Whether a reaction happens or not by xumiaomaio


									15   Thermodynamics:

What thermo tells us:

                 Whether a reaction happens or not.
By the time we finish this chapter, you will be able to look at reactions like the ones
in the table below and tell me all kinds of interesting things about the reaction,
including, most importantly, whether the reaction happens (whether the reaction is
spontaneous.) And since this is the start of getting really quantitative about
chemistry, you will be able to tell not only whether it happened, but also the extent
to which a reaction occurs.

Pretty exciting stuff from just a single chapter.
Reaction spontaneity: it’s not just for thermo anymore:
What is even more interesting is that except for Kinetics in Chapter 16, the rest of
the semester will be spent asking the question, did the reaction happen? And we
will use three different approaches to answering the question:

Did the reaction happen?   Thermodynamics         CH 15            DG < 0
Did the reaction happen?   Equilibria             CH 17-20          K >1
Did the reaction happen?   Electrochemistry       CH 21             E >0
                           Thermodynamic reaction worksheet

Predict the vales for DH, Dngas, w, DS and DG. At the least provide a sign, but if you can offer a
guess of the magnitude without using a calculator, that is even better.

reaction                                                      DH      Dngas   w      DS       DG
CH4(g) + O2(g) ‡ CO2(g) + 2H2O(g)
2H2(g) + O2(g) ‡ 2H2O(g)
2H2O(g) ‡ 2H2(g) + O2(g)
C2H5OH(l) + 3O2(g) ‡ 2CO2(g) + 3H2O(g)
C2H5OH(l) + 3O2(g) ‡ 2CO2(g) + 3H2O(l)
C3H8(g) + 5O2(g) ‡ 3CO2(g) + 4H2O(g)
4H2O(g) + 3CO2(g) ‡ C3H8(g) + 5O2(g)
CCl4(l) ‡ C(s) + 2Cl2(g)
Ba(OH)2(H2O)8(s) + 2NH4NO3(s) ‡ Ba(NO3)2(s) +
2NH3(g) + 10H2O(l)
3O2(g) ‡ 2O3(g)
H2O(s) ‡ H2O(l)
CO2(g) ‡ CO2(s)
NH3(g) + HCl(g) ‡ NH4Cl(s)
2H2O2(l) ‡ 2H2O(l) + O2 (g)
 (You should be able to do this worksheet by the time we finish this chapter (if you really
understand the material.)
Course Notes on Thermodynamics

A timeout before we get started: If you compare the notes for Chapter 15 to the
review sheet on thermodynamics (entitled “Thoughts in Thermodynamics”), you
will see some differences. In particular, the review sheet first presents the concept
of DG, the free energy of the system, while the notes begin with a discussion of
DH, the heat in the system. This is because the notes follow the textbook, while
the review sheet follows the way thermo should be taught (IMHO). Plan on me
providing an overview of thermo using the review sheet in the first lecture, and
taking a more detailed look at the course notes in the next two lectures.
Thoughts on Thermodynamics: A Study Guide for Chapter 15
Nine Content Areas You Should Know

In organizing your study plans for thermo, it is essential to be able to distinguish the
seemingly endless stream of very similar concepts from each other. This is best
done by creating a hierarchy of important facts about thermodynamics—note that
the relative importance of this material doesn’t run from front to back, it is more
like from back to front.

Thus we start with DG, the first of none topic areas:
1. DG, The Gibb’s Free Energy.

The most important thing about thermodynamics is this: thermodynamics tells you
whether a reaction is spontaneous or not, whether it occurs or not. To do this we
look at the Gibb’s Free Energy, DG and ask the following:

     If DG positive (+)   the reaction is not spontaneous—it doesn’t happen

     If DG negative (-)   the reaction is spontaneous—it happens
2. So what is DG? It is the free energy of a system.

