Many chemical events occur spontaneously, meaning on their own, without the input of additional
energy. Other events only occur when there is some sort of intervening action. We can oftentimes
anticipate spontaneous occurrences, for example, ice melting outside of the freezer is no surprise, living
things grow older and die. We know that tomorrow is not yesterday, thus we already know the natural
direction of many events. The first law of thermodynamics tells us that tomorrow’s energy will be the
same as today’s which is the same as yesterday’s. From an energy standpoint, all days are equal, so
events might as well runs backwards or forwards, but they do not. Milk does not unspoil, meat does not
cook itself. Time is a one-way trip. Why?
Chemical equilibrium is defined as meaning that the forward chemical reaction or event occurs at the
same rate as the reverse chemical reaction or event. We examined equilibrium when we discussed vapor
pressures of gases. The rate of evaporation (a phase change from a liquid into a gas) equaled the rate of
condensation (a phase change from the gas into the liquid). We examined equilibrium when we looked
at chemical reactions, as well as solid dissociation. So we know that chemical reactions can be reversible
as well. When the system is NOT at equilibrium, then one of the reactions – either the forward reaction
(usually termed as written), is occurring at a faster rate than the reverse of the chemical reaction. For
example, when silver nitrate is added to sodium chloride what happens?
Concept Test:
silver nitrate + sodium chloride 
The rate at which the AgCl is forming is greater than the rate at which it dissolves. The faster reaction, in
this case the precipitation reaction is said to be spontaneous. When Ag+1 ions encounter Cl-1 ions they
come together and form a solid with no intervening assistance from anything or anyone else. Sometimes
it is said that there is a net-driving force in the forward direction (or as written). When a reaction is at
equilibrium, there is no net driving force in either direction.
A reversible chemical reaction is somewhat analogous to a tug of war between two children. If one child
is stronger than the other, then a net driving force exists and the rope (reaction) will move in the
direction of the stronger child. If the two children are evenly matched against one another, there may be
momentary shifts in one direction or the other but the rope will maintain its equilibrium position.
A spontaneous change in a system, whether a chemical, physical, or a change in location, is one that
occurs by itself under some set of conditions (e.g. temperature, pressure, etc) without the ongoing input
of energy. Water spontaneously freezes at -5oC ad 1 atm. It spontaneously melts at 80oC and 1 atm.
Some spontaneous changes need a little input of energy initially, but once they get rolling, it continues
without the need for this extra energy. Fires are an example of this type of spontaneous process. After
you take the match away from the wood in your fireplace, the fire will keep on blazing if there is wood
present.
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In contrast, a nonspontaneous change requires the constant input of some form of energy in order to
maintain the chemical, physical, or change in location that they system is undergoing. A book falls off
the table spontaneously (if left balancing precariously on the edge), but what are the chances that the
book will spontaneously put itself back on top of the table? Not so good.
It is important to know that: if a reaction or change is spontaneous in one direction, it is not spontaneous
in the reverse direction.
We have talked about the first law of thermodynamics and the energy of the reaction – specifically we
were looking at heat and work. But is the first law of thermodynamics able to explain which reactions
are spontaneous or not? The answer is no. It can only account for energy transfer, it cannot predict or
explain why or which direction the reaction will proceed. For example, ice melts in your hand. It
absorbs the heat energy from your hand and melts. So why doesn’t the melted ice give off that heat is
absorbed and refreeze in your hand? The first law says that the energy of the system will be conserved if
that would happen – so why doesn’t it happen? Obviously there must be other variables to consider.
It was believed, for many years, that if a reaction was exothermic or endothermic, that the sign of H
would predict whether or not a reaction was spontaneous. For many years, it was believed that
exothermic reactions were spontaneous while endothermic reactions were nonspontaneous. And while
many exothermic reactions are spontaneous, there are several real world simple examples of
spontaneous endothermic reactions. Ice melting for example. Ice placed in your hand removes the heat
from your hand and uses that heat energy to change from a solid to a liquid. Water vaporizing is also
an endothermic reaction. It takes energy for water to change from the liquid to the gas phase. However,
even at room temperature and normal pressures, some water molecules are vaporizing! Salts dissolving
in water (NaCl, NH4NO3) and turning into ions is also an endothermic reaction that is spontaneous.
