Lecture 24 - Thermochemistry III

Tuesday, April 16, 2024

1:30 PM

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Today (4/16)
Molecular basis of heat capacities 
Phase Changes and Heating Curves
Intermolecular Forces
Thursday (4/18)
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Molecular basis of heat capacities
﷟HYPERLINK "https://youtu.be/nqGtji3ZjoI?si=YXLrY7iuKeqwez7B"Molecular Temperature & Degrees of Freedom (https://youtu.be/nqGtji3ZjoI?si=YXLrY7iuKeqwez7B)

Molecular Temperature & Degrees of Freedom Press enter to activate, Machine generated alternative text:
The,temperat re is he verage 
kine@energy ersphe 
Degrees of Freedom - Different types of motions available to molecules (e.g., translation, rotation, vibration, electronic)
and the number of coordinates necessary to describe each type of motion
Translational Degrees of Freedom
Rotational Degrees of Freedom
Vibrational Degrees of Freedom
All monoatomic gases and molecular gases have 3 translational degrees of freedom
Untitled picture.png leuope1sueJI 
Monoatomic gases do not have rotational degrees of freedom. The moment of inertia of the nucleus is too small to be significant. 

150px-ProhibitionSign2.svg.png 
In addition to the previous types of motion, there is also vibrational motion. This is most important a temperatures greater than room temperature.

Untitled picture.png Vibrational
Untitled picture.png leuope1sueJI 
Each of these degrees of freedom contributes to the temperature of a gas to give the average kinetic energy per mole
Ēk = 3/2 RT
For a monoatomic gas we can associate Ēk with the thermodynamic energy U
ΔU = 3/2 RΔT
If we substitute this into the equation:
ΔH = ΔU + PΔV = 3/2RΔT + PΔV
and then use the ideal gas relationship 
(PΔV = RΔT)
ΔH =3/2RΔT + RΔT 
we get
ΔH = 5/2RΔT

The definition of heat capacity at constant pressure is 
Cp = ΔH/ΔT
Therefore for a monoatomic gas
Cp = 5/2RΔT/ΔT
Cp = 5/2R
Substituting in R = 8.314 J/mol·K
Cp = 20.78 J/mol·K

However, the measured heat capacity values for diatomic and polyatomic molecules are significantly larger than those for monatomic gases. The explanation must lie in the existence of molecular motions, or degrees of freedom, in addition to translation that can accommodate thermal energy.
150px-ProhibitionSign2.svg.png 

Diatomic or linear molecules have two degress of rotational freedom. The moment of inertia along the internuclear axis is too small to be significant.

Untitled picture.png ation 

Nonlinear molecules have three degrees of rotational freedom. 
Untitled picture.png Rotationa I 
Each rotational degree of freedom adds another ½RT to the internal energy. 
Adding in the rotational component
For linear molecules Cp = 7/2 R = 29.10 J/mol·K
For nonlinear molecules Cp = 8/2 R = 33.26 J/mol·K
Untitled picture.png Vibrational 

To determine how many vibrational motions there are for a molecule you take the number of atoms (N) and multiply by 3 coordinates each, then subtract the sum of the translational and rotational degrees of freedom.

Linear Molecules
3N-5
Nonlinear Molecules
3N-6

The important thing to note is that Cp values are temperature dependent. Any difference in measured Cp values is essentially the magnitude of the vibrational contribution.

Untitled picture.png Measured Values of cp 
for Selected Gases at 298 K 
and 1 atm Pressure 
Gas 
He 
Ne 
02 
H20 
cp (J mol-I K-1) 
20.79 
20.79 
20.79 
28.81 
29.36 
31.30 
33.54
However, the measured heat capacity values for diatomic and polyatomic molecules are significantly larger than those for monatomic gases. The explanation must lie in the existence of molecular motions, or degrees of freedom, in addition to translation that can accommodate thermal energy.