The free energy is a combination of three kinds of energy:

       • heat energy, q used to calculate the enthalpy, D H

       • work, usually a change in P or V (PDV) for a reaction involving gases

       • TDS, the entropy, which is a measure of the disorder in a system
3. We can combine these three forms of energy in a couple of ways:

  • DG = DH – TDS

          which is a really famous equation used to calculate free energy

  • DE = q + w = DH – PDV

          which describes the internal energy in a system
4. Enthalpy, DH:

The heat of reaction: Very simply, this term provides a quantitative measure of
how much heat is given off in a reaction.

• Endothermic reaction: DH is +.

        The surroundings get colder as the system gets hotter.

• Exothermic Reaction: DH is -.

        The surroundings get hotter as the system gets colder.
  5. Calculating DH:

There are two ways to do this.

• Experimentally using a bomb calorimeter:

     Calculate DH = mCDT.

     This is how they measure the calories of food products.

• Using tables and Hess’ Law.

                Hess’ Law Hrxno = S DHfo products – S DHfo reactants
6. Hess’ Law: its where you start and stop, not how you get there.

No matter how we calculate a change in a state function, it isn’t the path that
matters, it is where you start and where you end. Because of the first Law of
Thermodynamics, we know that energy is conserved. So if we want to calculate
the DH for a reaction, we can put together any combination of reactions that
cancel out to leave the DHrxn we want.

There are two ways we apply Hess’ Law in CH302 problems:

  heats of formation               and              bond energy:

• Heat of formation, DHfo values in Appendix K. DHfo is the heat of reaction
  when elements in their standard state react to make an product.

• Bond Energy: Rather than use Appendix K, we can apply Hess’ Law using
  average values for each and every broken and formed bond. We use tables of
  average covalent bond energies.
7. PV Work (bombs) (and why it often is ignored):

Defining PV work. Some reactions don’t just give off heat, they produce and use
up gases that change the pressure or volume of a system. For example, the pistons
in your car do this with combusted gasoline. Very simply, this is PV work that
occurs when either the pressure changes or the volume changes when a reaction of
gases occurs. If a reaction involves only liquids and solids, or no net change in the
number of moles of gas, there is no PV work and

                      DE = q + w = DH – PDV or DE = DH

Calculating PV work. If there is PV work it is easy to calculate from the ideal gas
                          PDV = DnRT

For example, for a one mole change in gas, PDV = DnRT = 2.5 kJ which is much
smaller than most combustion reactions.

A practical note about PV work: It is often approximately true that

           DE = q + w = DH + (something small) and again DE = DH
8. Entropy, DS: It takes a lot of energy to keep things neat.

Systems tend to become messy. This is because there is energy in the world above
absolute zero that makes bonds vibrate and molecules jump around. Before you
know it, the entropy, S is larger.

            •    When DS = -, the system is increasingly ordered.

            •    When DS = +, the system is increasingly disordered.

Examples of how to predict DS: On the exam you will be asked to predict system
that are ordered or disordered.

• Disorder occurs as phase changes happen and solids become liquids become

• Disorder occurs when the volume increases, when temperature increases, when
  the number of particles increases.

• The opposite conditions favor increased order and require that energy be
  removed from the system to the surroundings so all that jiggling will stop and
  things will freeze over.
9. Putting together S and H to make G:

Look at the equation
                             DG = DH – TDS

Note that to make a reaction spontaneous, DG must be negative. This is done by
making S more positive and H more negative.

For any reaction, one of four possible combinations of DH and DS occur:

• If DH is negative and DS is positive a reaction happens

• If DH is positive and DS is negative a reaction can’t happen

• If both DH and DS are positive or both are negative, it is possible for a reaction to
  occur and depends on the temperature.
A Study Hint:

Everything you need to study for thermo is founded on the 9 concept areas
presented on the review sheet. When working thermo problems, before actually
finding an answer, identify the material on the review sheet associated with the
problem. That way you know where to go for help in figuring out how to do them.