Examining these examples carefully shows us some things in common:
solid  liquid  gas
ionic solid and water  ions in solution
Think about the order and disorder associated with solids, liquids, gases and ions in solution. As we
proceed from the solid state to the liquid state to the gas state disorder increases (or order decreases). It
appears that order, or lack thereof might be important for determining if a reactions is spontaneous or
not.
Therefore, there must be other factors that we must consider when determining if a reaction is
spontaneous or not.
2
3
Which situation would you expect to happen spontaneously?
(I)
(II)
Adding food coloring to water shows the color dispersing (randomizing) its molecules into the liquid
water. The food coloring does not stay as one big blob in the water. By increasing it’s the disorder of its
molecules and the water molecules the color is able to permeate throughout the entire water solution.
Would you expect that the color food coloring molecules would ever conglomerate back together into a
blob in the water? Would this be creating order or disorder if the molecules did just that? Which
situation is favored – order or disorder?
Which situation would you expect to happen spontaneously?
(I)
(II)
4
Placing ice on the countertop shows the solid, ordered water molecules melting and turning into liquid
water molecules. Is this increasing or decreasing the order of the water molecules? Would you expect
the liquid water molecules to group back together on the countertop at room temperature and turn back
into solid water? If they did, would this be creating order or disorder out of the water molecules?
Which situation is favored – order or disorder?
Entropy is a measure of disorder, which is related to the number of arrangements that are possible for a
system. A disordered system has many ways of being arranged and a high probability of existing. An
ordered system has few ways of being arranged and a low probability of existing. For example, if you
spill a glass of water it is highly improbable that the water will fall to the floor in the shape of the glass
because many more arrangements of the water molecules are possible if the water spreads out over the
floor. The water molecules are more disordered on the floor than in the glass.
A spontaneous change has a natural tendency to occur without outside intervention. The driving force
for a spontaneous process is an increase in the entropy (and disorder) of the universe.
There is a natural tendency for nature to become more disordered. Think about all the work you have to
do to keep your room clean, to keep the kitchen clean. It would and is so much easier to just throw the
dirty clothes on the floor, to leave the dishes in the sink. It is so hard and takes so much work to actually
put the clothes away or in the dirty clothes hamper and to wash the dirty dishes and then put them
away. Too much work  These are two classic easy examples to illustrate that disorder – or more
disorder is favored. Nature constantly moves towards more disorder.
Second Law of Thermodynamics: all spontaneous changes are accompanied by an increase in the
entropy of the system/surroundings. This increase in entropy of the system/surroundings means that
the final entropy is greater than the initial entropy which means that the entropy of the universe also
increases!
Entropy, like enthalpy, has its own symbol to define it. The symbol used for entropy is S. The greater
the degree if disorder, the greater its entropy. Again, entropy is also a state function. The entropy of a
system has a unique value for a given pressure, temperature, and composition. The difference in
entropy, S, between two states, will also have a unique value.
Increasing Disorder/Entropy Occurs When:
solids melt into liquids
solids or liquids vaporize to form gases
solids or liquids dissolve in a solvent . . .
a chemical reaction produces an increase in the moles of gas
a substance is heated – increased temperature means increased motion
and motion is random so there is more molecular disorder
5
6
If disorder increases, then S will be greater than 0. If disorder decreases, then S will be less than 0.
Again, Ssystem = Sfinal – Sinitial.
What if we want to examine a quantitative value (a measurable amount/quantity). To quantify the
value of S we will examine the change in entropy under reversible conditions. Under reversible
conditions, the entropy is being exchanged between the system and the surroundings at some
temperature (in Kelvin).
