Changes in matter are accompanied by a change in energy. Remember, nature is lazy!! Physical
and chemical changes occur because that is the path of least resistance. The production and
usage of energy that occurs in reactions have enormous impacts on society. People are making
big bucks as they try to find new ways to harness energy (think of nuclear energy, solar energy,
and fossil fuels).
Thermodynamics refers to the study of heat and its transformations while thermochemistry is
the branch of thermodynamics that deals with the heat involved in chemical reactions. Where
does the heat change come from??
Chemical reactions are accompanied by energy changes. Some reactions release energy to the
surrounding, some reactions absorb energy.
Energy can be converted from one form into another. Sunlight falling on a surface is absorbed
and converted into heat energy. The kinetic energy of running water and the thermal energy of
steam are converted into electrical energy by turbines. The stored potential energy of fuel is
converted into mechanical energy by automobile engines. Conservation of Energy tells us that
energy is neither lost nor gained, it just changes from one form to another. In fact, we know that
systems cycle from potential to kinetic and back again. What was once kinetic, then becomes
potential again. Thus is the cycle of energy.
First Law of Thermodynamics: energy can be neither created nor destroyed, it can only be
converted from one form to another.
To make a meaningful observation about this conversion of energy, we must first make
assignments. What substance/species/situation are we studying? This becomes the system.
Everything else, are the surroundings. These are “arbitrary” assignments. It all comes back to
point-of-view. In order to explain our observations, we must choose a point of view. Therefore,
we make an assignment that this is the system. Everything else will be the surrounding. The
system could be the plant growing in your yard, and its surroundings would be the ground, the
air, and everything else around it. The system is usually the object or reaction that we are
studying. The surroundings would include the container, the atmosphere, and the observer.
The first law of thermodynamics tells us that the system and surrounding will interact such that
the energy lost by the system will be absorbed by the surroundings. Or the energy absorbed by
the system will be absorbed from the surroundings.
The internal energy of a system is the sum of the kinetic and potential energies of the individual
particles. Some types of energy that we are thinking about include:
energy of translation (A): the motion from one location to another
energy of rotation (B): molecules spinning in place
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energy of vibration (C): molecules vibrating back and forth about equilibrium positions
A
B
C
Unfortunately, it is oftentimes difficult to evaluate the internal energy of a system unless it is a
very simple one. As such, chemists are not all that concerned with internal energy as much as
they are concerned with the changes in internal energy, that is, they are interested in the
quantity:
E = Efinal - Einitial
Or the difference in internal energy between some initial and some final state. A change in the
system is always accompanied by an opposite change in the energy of the surroundings. A
chemical reaction can change the internal energy in one of two ways:
1. By losing some energy TO the surroundings, the final energy state will be lower than the
initial energy state:
Efinal  Einitial
E  0
2. By gaining energy FROM the surroundings, the final energy state will be greater than the
initial energy state:
Efinal  Einitial
E  0
initial
Energy lost to the
surroundings
E
final
time
2
final
Energy gained from
the surroundings
E
initial
time
There are several ways in which a system and its surroundings can exchange energy. The flow
of heat is an energy transfer that results from a temperature difference. Remember, temperature
is simple a measure of the average kinetic energies of the molecules in the sample. Heat always
flows from warmer regions to cooler ones. Energy can also be transferred as work, the action or
force through some distance. The first law of thermodynamics tells us that the difference in
energy can be exactly accounted for. Energy can therefore be thought of as a bank account, and
we can “pay” into this account by either heat flowing or work being done. The change in the
system’s energy is therefore the sum of both types of inputs:
E = q + w
The numerical values for q and w can be either positive or negative depending on the change
that the system undergoes. In order for everyone to define heat and work in the same manner,
the perspective is taken from the system. Energy coming INTO the system is defined as
positive, energy going OUT or leaving the system is defined as negative.
Heat flowing INTO the system or work done by the surroundings ON the system add energy
into the account. These are therefore defined as positive q and w values because they increase
the overall energy of the system.