And finally, the course notes:
Thermodynamics is the study of energy change,       D,      in chemical systems.
Here are the kinds of things that change in a system. Some of them you have seen

                DH, DE, DS, DG, DT, DV, DP
What thermodynamic changes can do for you:

           They will tell you if two things will react.
           They will tell you how much energy change occurs.
           They will tell you how far a reaction will go to completion.
              DH Reaction Categories: Exothermic and Endothermic

For thermodynamic purposes, chemical reactions fall in two categories, the ones
that happen and the ones that don’t.

     Spontaneity favored   =     exothermic       =    negative DH

                NaOH + HCl -----> H2O + Na+ + Cl- + heat
                    (temperature of surroundings increases)

     Spontaneity not favored     =     endothermic     =     positive DH

Ba(OH)2.8H2O + 2NH4NO3 + heat ----> Ba(NO3)2 + 2NH3 + 10H2O
                 (temperature of surroundings decreases)
Thermodynamic concepts that you should know:
 system                         the substances involved in the chemical system
 surroundings                   everything else in the universe
 universe                       system and surroundings
 first law of thermodynamics    energy is conserved in chemical reactions
 second law of                  entropy in the universe is always increasing
 state functions                parameters that define the current state of a
                                chemical system: P, V, T, H, S, G, composition
 change in state functions      all we care about in thermo: D = (what we end with
                                - what we start with)
 change in enthalpy (DH)        change in heat content of system
 calorimetry                    experimental technique to find DH
 Hess’ Law                      The path to a change does not matter, only where
                                you start and where you stop
 B.E. Bond energy calculation   Path through gaseous atoms to determine DHrxn
 DHfo Heat of formation         Path through elements
                                in std. states to determine DHrxn
     Time out for Specific Heat
I know that everyone remembers specific heat and heat capacity since you
learned about them in Chapter 1-13, but I thought we could use a breather before
things get hard.

specific heat:   The amount of heat required to raise the temperature of a gram of
                 substance one degree C.

                                  (amount of heat, J)
specific heat has units of: C =

                                  (mass of substance, g)(DT)

heat capacity: Amount of heat required to raise temperature of an object one
degree C. (You get the heat capacity just by multiplying the mass of an object
times the specific heat.)

molar heat capacity: Of course everyone’s favorite mass is the mass of a mole of
things which yields the idea of molar heat capacity
A few constants for those who don’t want to peruse Appendix E :

substance                   specific heat               molar heat capacity
aluminum                    0.90 J/goC                  24.3J/moloC
iron                        0.444 J/goC                 24.5 J/moloC
benzene                     1.74 J/goC                  136 J/moloC
water                       4.18 J/goC = 1.0 cal/ goC   75.3 J/moloC

Of these, the one that matters for bomb calorimetry measurements is water.

Rearrange the definition of specific heat above to yield:

                 DH = mCDT        fl the experimental way to find DHrxn

Example: How much heat is necessary to raise the temperature of 205 g of H2O
from 21.2oC to 91.4oC?

DH = mCDT = (205g)(4.18J/goC)(70.2oC)                       = 60,154 J = 60.1 kJ

Wait, this is how to do the problems back in Chapter 13!!
How to calculate DH experimentally using a bomb calorimeter:

                Applying the Concept of Conservation of Energy.

The problem. Those values for calories on the back of a candy bar were obtained
by using a calorimeter. A calorimeter is a device that captures all of the heat
produced in a combustion reaction. Now you might wonder why we would go to
all of that trouble. After all, if
                                   DH = mCDTcandy bar

then if we know the values for m, C, and DTcandy     bar   for the candy bar, we just
calculate DH for the candy bar.