S =
q rev
T
To make S a state function we have to specify a path that the reaction travels. We know that heat is not
a state function and IS dependent on the pathway that the reaction undergoes. By specifying a reaction
path, a unique value, and by specifying a temperature, we set the S value for that reaction. Recall that
for a constant pressure process, q = H. So we can make a new equation that represents this:
S =
H
T
The third law of thermodynamics: the entropy of a pure, perfect crystal at 0 K will be 0.
We can only measure changes in enthalpy, not specific enthalpy values because we have no baseline
place where we can begin our measurement. Knowing that the entropy of a pure crystal = 0 at 0 K gives
us a starting point from which all other entropy measurements can be made by increasing the
temperature. Plotting changes in entropy as the temperature increases actually shows some interesting
characteristics. There is a sharp increase in entropy at the melting point of the solids and there is an even
sharper increase in entropy at the boiling points of liquids. This suggests that the greatest disorder
comes from converting the liquid into the gas phase.
In general, the more atoms in its molecules, the greater the entropy of the substance will be.
The magnitude of S is dependent upon the amount of substance present. More/less moles of reactants
and products will mean more or less disorder for the chemical reaction. Again, there is a table of S
values for the standard state of the elements/compounds in the back of the book. This is the value for 1
mole of substance at a standard temperature and pressure.
The standard molar entropy values will be used to determine the S value for a particular chemical
reaction in the same manner that H values were calculated.
As we are examining a change in entropy for a chemical reaction, there are many similarities between
the reactions that are used to calculate the entropy state function values and the reactions used to
calculate the enthalpy values for chemical reactions:
So =  n p S product -  n r S
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reactant
We can not use H as the sole criterion for determining if a reaction is spontaneous or not. Can we just
use S? You can, in fact, determine if a reaction is spontaneous or not based on this value alone if you
remember that spontaneous reactions increase the total entropy of the universe. The system does NOT
equal the universe!! In fact, Suniverse includes what happens in both the system and the surroundings,
such that:
Suniv = Ssys + Ssurr
Why does water freeze at -15oC? Freezing is decreasing the disorder of the system. Decreasing disorder
should be unfavored – and unfavored reactions should not be spontaneous. Yet we all know that if you
put liquid water in the freezer, it turns into a solid. Do not forget that when water freezes, it gives off its
heat to the surroundings. Because the heat is absorbed by the surroundings, it increases the disorder of
the surroundings. By calculations that are beyond the scope of this class, Ssurr > Ssystem. This means
that Suniverse is positive for the freezing of water which makes the chemical event spontaneous. But that
is really difficult to rationalize using Suniverse!! We KNOW that ice freezes, so obviously something is
going on – but there is an easier way to define reaction/event spontaneity.
First, here are some examples of situations to remind you of the disorder/order associated with them:
1. temperature changes affect S values. Increasing the temperature of the system cause more
molecular motion – even when no phase change occurs, such that S will increase as temperature
increases.
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2. phase changes: S of the system increases as the sample becomes more disordered such that as
the sample undergoes a phase change from a solid to a liquid  to a gas, the S of the system
value increases. Conversely, if the phase changes occur from a gas  to a liquid  to a solid the
S of the system will decrease.
3. dissolution of a solid or a liquid: dissolving solids or other liquids into a solution increases the
randomness of the molecules and thus increases the entropy. However, when some ionic solids
dissolve in water, the hydration of the ions by the water molecules actually causes the system to
become more ordered! This makes a negative contribution to the overall entropy of the system.
For molecular compounds (ethanol, methanol for example) dissolution in a solvent does not
cause the extensive ordering of water molecules that ions do, therefore the entropy of the system
is increased.
4. dissolution of a gas: actually decreases the order of the gas molecules that were once free to roam
wherever they pleased. In a solution, the molecules are contained in the liquid and the entropy
of the dissolved gas is less than the entropy of the free gas. When a gas dissolves in another gas
however (the mixing of two gases!) the total entropy increases as the number of moles of species
is increased and the motion of the molecules is unimpeded.