Untitled picture.png Measured Values of cp 
for Selected Gases at 298 K 
and 1 atm Pressure 
Gas 
He 
Ne 
02 
H20 
cp (J mol-I K-1) 
20.79 
20.79 
20.79 
28.81 
29.36 
31.30 
33.54 


Untitled picture.png 60 
so 
— 40 
30 
20 
C02 
7 
Rotation 
(linear) 
Vibration 
02 
Vibration 
Rotation 
(nonlinear) 
Translation 
1000 
2000 
Temperature (K) 
3000 

Phase Changes
Phase Changes
Untitled picture.png 

Process
Before
Energy Change
After
Endo/exothermic
Melting Ice








Freezing Water








Heat transfer during Phase Transitions (Heating and Cooling Curves)
 
During the process of melting energy is needed to change a substance from solid to liquid

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At constant pressure this is known as the enthalpy of fusion (ΔHfus) and is typically given in kJ/mol 

The opposite process is known as enthalpy of freezing and the energy change is equivalent to -ΔHfus


A related process that requires much more energy is the enthalpy of vaporization liquid to gas (ΔHvap)

The opposite process is known as enthalpy of condensation and the energy change is equivalent to -ΔHvap


These are latent heat processes meaning the phase change occur at constant temperature

Substance
Molar Heat of fusion
(kJ/mol)
Melting point
(°C)
Molar Heat of vaporization
(kJ/mol)
Boiling point
(°C)
Ammonia (NH3)
5.6572
−77.74
23.32
−33.34
Carbon dioxide (CO2)
8.10
−78
25.3
-78.46
Hydrogen
0.12
−259
0.917
−253
Lead
4.8
327.5
180
1750
Methane (CH4)
0.95
−182.6
8.20
−161.6
Nitrogen
0.720
−210
5.60
−196
Oxygen
0.445
−219
6.82
−183
Water
6.02
0
40.80
100

How much energy is required to melt 20.0 g of lead at 327.5 °C?





How much heat is released by the condensation of 2.88 mols of oxygen at -183 °C?




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When studying the energy required to transition between phases, it is useful to plot the temperature versus energy. 

When energy is absorbed by the system this is known as a heating curve

and when it is released by the system it is a cooling curve


Untitled picture.png Machine generated alternative text:
heat (J) 



Find the energy required for 20.0 g of ice at -10 °C to change to steam at 130 °C.




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Intermolecular Forces - Attractions between molecules

Type of Force
Particle Arrangement
Example
Strength
Explanation
Between Atoms or Ions




Ionic bonds





Na+Cl–

strongest
The strong intermolecular forces between positive and negative charges creates solids with very high melting points.
Covalent bonds
 (X = nonmetal)



Cl−Cl
strong
The sharing of valence electrons results in an overlap between atoms that require energy to break.
Between Molecules




Hydrogen bonds
(X = N, O, or F)



Hδ+−Fδ− ••• Hδ+−Fδ−

medium
Polar molecules can form an attraction between the partially positive hydrogen atom in one molecule and the partially negative atom (F, O, and N) in another. 
Dipole-Dipole Attractions
(X and Y = nonmetals)



Hδ+−Clδ− ••• Hδ+−Clδ−

Weak-medium
For polar molecules, these attractions occur between the positive end of one molecule and the negative end of another. 
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Weak-medium
Dispersion Forces (London Forces)
(temporary shift of electrons in nonpolar bonds)


Fδ+−Fδ− ••• Fδ+−Fδ−

weakest
Very weak attractions that result from a temporary dipole that align the molecules so the positive end of one is attracted to another. Although, weak they allow for nonpolar molecules to form liquids and solids and are important for protein folding



A piston compresses a gas from a volume of 4.10 L to 1.65 L against a constant pressure of 1.95 atm. There is a heat gain by the gas of 175 J. What is ∆U for the gas in J?













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The combustion of hydrogen-oxygen mixtures is used to produce very high temperatures (ca. 2500 °C) needed for certain types of welding operations. Consider the reaction to be 
H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -241.8 kJ/mol 

What is the enthalpy change, in kilojoules, when a 180 g mixture containing equal parts of H₂ and O₂ by mass is burned?

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A gas expands against a constant pressure of 0.750 atm while it absorbs 115 J of heat from the surroundings. The final volume is 39.2 L, and the internal energy of the gas has decreased by 255 J. What was the initial volume of the gas in L?

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In 2020 in the Port of Beirut in Lebanon, a deadly explosion occurred when 2750 metric tons of ammonium nitrate fertilizer detonated during a fire in a harbor warehouse. In the Beirut explosion, 3.436 × 10⁷ moles of NH₄NO₃ produced 4.88 × 10⁹ liters of gas, The detonation reaction of ammonium nitrate is represented by:
 2 NH₄NO₃(s) → 2 N₂(g) + O₂(g) + 4 H₂O(g). 
The standard enthalpy of formation of ammonium nitrate is -365.6 kJ/mol and the standard enthalpy of formation of H₂O(g) is –241.82 kJ/mol. 
Determine the enthalpy, in kJ, from the Beirut explosion under standard conditions













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