The difficulty, of course is finding DT, I don’t know about you, but determining the
DT for a candy bar when you catch it on fire is not an easy thing.
A solution: In a closed system, the first law of thermodynamics states that:
                             DHsystem = DHsurroundings

So what about placing the candy bar into a container in the middle of a vat of water
(a calorimeter). Then is must be true that

     DHsystem (candy bar) = mCDT candy bar = DH water (surroundings) = mCDT water

And since it is much easier to measure temperature change for a container of water,
we use values of m and C for water to get DH for the system (the burning candy
bar) indirectly.
Nothing is perfect. Of course, not all the heat from an exploding candy bar will end
up in the water. Some will transfer to the calorimeter itself. So we need to add a
correction for the lost heat and we get the true calorimeter equation:

DHsystem (candy bar) = mCDT candy bar = DH water + DH container = mCDT water + mCDT container
Working problems: Now with this big old equation, we can ask you to find just
about any value by giving you all the other data. The big trick: keeping track of
what is the system and what is the surroundings.
Thermodynamic Equations
A balanced equation that includes a value for DH is a thermochemical equation.

Example, the combustion of pentane:

C5H12(l)+ 8 O2(g) -----> 5CO2(g) + 6 H2O(l) DH = -3523kJ/mol rxn
1 mole    8 moles        5 moles    6 moles
Things to note about pentane combustion:

1. This is an exothermic combustion reaction (the DH is negative).

2. The reaction could be made endothermic by reversing reactants and products.

5CO2(g) + 6 H2O(l) ----->          C5H12(l)+ 8 O2(g) DH = +3523kJ/mol rxn

3. The physical state (gas, liquid or solid) of the materials matters in assigning DH.

4. The assumption is made that this is a “mole reaction” in which the amount of
energy released is for one mole of reaction. If there is a change in the amount of
material, then DH must be changed proportionally.
Getting ready to work Thermo problems:
                  Standard States
thermochemical standard state: the most stable state of a substance under standard
pressure (one atmosphere) and temperature (usually 25 oC)

Examples of standard states:
water                                     clear liquid, H2O (l)
mercury                                   silvery liquid, Hg (l)
hydrogen                                  diatomic gas, H2 (g)
limestone                                 chalky solid, CaCO3 (s)
We denote that a change occurs under standard conditions with a superscript zero,

standard enthalpy change, DHrxno : This is DH when reactants in standard states
are converted to products in standard states

A special case of DHrxno is:
heat of formation, DHfo :This is the case where DH is for one mole of substance
formed under standard conditions from its elements in their standard conditions. By
definition, DHfo of an element is zero.
Examples of DHfo:
CH4 ,(g)                               -74.21 kJ/mole
C 6H6 (l)                              49.03 kJ/mole
CO2 (g)                                -393.53 kJ/mole
H2O (l)                                -285.8 kJ/mole
Br2 (l)                                0 kJ/mole
Br2 (g)                                30.91 kJ/mole

There are plenty more in Appendix K.
Example: Calculate the enthalpy change when for the reaction in which 15.0 g of
aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.

                4Al(s) + 3O2(g) -----> 2 Al2O3(s)     DH = -3352KJ/mol rxn

kJ = (15g Al)(1mol Al/27g Al)(1mol rxn/4 mol Al)(-3352kJ/mol rxn)

                     = -465 kJ released to surroundings
Hess’ Law
Hess’ Law:      The DH for a reaction is the same no matter what the pathway is
between reactants and products.

or stated as a nice equation to be memorized for working dozens of problems:

Hess’ Law:
                DHrxno = S n DHfoproducts -            S n DHforeactants

What this law allows us to do is combine thermodynamic equations any way we
want to fine new thermodynamic equations for unknown reactions.
Bond Energies
bond energy: Energy necessary to break one mole of bonds in a gaseous
substance. Note the substance has to be in the gas phase to use bond

Bond energies can be used in a special case example of Hess’ Law to give
a rough estimate of DH rxno when all else fails. Consider a reaction in
which the H-H bond for hydrogen in its standard elemental state is broken
to form atoms.