5. atomic size and molecular complexity: the larger the atom or the more complex the molecule the
greater the disorder if the species are in the same phase. Molecular complexity includes
compounds with more atoms involved as long as the species are in the same phase. This means
that carbon has a larger entropy value than lithium, and AlCl3 has a larger entropy than NaCl. A
larger molecule has more ways to move and rotate than a smaller molecule. Think of all the
permutations that C6H12 (l) can undergo compared to CH4 (l) (draw Lewis structure if you have
to!). Keep in mind the substances must be in the SAME phase. If you start comparing different
phases, the physical state usually dominates over molecular complexity.
Most of what we have discussed this far has been focusing on the system. But S has two
components, the system and the surroundings. In order for a reaction to be spontaneous, the S of
the universe has to be positive. This means that we really need to consider the S of the
surroundings!
The second law of thermodynamics tells us that if the S of the system decreases for a spontaneous
process then the S of the surroundings must increase. The surroundings act a lot like a heat sink or
a heat source for the system.
In an exothermic change, heat lost by the system is gained by the surroundings. The gain of the heat
by the surroundings increases the molecular motion of the molecules in the surroundings, thus the
entropy of the surroundings increases:
qsystem < 0 surroundings > 0 and Ssurr > 0
In an endothermic change, heat will be gained by the system and that is the same heat that is lost
from the surroundings. Thus, the loss of heat reduces the entropy of the surroundings.
qsystem > 0 surroundings < 0 and Ssurr < 0
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The change in the entropy of the surroundings is proportional to the quantity of heat that is transferred
into or out of the system. The change in entropy of the surroundings is proportional to an opposite
change in heat of the system. Remember, the system and the surroundings are opposite to one another!!
Therefore, we can generate a new relationship of these variables:
Ssurr = - Hsystem
T
This means that we can calculate the Ssurroundings by measuring the Hsystem. We could also write this
equation looking for the Ssystem if given the Hsurroundings. Knowing that for a spontaneous process to
occur that the total Suniverse must be positive, if either Ssystem or Ssurroundings is negative, the other value
MUST be more positive to make Suniverse positive  Yikes!
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Concept Test:
At 298 Kelvin, the formation of ammonia has a negative Ssystem
N2 (g) + 3H2 (g) → 2NH3 (g) Ssystem = -197 J/Kelvin
Calculate the Suniverse for the above reaction and determine if the reaction is
spontaneous as written at this temperature
Things to think about: what must the Ssurroundings value be in order for the reaction to be spontaneous?
How will you calculate Ssurr?
How will you calculate the H value needed?
As a chemical reaction proceeds along, we can imagine it going in the forward direction. As the reaction
begins to shift and reach equilibrium, the reaction will now be proceeding in the opposite direction.
What happens to S when equilibrium is reached? The reaction is proceeding in neither direction, the
net change in entropy is going to be 0. But remember, on the itty bitty teeny tiny scale (the nanoscopic
level) the forward reaction happens and then is counteracted with the reverse, so the reaction has some
tiny oscillations at some frozen instant in time. If the entropy =0 at equilibrium then that must mean that
the entropy of the forward direction exactly equals in magnitude the entropy of the reverse direction but
that they are opposite in sign to one another. We previously talked about the reaction being
spontaneous in one direction, it must therefore be nonspontaneous in the other!
In an exothermic reaction: Hsystem < 0 heat is given off by the system. That means that the heat is
gained by the surroundings which increases the disorder of the surroundings (temperature change
makes the molecules move more – more craziness!) If the reacting system yields products that are more
reactive than the reactants then the Suniverse will be >0 and the reaction will be spontaneous. If the
products are more ordered than the reactants, then Ssystem becomes more ordered and in order for the
reaction to be spontaneous, the heat lost by the system must make the Ssurroundings more disordered to
overcome the -Ssystem. If the Ssurroundings is NOT more positive (more disordered) than the system, the
reaction will be nonspontaneous.
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In an endothermic reaction: Hsystem > 0 as heat is lost by the surroundings and gained by the system.
