The molar heat of fusion of aluminum metal is 10.79 kJ / mol , whereas its heat of vaporization

The molar heat of fusion of aluminum metal is 10.79 kJ / mol...

The molar heat of fusion of aluminum metal is 10.79 $\mathrm{kJ} / \mathrm{mol}$ , whereas its heat of vaporization is 293.4 $\mathrm{kJ} / \mathrm{mol}$ . a. Why is the heat of fusion of aluminum so much smaller than the heat of vaporization? b. What quantity of heat would be required to vaporize 1.00 $\mathrm{g}$ of aluminum at its normal boiling point? c. What quantity of heat would be evolved if 5.00 $\mathrm{g}$ of liquid aluminum freezes at its normal freezing point? d. What quantity of heat would be required to melt 0.105 mole of aluminum at its normal melting point?

The molar heat of fusion of aluminum metal is 10.79 $\mathrm{kJ} / \mathrm{mol}$ , whereas its heat of vaporization is 293.4 $\mathrm{kJ} / \mathrm{mol}$ .
a. Why is the heat of fusion of aluminum so much smaller than the heat of vaporization?
b. What quantity of heat would be required to vaporize 1.00 $\mathrm{g}$ of aluminum at its normal boiling point?
c. What quantity of heat would be evolved if 5.00 $\mathrm{g}$ of liquid aluminum freezes at its normal freezing point?
d. What quantity of heat would be required to melt 0.105 mole of aluminum at its normal melting point?

Key Concepts

Molar Quantities and Unit Conversion

Molar quantities are used to express the amount of substance in terms of moles, linking energy per mole information to practical quantities of material. Converting between grams and moles using the molar mass is essential for calculating the total energy changes involved in phase transitions.

Energy Calculations in Thermodynamic Processes

Energy calculations in thermodynamics involve using the molar heats of phase transitions (fusion or vaporization) to determine the total energy required or released during a phase change. This includes converting a given mass of a substance to moles and multiplying by the appropriate latent heat value.

Heat of Fusion

The heat of fusion is the amount of energy required to change a substance from its solid phase to its liquid phase at its melting point. It reflects the energy needed to overcome the forces that maintain the organized structure of the solid, which is generally less than what is required to transition to a gas phase.

Phase Changes

Phase changes refer to transitions between different states of matter, such as solid, liquid, and gas. These transformations involve latent heat, meaning energy is either absorbed or released to overcome or establish intermolecular forces without a change in temperature. This concept is central to understanding processes like melting and vaporization.

Heat of Vaporization

The heat of vaporization is the energy necessary to convert a liquid into a gas at its boiling point. This energy is significantly higher than the heat of fusion because it must completely overcome the intermolecular forces holding the liquid together, allowing the molecules to enter the much less interactive gaseous state.

Transcript

00:01 All right, so for part a, the reason why our heat of fusion is smaller than our heat of vaporization is because when we're melting our aluminum, we're simply just overcoming the forces that are holding together our crystal lattice in our solid aluminum form.

00:18 And that doesn't need nearly as much energy as when we are vaporizing our aluminum from liquid into gas, which requires overcoming all of our inner molecules.

00:30 Forces.

00:32 So that's why our heat of vaporization is so much larger because that action requires more energy to overcome all those forces.

00:41 Okay, so for part b, the equation we're using is q equals n times our heat of vaporization.

00:49 And so q is going to be the heat we add in.

00:52 N is our number of moles.

00:54 In delta h vaporization will be our molar heat of vaporization.

00:59 So the first thing that i did, is converting our one gram of aluminum into moles.

01:06 So how you do that is just dimensional analysis.

01:08 So we start with our one gram of aluminum here.

01:11 We multiply it by our molar mass of aluminum, which is 26 .98 grams of aluminum.

01:16 We see our grams of aluminum cancel out, and we're left with just moles, and we have 0 .0371 moles of aluminum.

01:24 So now we can plug that back into our equation.

01:27 So we're going to have q.

01:28 It's equal to our 0 .0371 moles of aluminum.

01:32 Times our heat of vaporization, which is 293 .4 kilojoules per mole.

01:38 And our answer is that the heat we added in is 10 .9 kilojoules.

01:42 Now moving into part c, this is going to be similar, but we have a really key difference here...

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