              H2(g) ------> 2H (g)     DHrxno = +435 kJ/mol

This is the simplest example of a bond energy.
Consider the more complicated case of successive cleavage of a methane

    CH4 --->       CH3(g)    +    H(g)      DHo = +427kJ/mole
    CH3 --->       CH2(g)    +    H(g)      DHo = +439kJ/mole
    CH2--->        CH (g)    +    H(g)      DHo = +452 kJ/mole
    CH --->        C (g)     +    H(g)      DHo = +347kJ/mole

If you think about it, you can cleave a C-H bond in any kind of organic
molecule and the DHo will be a little different each time. For this reason
we need to use an AVERAGE bond energy. The good thing is that we can
include all the average bond energies in a single table.
A nice table lifted from Davis, p561 gives us some single bond values.
We can also construct a table of average multiple bond energies.
Why go to this trouble? Because in a variation on Hess’ Law we can find
the DHrxno for any reaction IN THE GAS PHASE by

         DHrxno = S B.E. reactants - S B.E. products

Example: Use the bond energies in the two table above to estimate the
DHrxno for

         CH4 (g) + 2O2 (g) ---->    CO2 (g)    + 2H2O(g)

DHrxno   = [4BEC-H + 2BEO=O ] - [2BEC=O + 4BEO-H]

         = [(4)(414) + (2)(498)] - [(2)(741) + (4)(464)]

         = -686 kJ/mole
Notes about BE:

  • Using bond energies we found a value for DHrxno and we never had to
    access thermochemical data like other DHrxno or DHfo.

  • But... we could only look at gas phase reactions.

  • And...the  answer is only approximate and is only close when all
    reactants and products are gases. There are huge errors when using
    BE for other phases.

  • The advantage:    all the data you need for the calculation can be found
    in a small table of individual bond energies.
  • Internal Energy Change: An Intro to DE

We’ve examined DH for a system and talked a lot about changes in
energy. But we’ve not discussed DE. What is the relationship between
the two? The fundamental equation relating DH and DE is

                      H = E + PV
PV is pressure-volume     work.
It is work in which during the course of a reaction, there is a change in
pressure or in volume of the system. As we will see, on many occasions
there is no DP or DV for a reaction and DH = DE. Very often even when
DP or DV are non-zero they are negligible and the approximation that DE
= DH holds. To understand the experimental conditions under which these
cases occur, it is necessary to better understand PV work.
DE defined:

         DE = Eproducts- E reactants = q + w

where q is the heat absorbed or released by the system and w is work done
on or by the system. We will see that w is PV work.
But first, we have to get our signs straight:

According to Davis (and therefore all the problems you will do)

     q is positive:   when heat is absorbed by the system
     q is negative:   when heat is released by the system
     w is positive:   when work is done on the system
     w is negative:   when work is done by the system

In this convention whenever energy is added to the system, the energy of
the system increases.
PV work
Why is PV work, work anyway? Look at the units:

                  P        x           V
                 F/d2 x        d   3
                                       = Fd = w !!!

PDV Work: Our first example of PV work will be the case where we
keep P constant and vary V. This is PDV work. Actually, most of the
reactions you do will involve PDV work because they are done at
atmospheric pressure.

Thus                       DE = q - PDV
But wait. Why is there a minus sign? Well consider what happens when
we have a reaction in which the amount of gas in the system increases
compared to a reaction in which the amount of gas decreases in the
In the picture above, a reaction occurs in which the volume of gas
decreases. Work is being done by the surroundings to the system. From
our table of signs above, this means w is positive. But the volume
decreased, i.e., DV is negative. To make the equality hold,
                          +        -           -
                          w =                PDV
we must add a negative sign to the equation. hence the general equation

                        DE = q - PDV

    Note that if the volume of a reaction doesn’t change, then

                             DE = qv
The energy of the system is just the heat released or absorbed by the
system. When does this happen? Well to a good approximation, in the
following three cases.

When can we ignore PDV work?