That means that the heat lost by the surroundings has lowered the Ssurroundings. The only way for the
reaction to be spontaneous is for the Ssystem to be more disordered and have a more positive value than
the Ssurroundings. The system gained the heat, so that will make the molecules more disordered.
However, this one is a little bit harder to intuitively answer. If the products are a gas, then the heat
absorbed by the surroundings and the formation of the gas make Ssystem >> 0. If the products are a
liquid (or aqueous) from a solid – again there is an increase in the disorder. But if the products become
more ordered – the heat gained by the system – which contributes to order, may or may not be enough
to overcome the -Ssurroundings to make Suniverse > 0. So the reaction MAY not be spontaneous.
By making two different independent measurements for two different S variables – the system and the
surroundings – we are making life very difficult for ourselves. Isn’t there an easier way? Wouldn’t it be
nice if spontaneity could be defined by ONE variable instead of taking into consideration Ssystem and
Ssurroundings and H?? There IS such a variable – and it mathematically takes into consideration both H
and S at the same time for a given temperature which means we do not have to think so hard 
This variable is called Gibbs free energy – or simply free energy – and is given the symbol G.
And the change in the free energy of the system is related to the enthalpy component and the entropy
component at a temperature of the system:
Suniv = Ssys + Ssurr
Suniv = Ssys - Hsys
T
TSuniv = TSsys - Hsys
-TSuniv = Hsys - TSsys
The free energy change G for the system = H of the system minus temperature x Ssystem
Gsystem = Hsystem - TSsystem = -TSuniv
The sign of G is going to tell us if the reaction is spontaneous or not. And now we do not have to worry
about Ssurroundings anymore! Gibbs free energy is just another way of examining the S of the universe
under some temperature condition and determining if the reaction is spontaneous. In fact,
Gsystem = -TSuniverse So that if Suniverse is a positive #, G will be negative (all temperatures are given in
K, so the temperature value is always a positive component). And if Suniverse is negative, then G will be
positive. We know that spontaneous processes occur when Suniverse is positive, therefore spontaneous
processes must have a -G value. We know that nonspontaneous processes have a -Suniverse which
means nonspontaneous processes must have +G values. And at equilibrium, Suniverse = 0. This means
that at equilibrium, G must also equal 0.
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The sign of G will tell us if a reaction is spontaneous or not: The second law of thermodynamics tells
us:
Suniverse > 0 means that the reaction is spontaneous as written under those conditions
Suniverse < 0 means that the reaction is nonspontaneous as written under those conditions
Suniverse = 0 means that the reaction is at equilibrium
Since G = -TSuniverse
G < 0 means that the reaction is spontaneous
G > 0 means that the reaction is nonspontaneous
G = 0 means that the reaction is at equilibrium
An important point to keep in mind is that if the reaction is spontaneous in one direction, it will be
nonspontaneous in the reverse direction. Just like H values reverse their sign when the reaction is
reversed, so do G values.
Concept Test:
4KClO3 (s) → 3KClO4 (s) + KCl(s)
Grxn = -133 kJ
What is the Grxn for the following reaction?
3KClO4 (s) + KCl(s) → 4KClO3 (s)
By using G instead of Ssystem and Ssurroundings we are able to define spontaneity with one variable rather
than 2. The degree of spontaneity (the sign and magnitude of G) tell us nothing about how fast a
reaction occurs – it tells us nothing about the rate of the reaction. Just because some reaction has a large
G value does not mean that the reaction occurs quickly!
G is a state function along with H, S, and temperature. State-function variables are designated with
a small o sign next to the letter. Using this our state function equation becomes:
Gosys = Hosys - TSosys
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Knowing two of the central thermodynamic variables – or being able to solve for two of the
thermodynamic variables, we can then calculate the third. Temperatures will usually be stated.
Remember, when given tables with G, H, and S values and a reaction, you should be able to solve
chemical equations for the Grxn, Hrxn, and or Srxn.