Heat flowing OUT of the system, or work done BY the system on the surroundings are
withdrawals from the energy bank. Such withdrawals remove energy from the system and are
therefore defined as negative q and w values because they decrease the overall energy of the
system.
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Concept Test:
An expanding gas does 1500 kJ of work while it absorbs 800 kJ of heat. What are q, w,
and E for the gas?
Sometimes systems do NO work. Therefore w = 0 and the equation simply becomes E = q.
There will then be a direct correlation between the heat lost or gained and the energy if the
system. In just the same manner, sometimes systems will have no heat exchanged, and will
only transfer energy as work. In this case, E = w. And the direct relationship is now between
the work and the energy of the system.
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The SI unit of Energy is the Joule which we have seen previously with our kinetic energy
calculations from CH131.
1J=1
kg m 2
sec 2
The calorie is an older unit that was defined originally as the amount of energy needed to raise
the temperature of 1 gram of water 1oC. The calorie is now defined in terms of the Joule.
1 cal = 4.184 J or 1 J = 0.2390 cal
As the quantities of energy can get quite large, chemists also use the kJ and kcal to define the
energy of the system. (The nutritional Calorie associated with food (note the capitol C) is
actually a kilocalorie).
The state of any system is described by properties such as pressure, volume, temperature,
density, composition, and so forth. Some, or all, of these properties will change as the state of
the system changes. For example, if 1.00 L of gas is warmed from 25oC to 100oC at a constant
pressure, the volume increases. In this example, the volume will increase to 1.25 L. There is a
change in the volume such that V = Vfinal –Vinitial = 1.25 L – 1.00 L = 0.25 L, and the change in
temperature T = Tfinal – Tinitial = 100oC – 25oC = 75oC. These changes are simply the differences
between the final and initial states. It does not matter if it took 3 days to raise the temperature
or 3 seconds, the final volume will still be 1.25 L. Any property whose change depends only on
the initial and final states is called a state function. Volume and temperature are state functions.
Density, pressure, and energy are also state functions.
E does not depend on HOW the change takes place. In other words, for a given change, the
value of E stays constant. Heat and work, however, are NOT state functions. The amount of
heat and work required to go from some initial state to some final state will depend on the
pathway – the way the change is carried out – and many different pathways are sometimes
possible, but the different pathways ultimately lead to the same net change (E).
The heat released or absorbed during a chemical reaction is called the heat of the reaction. An
exothermic reaction releases heat from the system to the surroundings. Heat is given off. This
means that q is negative. An endothermic reaction absorbs heat from the surroundings. The
heat of the reaction (q) is given by rearranging the equation given above:
q = E – w
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There are several types of reactions to be considered. The first is a constant volume reaction, or
one that takes place in a sealed container with rigid walls that do not expand or contract. The
pressure of the system in a sealed container like this will generally change. If there is an
increase in the number of moles of gas, then the pressure will increase. If there is a decrease in
the number of moles of gas, then the pressure will decrease. In a constant pressure reaction, or
a reaction that takes place under a constant external pressure, it is the volume that usually
changes. Production of gases will cause the volume to increase. Consumption of gases in the
chemical reaction will cause the volume to decrease. Most chemical reactions are carried out
under constant pressure conditions.
In the constant pressure reaction, the volume change leads to an exchange of work with the
surrounding. If a gas is produced, the system will do work on its surroundings by expanding
and pushing away air molecules. If the gas is consumed, the surroundings do work on the
system, the atmosphere can press in on the system, shrinking its total volume.
The work associated with this change in volume is called Pressure-Volume work – or PV work.
The system is kept under constant pressure by the presence of the piston.
If the system expands, the piston is pushed upwards by some distance h.
Volume = area x height so V = Area x h.
Pressure = Force/Area
work = force x distance
work = PV
The work done on the system is the negative of the work done BY the system
w = -PV (system does the work)
P = the pressure of the reacting system and V = Vfinal – Vinital or the volume change.