1. In reactions involving only solids and liquids

2. In reactions involving gases in which an equal number of moles of gas
          reactants and products are involved.

3. In a bomb calorimeter. That is right, our friend the bomb calorimeter
measures only qv because the volume is held constant and no PDV work is
done so
                                 DE = qv

and we can do exactly the kind of calorimetry problems we learned about
at the beginning of the chapter.
A little more on DE and DH.
Now that we better understand PV work, we can better understand the
relationship between DE and DH. Remember by definition

                          H = E + PV

And at constant T and P

                          DH = DE + PDV
Example: Can we show that at constant pressure,

                      DH = qp
we know       DH = DE + PDV and DE = q + w

so,       DH = q + w + PDV (constant T, P)

at constant P, w = -PDV

so,       DH = q + (-PDV) + PDV = qp
How does DH compare to w in magnitude?

Thus the difference between DE and DH is PDV and if we have no gas
present or if equal moles are exchanged then DH = DE.

But what about when we do have a gas present. Assume an ideal gas law
                   PDV = DnRT
so                DH = DE + DnRT

How significant is this DnRT term?
Sample calculation of w

Example: If a reaction occurs at standard temperature in a bomb
calorimeter and one mole of gas is produced, how much PV work is done?
                   w = -PDV = -DnRT

w = -(1mol)(8.314J/mol K)(298K) = 2.5 x 103 J

so, PDV work is a couple of kJ/mole at room temperature

Note we get a value of a couple of kJ/mole. How does this compare with
typical combustion reaction DHrxno? Consider for example the DHrxno of
methane which is -890 kJ/mol, PDV work is much smaller. For many gas
reactions, then, the PDV contribution to the energy of the system is
negligible and again,

                            DH ª DE
Spontaneity and Entropy
I lied. Actually I approximated. On page one of the notes it is suggested
that an exothermic reaction, -DH, is synonymous with a spontaneous
reaction while an endothermic reaction, +DH, is synonymous with a non-
spontaneous reaction. This is not always true. Usually it is, but not
always--because in addition to the energy change associated with a
reaction, there is a second parameter to consider in determining

              the change in entropy, DS
Entropy and the Second Law—the universe is getting messier

Now you all know a bunch of jokes about entropy, and it may be kind of
easy to tell you pat definitions about entropy like

The second law of thermodynamics: In spontaneous changes in chemical
systems, the universe tends to a state of greater disorder.

But actually, entropy is quite difficult to discuss on a quantitative level.
Thus whereas you will work hundred of problems involving energy
changes, you will work only a few involving entropy changes.

The problem is that to describe entropy mathematically, you have to
understand probability theory and be a whiz at statistics. I would bet most
of you don’t and aren’t. So we will stick to a kind of hazy overview in
which we hand wave about entropy being an increase in disorder of a
system--we’ll save the quantitative details for when you take a statistical
mechanics course, that’s CH354L for those who want to plan ahead.
Entropy and signs: When we talk about increasing disorder in a system,
we are describing an increase in S. In other words, DS is positive. For
example, consider a phase change as molecules go from solid to liquid to
gas. There is increasing disorder so DS for the change in the system if
Spontaneity and entropy: According to the second law, if entropy
increases, the spontaneity of the process is favored. But, this doesn’t
mean the process will be spontaneous--remember, we have to consider DH
as well.
Entropy and perfection.

Suppose we try to increase the order of a chemical system, make it
perfectly ordered. This is the same as saying that entropy of the system is
zero. Can this happen? No. This is the basis of the third law of

Third law of thermodynamics: The entropy of a perfectly ordered
crystal is zero at absolute zero (0K).

Note in the picture that were a perfect crystal to be created, then by adding
heat to the system (raising the temperature), the bonds in the crystal will
vibrate and entropy increases.

We will see this at play when we look at kinetic theory in the study of
gases, liquids and solids.
A Hess’ Law calculation using S data from Appendix K

Standard Entropy Changes.