Chemical reactions occur because they are following fundamental laws of nature or because something
forces them to react. Thus, when certain reactants are put together two options present themselves, they
react or they may not.
The two fundamental laws of nature that are particularly important are:
 systems tend to attain a state of minimum energy
 systems tend to attain a state of maximum disorder
A system can be anything you designate as a system. It could be a solution in a 250 ml Erlenmeyer flask,
a test tube you are heating, or your small intestines. Everything else in the universe would be
considered the surroundings. So, systems tend to give energy to their surroundings and become less
organized.
The energy of a system is termed enthalpy. The abbreviation is H. (unit kJ/mol)
The disorder of a system is termed entropy. The abbreviation is S. (unit kJ/molK)
Gibbs’ had formulated an expression that incorporated both laws that drive reactions, systems tend to
attain a state of minimum energy and systems tend to attain a state of maximum disorder. This equation
leads to units of kJ/mol for Gibbs.
If you think about these two laws, they follow common sense. What happens to something hot that you
let sit around? It gets cold it loses enthalpy. Hot coffee can go cold in minutes. What happens to
something organized if you don’t do anything to it? It falls apart. A building not tended to will fall apart
in months.
From these laws a system will want to lose H and gain S.
You may ask yourself, how do I know H and S? These values are found experimentally and will be in
the CRC of Chemistry and Physics. Just as people experimented to determine the density of thousands
of different substances, others have experimented to determine the S and H of other substances.
Another way to calculate G is to use an equation that uses the standard free energy of formation
constants that are provided in the back of chemistry textbooks everywhere. This equation should look
vaguely familiar to you! The Gf value is the free energy change that occurs when 1 mole of compound
is made from its elements (VERY similar to Hf if you remember!!) As G is a state function (as are H
and E and S) it does not matter the pathway that the species are on to get to the final state, G can be
determined from the sum of each step along the way as well.
Grxno = npGfoproducts - nrGforeactants
Gfo values are very similar in nature to Hfo values in that the Gfo value for an element in its standard
state = 0.
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The coefficients in the equation (the moles) of each species is multiplied by each individual Gfo value to
determine the sum for the Gfo values on the reactant and product side.
Free energy is related to the amount of work that a particular reaction can do. The free energy is
therefore sometimes stated as the maximum amount of work that a particular reaction can possible do.
Whether the reaction actually does that amount of work is quite another story. The actual amount of
work that a particular reaction can do will depend on how the free energy is released . Ultimately, what
it comes down to is that some energy is going to always be lost – and therefore is not harnessed into
usable work. This means that the actual work that a reaction can do will be less than the theoretical
amount of work that the reaction could do. For example, a light bulb only converts 5% of incoming
electrical energy into light – the rest is lost as heat. THAT is a huge waste of energy – a huge loss!!
Remember back when you were in elementary school and it was really really warm out and the teacher
turned off the lights to cool down the room – well turning off the lights really does cool down a room!
A spontaneous reaction (Grxno < 0) will occur and will DO work on the surroundings. In
any real machine, the work obtained from the reaction is always less than the maximum
because some of the Go released is lost as heat.
A nonspontaneous reaction (Grxno > 0) will not occur unless the surroundings do work
on the system (input energy). Since some of the energy put in will be lost as heat, you
actually need to exceed the minimum amount of energy in order for the reaction to occur.
A reaction at equilibrium (Grxno = 0) does no work at all
Because three variables, Ho, So, and T are present in the equation for Grxno each can affect whether or
not the reaction is spontaneous. Thus, signs (positive or negative values) as well as magnitude will be
the determining factors in whether or not the overall reaction is spontaneous. However, all we simply
need to do is “plug” numerical values into the equation to determine Grxno and if Grxno is positive we
can say the reaction is spontaneous, if Grxno is negative we can say that the reaction (as written) is
nonspontaneous. BUT, we could do a little thinking beforehand and generalize about each variable and
THINK or predict what we would expect Grxno to be.
