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QUESTION

# Methane (CH4) has a normal boiling point of -161.6 C. At this temperature, the delta H condensation= -8.17KJ/mol. If 16.5g of the liquid methane vaporize, how much energy is absorbed. ? Thank you

"8.40 kJ"

Before doing any calculations, make sure that you understand what a substance's of condensation, DeltaH_"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 -> liquid takes place (condensation), you can say that heat will be absorbed when a liquid -> 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

color(blue)(DeltaH_"vap" = -DeltaH_"cond")

This means that you can use DeltaH_"vap" as a way to express both the enthalpy of vaporization and the enthalpy of condensation

DeltaH_"vap" = color(red)(+)"8.17 kJ/mol" -> enthalpy of vaporization

DeltaH_"vap" = color(red)(-)"8.17 kJ/mol" -> enthalpy of condensation

Now, 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")

DeltaH_"vap" = +"8.17 kJ/mol"

All you have to do now is figure out how many moles are contained in that "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 "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 .

Therefore, you can say that in order to convert "16.5 g" of methane from liquid at -161.6^@"C" to vapor at -161.6^@"C", you need to provide it with "8.40 kJ" of heat.