We can actually treat calculations of DSrxno as we would DHrxno and use a
Hess’ Law equivalent to take known DS rxno values from Appendix K to
find unknown DSrxno values. The equation we use is

          DSrxno = S n DSfoproducts -   S n DSforeactants

and the problems are worked identically to those for DHrxno.
Predicting the sign of DS.

We can’t do the math right now, but we all have an intuitive feel for
whether disorder increases or decreases in a system. Can we predict the
sign of entropy changes that occur in molecular systems?

Phase changes: This is an easy one. As we move from solid to liquid to
gas, molecules have increased freedom of motion and hence the
opportunity for increased disorder, so DS is positive. Conversely, as
molecules are cooled from gas to liquid to solid, DS is negative as order
Temperature changes: The argument for phase changes holds here.
Adding heat to a system gives the molecule greater ability to vibrate,
rotate or travel. Hence DS is positive.

Volume changes: If we increase the volume of a system, the molecules
have more places they can go, hence DS is positive. If the volume
decreases, DS is negative.
Increase in number of particles.

Obviously the more particles we have in a system, the greater the
opportunity for disorder, hence DS is positive. A particular application of
this is to chemical reactions. Consider the dissociation of a diatomic

         H2 (g) ----> 2H (g)       DSrxno = +98.0J/molK
Example of entropy prediction

In general we can look at a chemical reaction and evaluate whether there
is an increase in the number of reactant species or an increase in the
number of product species. If the number of products increases, DS is
positive. If the number of products decreases, DS is negative.

For example,
     2H2(g) + O2(g) ----> 2H2O (g)

Because Dn for the system goes down, there is more order and we would
predict a decrease in entropy

And sure enough, DSrxno = -89J/molK from App K data.
                 Turning entropy into energy units: TDS

Where do we go from here? Well it isn’t too much of a stretch to see that
we will want to combine what we know about DS and DH. But before we
do, let’s make a guess at how to relate the magnitude of D S to DH.
Consider for example the 89 J/moleK just calculated for the diatomic
molecule above. How do we convert it into something to compare to DH?
Why not just multiply by T to get J/mol? At room temperature:

          TDS = (298K)(89J/moleK) = 26 kJ/mole

Compare this value with some of the DHrxno in Appendix K. Note that in
many cases, the entropy term for a reaction will be relatively insignificant
in determining the spontaneity of a reaction.
         DG (Gibbs Free Energy) and Spontaneity
Now it is time to put all we have learned about DH and DS together. As
noted at the beginning of the chapter, spontaneity is actually determined
from both DH and DS. This is because when energy, DH, evolves from a
system some of it may do work, but some may also be used to alter the
entropy of the system, DS. A state function was developed by Willard
Gibbs to account for this.
Gibbs Free Energy, DG:
                   DG = DH - TDS (constant T,P)

Note that DG is increasingly negative if :
          i. the system gives off more heat, larger negative DH
          ii. the system becomes more disordered, larger positive DS
          iii. T for the system increases

If you think about it, these are the kinds of things that happen to make a
reaction more likely to occur. In general we can determine spontaneity of
a reaction from a determination of DG.

     DG                         spontaneity
     positive                   nonspontaneous
     0                          at equilibrium
     negative                   spontaneous
DG and the Equalizer Bunny: As we will learn, later in the course, if
DG is large negative then we have a system capable of doing a lot of work,
for example, a brand new battery. If DG = 0 we are at a special place in
the reaction called equilibrium where nothing happens--DG = 0 is the
scientific definition of a dead battery.
The special case of a standard reaction:

Consider the case for a thermochemical reaction that occurs at 1
atmosphere pressure, at 25oC, and with stoichiometrically correct molar
quantities of all the reactants and products. We will have created the
special case of a standard reaction with a DG rxno value found right in
Appendix K.