The equation is as follows:
G = H - TS
From this equation, you do not actually have to know the values of H and S: only their sign and the
temperature.
H
+
+
+
-
T
n/a
n/a
high
low
high
low
S
+
+
+
-
-TS
+
+
+
Description
always spontaneous regardless of temperature
never spontaneous regardless of temperature
spontaneous at high temperatures
nonspontaneous at low temperatures
nonspontaneous at high temperatures
spontaneous at low temperatures
G
+
+
+
-
When the signs of H and S are the same, then temperature becomes the determining factor in whether or
not the reaction is spontaneous. Thus, at one temperature the reaction may be nonspontaneous, but it
you raise or lower the temperature you can suddenly create a spontaneous reaction. That is WHY we
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state conditions when we speak of a reaction being spontaneous or not – it is spontaneous at certain
conditions. When/if we change those conditions we really can control the overall reaction. You can
determine the exact temperature, which will be known as the crossover temperature by using the
following equation:
T = Ho
So
It is important to remember what this temperature means, at any value above or below this
temperature, the reaction under one set of conditions will be spontaneous and at the other,
nonspontaneous.
Examining the sign of H
How do you know if the system has lost enthalpy? Enthalpy is really the sum of the heat in a system
and the work that the system could do. An example of a systems ability to work can include its position,
is it water above the dam or water at the bottom of the dam. Another example is a fluid under pressure;
it could move something like a pellet in an air rifle.
Examining the sign of S
How do you know if a system has gained entropy? Entropy is a measure of disorder of a system, so is
your system more or less organized. Ice turning to water is an example of a loss of order. Liquids are
more disordered than solids. Along that same line, gases are more disordered than liquids. So, an ice
cube melting in your home is an example of an increase in entropy.
The ability of plants to create sugar and oxygen from carbon dioxide and water with the input of light is
one of the most important reactions on the planet. Is this vital reaction spontaneous or nonspontaneous?
CO2(g) + H2O(l)
light


C6H12O6(s) + O2(g)
It is nonspontaneous. Light is needed to drive the reaction, without light photosynthesis cannot occur.
The system takes in energy; this increases the enthalpy, giving positive H, which is not favorable to a
spontaneous reaction. Now let’s look at the entropy. A gas and liquid combine to form a gas and a
solid. These products are more organized than the reactants, giving a negative S, which is not favorable
to a spontaneous reaction. If you look at the table above a +H with a –S will never be spontaneous.
But what about the reverse reaction?
C6H12O6(s) + O2(g)


CO2(g) + H2O(l) + energy
This reaction is spontaneous. This reaction is respiration of glucose/blood sugar. This system gives off
energy, which is used by your cells, so H is negative that is favorable for spontaneous reactions. The
reactants are solid and a gas and they are converted into a liquid and a gas. This gives a positive S. If
you look at the table above a –H with a +S will always be a spontaneous reaction. You blood holds the
glucose and you breath in O2(g) and exhale CO2(g).
Many spontaneous reactions occur in the body, here is another. The decomposition of organic matter is
a spontaneous reaction. The result is sometimes an unfortunate spontaneous release of gas, the smelly
kind.
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CxHyOz(s)  CH4(g) + H2O(l)
This is a decomposition reaction, one reactant becomes two products with the reactant being a solid and
the products are a liquid and a gas. The entropy of the system has increased considerably.
Many times, in a series of chemical reactions, where one reaction is tied to another, these reactions are
said to be coupled. If we examine each set of reactions independently, we might see that some reactions
are spontaneous while others, actually are not. Yet the series of reactions occurs – without any outside
intervention. Under these circumstances, the nonspontaneous reaction is actually driven, or helped
along, by the spontaneous reaction. One spontaneous reaction supplied enough energy for the other to
occur – just like gasoline combustion supplies enough energy to keep your car moving.
When considering the overall reaction (the target reaction) we can examine G values in the same way
we examined H values for step-wise reactions:
Concept Test:
Calculate the G for the following reaction:
Cu2O(s) + C(s)  2Cu(s) + CO(g) G = ???
