Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109

General Chemistry

Summer 2002

Lecture Notes:: 25 June

© R. Paselk 2002
 
     
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Thermochemistry

Hess's Law

Hess's Law states that changes in enthalpy in any process depend only on the nature of the reactants and products, and is independent of the number of steps in the process or the pathway taken. Hess's Law is thus a result of the fact that enthalpy is a state function.

Hess's Law turns out to be extremely useful for determining the energy of various processes, and thus the conditions necessary for reactions to proceed. The pathway independence is particularly nice, because we can look at processes that have never been observed to occur in a laboratory, and reasonably discuss the thermochemistry of processes that might occur at the Earth's core or the heart of a comet etc.

In order to use Hess's Law we need to keep some properties of enthalpy in mind.

 

Standard Enthalpies (Heats) of Formation

Formation Reactions and Standard Enthalpies of Formation: If we consider a reaction in which compounds are formed from elements in their standard states then the value of DH is the standard enthalpy (heat) of formation.

Standard States:

With this information we can now find DH for any chemical reaction!

Example: Find the value of DH for the reaction:

2 CO2 + 7 H2 Æ C2H6 + 4 H2O(g)

From Table find DH values:

Often just list compounds, since known to be from elements.
 C + O2 Æ CO2  DH = -393.5 kJ mol-1
 2 C + 3 H2 Æ C2H6  DH = -84.6 kJ mol-1
H2 + 1/2 O2 Æ H2O(g)  DH = -241.8 kJ mol-1

now we can put the reactions together, adding the enthalpies for products, and subtracting enthalpies for reactants (since the reaction directions are reversed), and multiplying enthalpy values by the coefficients of the balanced equation (since all of the formation reactions were based on coefficients of one).

-2 (DHCO2) -7 (DHH2) + (DHC2H6) + 4 (DHH2O) = DHrxn

-2 (-393.5 kJ mol-1) - 7 (0) + (-84.6 kJ mol-1) + 4 (-241.8 kJ mol-1) = -2.648 x 102kJ mol-1

 

Atomic Structure & Electromagnetic Radiation (Light)

Electromagnetic Radiation comprises the various types of forms of radiation which propagate through space not associated with mass. The visible spectrum encompasses a very narrow region of the overall electromagnetic spectrum as seen on figure 7.2 on p 293 of your text. [overhead 55] Electromagnetic radiation behaves in most circumstances as waves [overhead 54, Figure 7.1 p 292] and can thus be characterized as waves.

Three parameters determine a wave:

These parameters are related by the the expression:

v = nl

For electromagnetic radiation (light) the speed is defined in a vacuum: v = c = 2.9979 x 108m/s

c = nl in vacuo

 

Quantum Reality

It turns out that the energy associated with light is not continuous like an ocean wave, but rather is "packaged' as we would expect with a particle, and proportional to the frequency. Thus the energy per packet, or photon is:

Ephoton = hn = hc/l

where E is the energy per photon in J when l is expressed in meters and

h = Plank's constant = 6.626 x 10-34J*s

Example: What is the energy of a 400.0 nm photon?

Ephoton= hc/l

Ephoton= (6.626 x 10-34J*s)(2.9979 x 108m/s) / (4.000 x 10-7m) = 4.966 x 10-19 J

As particles photons also exhibit momentum (mv), that is, in reflecting off a surface they act as if they were particles with a certain mass and velocity. Though photons themselves don't have "mass" per se, their energy can be interconverted to mass via Einstein's famous equation:

E= mc2

Example: What is the energy of 1 g of matter?

E= mc2

E = (0.001 kg)(2.9979 x 108m/s)2

E = 9 x 1013kJ (Note a Joule = kg m2s-2, this corresponds to about 22 megatons of TNT!)

Since light has matter-like properties, perhaps matter has light-like (wave) properties. In the 1920's de Broglie noted that one could set the energy relationships above equal to each other and postulate a wavelength for matter particles:

E= mc2= hc/l

dividing both sides by c2,

m = hc/lc2 = h/lc for photons, but since matter moves at less than c,

m = h/lv, and

l = h/mv

It turns out that small particles such as electrons do indeed behave as if they are waves under certain circumstances. Thus for an electron with mass = 9.11 x 10-31kg traveling at 4 x 104m/s:

l = h/mv

l = (6.626 x 10-34J*s) / (9.11 x 10-31kg)(4 x 104m/s)

= (6.626 x 10-34kg m2s-2*s) / (9.11 x 10-31kg)(4 x 104m/s)

l = 1.8 x 10-8m = 18 nm

Compare this to the wavelength of visible light (400 - 700 nm). This is why electron microscopes are so valuable - they have much greater resolution since the wavelengths are much shorter. Other particles, such as neutrons have higher masses and thus shorter wavelengths so they should have even greater potential resolution. Today folks have even created systems where atoms may be used as probes using their wave-like properties.


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