We can create a Hess’ Equation equivalent for DG rxno standard reactions
that allows us to mix and match other thermochemical reactions just like
we did for DHrxno and DSrxno

         DGrxno = S n DGfoproducts - S n DGforeactants
Simple calculation of DGrxno using Appendix K

Example. Calculate DGrxno for the reaction below. Is the reaction
C3H8(g) + 5O2 (g) -----> 3CO2 (g) + 4H2O (l) from App.K
-23.49     0             3(-394)   4(-120.4)

DGo = 4(-120.4) + 3(-394) - (-23.49) = -1638 kJ/mol

In the example above, DGrxno is negative so the reaction is spontaneous
(which you already knew, because you know that combustion reaction are
spontaneous which is why your car works.) Note that the reverse reaction
is endothermic because DGrxno would be positive. This also makes sense
because you can’t make gasoline by blowing into a bowl of H2O.
The Real World and DG.

The thing is, DGrxno is valid only at 1 atm and 298K for a molar reaction.
This is not a value we would use in real life where the concentrations are
not often 1M and partial pressures are not often 1 atmosphere. In the real
world we evaluate spontaneity by examining DGrxn (no tiny superscript o)
which differs from DGrxno by an amount dependent on concentration

         DGrxn = DGrxno +    something that accounts
                             for concentrations not = 1M

We will learn about this equation later. For now we will just play around
with DGrxno.
Temperature and spontaneity.

Look at the Gibbs equation:

                        DG = DH - TDS

The temperature term is a nice easy experimental variable just begging to
be manipulated. Try sticking various values of DH and DS into the
equation while you vary T. You’ll note the following trends:

   class         DH     DS       DG
   one           -      +        spontaneous (-) at all T
   two           -      -        spontaneous (-) below a specific T
   three         +      +        spontaneous (-) above a specific T
   four          +      -        nonspontaneous (+) at all T
Combinations of DS and DS that determine DG

Consider an example from each class. Try thinking about the reactions in
terms of real world experiences.

 class   Reaction Example                    DH       DS       sponta
                                             kJ/mol   J/molK   neity?
 1       2H2O2 (l) --->2H2O (l) + O2 (g)     -196     +126     all T
 2       NH3 (g) + HCl (g) ----> NH4Cl (s)   -176     -285     below
 3       CCl4 (l) ---> C (s) + 2Cl2(g)       +136     +235     above
 4       3O2 (g) ---> 2 O3 (g)               +285     -137     never
Thinking for yourself.

For each of these examples, think about the conditions under which a
spontaneous reaction is to occur. Also, could you predict whether DH and
DS would be + or - from what we’ve learned, and could you then use these
values to look at any reaction and make a reasoned guess as to whether it
is spontaneous?
Rearranging an DG = DH – TDS to determine when spontaneity

Example. The answer is given above, but why don’t you calculate from
thermochemical data the temperature at which the reaction written below
becomes spontaneous. As a hint, remember, a reaction changes from
spontaneous to nonspontaneous when DGrxno = 0.
              NH3 (g) + HCl (g) ----> NH4Cl (s)

                  DG = DH - TDS = 0
                   DH     -176
              T = ææ = æææ = 618 K
                  DS    -.285
    Rearranging an DG = DH – TDS to calculate boiling points

To top it all off, use the same idea to determine the boiling point of water,
thermochemically, when you don’t have a thermometer around.

Example. Use the thermochemical data in Appendix K to estimate the
boiling point of H2O

          Consider the reaction H2O (l) ----> H2O (g)

This is the temperature at which liquid and gas exist in equilibrium at 1
atmosphere for standard conditions. This means DGo = 0.

DHorxn = DHf H2O(g) - DHf H2O(l) =
        = -241 kJ/mole - (-285 kJ/mole) = 44 kJ/mole

DSorxn = SoH2O(g) - SoH2O(l) =
       = 0.188 kJ/moleK - 0.69 kJ/moleK = 0.119 kJ/molK

                            DH    44 kJ/mol
DG = 0 = DH - TDS         T = ææ = æææææ               = 370K = 97oC
                            DS   0.119 kJ/molK

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