161.6}^{\circ} \text{C}$, you need to provide it with $\text{8.40 kJ}$ of heat.