Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec 7_10 September	© R. Paselk 2003
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Thermodynamics, cont.

Spontaneity and Entropy, cont.

We must consider the system and its surroundings to determine the overall change in entropy and thus the spontaneity of a given process. In most cases we need only analyze the local surroundings to understand what is happening, though ultimately it is the entire Universe which determines spontaneity.

To make things even more complex, the net entropy change, and therefore spontaneity of a reaction depends also on the temperature. To understand the temperature dependence, consider the vaporization of water as an example:

• First note that the vaporization of water is endothermic (heat must be added) - it is opposed by enthalpy.
• On the other hand the vapor is much more disordered than the liquid - vaporization is favored by entropy.
• Observationally we see:
  • The liquid state is favored at temperatures below 100°C.
  • The vapor state is favored at temperatures above 100°C.

Why should such a change occur with temperature? At either side of 100°C DH will be about the same as will DS, within the system.

Apparently the change must be due to changes in the surroundings. Let's look at an exothermic process because its a bit easier to follow. Essentially, the fractional change in entropy in the surroundings due to enthalpy will be greater at low temperatures. Generally we can therefore say:

1. DS_sur depends on the direction of heat flow. For exothermic processes DS_sur is positive at constant T.
2. The magnitude of DS_sur depends on T - the magnitude increases as T decreases.

Thus DS_sur depends on heat/temperature (with temperature in Kelvins):

• Exothermic: DS_sur = + heat/T
• Endothermic: DS_sur = - heat/T

At constant pressure we use DH for heat exchange. We also generally look at the enthalpy and entropy of the system, so an exothermic process, for which the DH of the surroundings is +, the DH of the system is -, and for endothermic DH is +. Thus

Note that the (-) sign occurs because we are looking at the surroundings, but DH is defined for the system:

-DH_sys = +DH_sur

We can now begin to see why the vaporization of water changes in spontaneity with T:

DSsys DSsur DSUniv   Under 100°C + - - DSsys < DSsur Over 100°C + - + DSsys > DSsur

Note that this represents a continuum - that is the % liquid vs. vapor changes continuously as the vapor pressure increases, until at 100°C the vapor pressure = atmospheric pressure and all of the water evaporates.

Summarizing overall:

DSsys DSsur relative magnitude  DSUniv Spontaneous Processes + + + + - |DSsys| > |DSsur| +  - + |DSsys| < |DSsur|  + Non-spontaneous Processes - - |  -   + - DSsys| < |DSsur| -   - + DSsys| > |DSsur| -

Free Energy

To predict whether a process is spontaneous then, we need to determine the DS of the Universe. This is not always convenient however, so another expression is more often used in chemistry and the life sciences, the "free energy" = G, defined as:

The free energy is particularly useful because it tells us the energy from a process available to do useful work at constant pressure - often an important consideration in life!

For a process occurring at constant pressure then,

Thus a negative DG must correspond to to an increase in the entropy of the Universe. This can be seen if we divide both sides of the free energy expression by T:

but we showed earlier that:

substituting:

So

Thus a process with a negative DG is spontaneous!

Discussion of DG vs. negative and positive values of DH and DS.

 

Free Energy and Chemistry

First let's review some important aspects of free energy.

• DG, DH, and DS are all state functions, they are pathway independent , depending only on the initial and final states.
• What about the problem of using DG = DH -TDS to look at the operation of reciprocating heat engines such as steam engines or gasoline engines in cars? After all, the pressure changes in the pistons as the chemistry takes place.
  • However we can still use this expression if we construct the problem carefully, looking at DG at the beginning and end of the cycle where the cylinder is open to the atmosphere and thus at the same pressure. Under these defined conditions DH is completely legitimate.
• So, one of the things to keep in mind for solving thermodynamics problems is to make sure the conditions at the beginning and end of the process are the same!

As an example of picking conditions to solve a problem let's look at a variation of the example on p 805 in your text.

Find the entropy of vaporization of Br₂ given the enthalpy of vaporization, DH, and a table of vapor pressure versus T. DH = 3.1 x 10⁴J/mol. So how do we approach this problem?

1. Write the equation:

   DG = DH -TDS

   rearranging

   -TDS = DG - DH

   TDS = DH - DG

   DS = (DH - DG)/T

   • We know DH and T, but what about DG? Let's think about the process. Remember that DG tells us about spontaneity. Is there a condition where the system has no net tendency to either vaporize or condense? If the system doesn't tend either direction then DG must be 0! This occurs in any system at equilibrium. An obvious equilibrium position is at the boiling point where the vapor pressure = atmospheric pressure. So all we need to do is look up the boiling point and set our calculations to the appropriate temperature! T = 333K.
2. We can now place values into our equation:

   DS = (DH - DG)/T = (3.1 x 10⁴J/mol - 0)/333K

   DS = 93.0 JK^-1mol^-1

 

Free Energy and Chemical Reactions

We very frequently want to find DG for a chemical process. Last semester we saw that we can determine the values for DH for chemical reactions using tabulated heats (enthalpies) of formation (DH°). Similarly we can find DS from tabulations of Standard Molar Entropy values, DS°.

Example: Calculate DS for the complete combustion of ethane (C₂H₆) with oxygen (O₂).

First we need to write the chemical equation:

C₂H₆ + O₂ Æ CO₂ + H₂O

Balancing by inspection:

C₂H₆ + 7/2 O₂ Æ 2 CO₂ + 3 H₂O

(Notice that we left the coefficients in fractional form. This is acceptable for thermo problems because we are attempting to find the molar values of DH, DS, DG, etc. and giving whole number coefficients will add another step and thus another chance for error.)

We'll finish this example next time!

© R A Paselk

Last modified 12 September 2003