Given the following information:
Cu2O(s)  2Cu(s) + ½ O2 (g) G = 140.0 kJ
C(s) + ½ O2 (g)  CO (g)
G = -143.8 kJ
If the two reactions were added together, we would generate the
target equation, thus we add the two G values to get the
target G value
We have previously discussed reactions as either being reactant favored (we called these no-reactions) or
product favored (reactions that “go to completion”). Reactions that were NR had very small K values
(meaning that the concentrations of products was small) while reactions that “go to completion” have
large K values (meaning that the concentration of products is large). The equilibrium constant, K, can be
related to the free energy of a reaction, G.
Remember the general setup to determining Q (or K!):
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aA + bB  dD + eE
Q = [D]d[E]e
[A]a[B]b
Recall that if you know what the reaction is doing at equilibrium (K) then you can examine Q and
determine if you need more product, less product, more reactant, less reactant – meaning you can
predict which direction the reaction will proceed based on examining Q and K (mini-exam 1 and in-class
exam 1 if you want two example problems!!)
 If Q < K : If Q is less than K that means that the reaction is not at equilibrium and in order
to reach equilibrium the numerator must get larger – which means the reaction will
proceed to the RIGHT
 If Q > K : If Q is greater than K that means the reaction is not at equilibrium and in order
to reach equilibrium the numerator must get smaller – which means that the reaction will
proceed to the LEFT
 If Q = K : If Q is the same numerically as K then the reaction is at equilibrium
And now we know that we can predict spontaneity for a reaction as written based on the sign of G.
Sometimes we saw that the reaction as written was spontaneous (-G) or perhaps the reverse reaction
was spontaneous (+G)
For example:
Cu2O (s) → 2Cu (s) + ½ O2 (g) Go = +140.0 kJ reaction nonspontaneous as written
2Cu 9s) + ½ O2 (g) → Cu2O (s) Go = -140.0 kJ reaction spontaneous as written
The standard free energy change for a reaction, Go, monitors the increase or decrease of the free energy
as the reactants in their standard states are converted completely to products in their standard states.
But, we know that this idea of “complete” conversion is not really the truth for many chemical reactions.
A product-favored reaction proceeds towards the formation of the product, but if the reaction is able to
attain equilibrium, some reactant will remain! A reactant-favored reaction means that very little product
is formed, but we now know that SOME is formed (remember Ksp??!!). As you can imagine, the two
ways of predicting spontaneity are related to one another!
When reactants are mixed together, the system will proceed spontaneously to a position of lower energy
(lower free energy), and the system will eventually reach equilibrium. At any point along the way from
reactant to product, the species are not at standard conditions, thus a “new” equation can be used to
describe the system as the reaction proceeds:
G = Go + RTlnQ
G means non-standard conditions
Go means standard conditions
R = 8.314 J/moleK
T = temp in Kelvin
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Q = reaction quotient
This equation allows us to monitor the change in free energy that occurs as the reaction proceeds (as we
have incorporated Q – which we know already allows us to monitor the changes in concentration of
species as the reaction moves from reactants → products). At a given temperature, the difference in free
energy between that of the pure reactants and a mixture of reactants and products is determined by the
values of Go and Q. As long as G is negative, we can say that the reaction is spontaneous.
Eventually, the system will reach equilibrium. No further change in reactants or products is seen at this
point in time. This means that there is also no further change in the free energy of the system: G = 0.
In effect, the system has released all of its free energy in the process of attaining equilibrium.
G = 0 = Go + RTlnK
(at equilibrium Q = K)
0 = Go + RTlnK
Rearranging the equation:
Go = -RTlnK
From the above equation, you should notice that if Go is negative, indicating a product favored
reaction, K must be greater than 1, and the reaction is spontaneous as written.
If Go is positive, this indicates a reactant favored reaction, K must be less than 1, the reaction is
nonspontaneous as written (but is spontaneous in the reverse direction!).
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