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Substituting in the value for w into the original equation we get that at a constant pressure:
qp = E + PV
For reactions involving solids and liquids, the volume change will be negligible, for reactions
involving gases, this V value becomes very important. For reactions that occur at a constant
pressure, the heat of the reaction is also called the enthalpy of the reaction (H), such that q p =
H. The heat of reaction at a constant pressure equals the change in enthalpy of the reacting
system. Enthalpy is also a state function.
H = E + PV
Thus, for reactions involving solids and liquids (and even gases where there is no overall
volume change, H = E.
If heat is absorbed by the system, q is positive and H will be positive. If heat is given off by the
system, q is negative and H will be negative.
H is always positive for endothermic reactions : H  0
H is always negative for exothermic reactions : H 0
Enthalpy or heat of reaction changes are often given specific names depending on the type of
chemical reaction. For example, the heat (or enthalpy) of combustion, the heat (enthalpy) of
neutralization, or the heat (enthalpy)of formation.
Concept Test:
When ammonia gas is formed from its elements, its heat of formation is -46.1 kJ/mole
When nitrogen dioxide forms from its elements, its heat of formation is +33.2 kJ/mole
Are the reactions endo or exothermic?
It is sometimes very helpful to think of heat as a reactant or product. If reactants come together
and react and give off heat, then heat was a product. Heat given off is exothermic. Exothermic
reactions have -H values. If it takes heat in order for the chemical reaction to occur, meaning
you have to put heat in as a reactant, then heat is used by the system and this signifies an
endothermic reaction which has a +H value.
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Certain chemical reactions, as mentioned above, have their own special “names” and are
defined in a special manner:
When 1 mole of a substance is combusted in oxygen the heat of reaction is defined as the heat of
combustion: (Hcomb)
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O(l)
Hcomb = -890.4 kJ
When 1 mole of compound is produced from its elements, it is defined as the heat of formation
(Hf)
Hg(l) + Cl2 (g)  HgCl2 (s)
Hf = -224.3 kJ
When 1 mole of substance melts, it is defined as the heat of fusion (Hfus)
When 1 mole of substance vaporizes, it is defined as the heat of vaporization (Hvap)
A thermochemical equation is an equation EXACTLY like you all have been writing chemical
equations (phew!) except off to the right of the equation you include the H value for the
reaction. Some assumptions are made in terms of pressure and temperature such that if not
stated, P = 1 atm and T = 25oC. Typically H values are given in tables at 25oC. Enthalpy
changes for the reaction are given as the equation is written. However, the values are
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sometimes given for the formation of 1 mole of product. This means that sometimes you now
WILL have fractional coefficients.
2H2 + O2  2H2O
H = -571.7 kJ
becomes
H2 + ½ O2  H2O
H = -285.8 kJ
H is related to the molar amounts of reactants and products present. Thus, if we wanted to
mathematically manipulate our chemical reaction, for fun or, as you will see later, as a necessity
in Hess’ law reactions, you must manipulate the H value as well. Above is an example of this.
In the first equation, we formed 2 moles of H2O and have a H value of 571.7 kJ. In the second
equation, we formed half as much water using half as much reactant material and t he H value
was half the size. What if we wanted to know the H value for the formation of 10 moles of
H2O. This means that will will be multiplying the ENTIRE equation by 10 – and we MUST
MUST MUST multiply the H value by 10 as well.
10(H2 + ½ O2  H2O H = -285.8 kJ
10H2 + 5O2  10H2O H = 2858 kJ
Another cool fact is that H represents the reaction. When 1 mole of H2 is mixed with 0.5 mole
of O2 to form 1 mole of water, 285.8 kJ of heat is given off. BUT, if 1 mole of water is
decomposed into its elements giving back 1 mole of H2 and 0.5 mole of O2 then 285.8 kJ of heat
are absorbed from the surroundings. Thus the value of H stays constant for a particular
chemical reaction (for H2O, H2 and ½ O2 the value is 285.8 kJ) it is just the sign of the H value
that changes.
H values are also related to the stoichiometric amounts of the reactants and products and can
be used as conversion factors. Again you should think about the H value as a reactant or
product. Such that:
H2 + ½ O2  H2O H = -285.8 kJ
We can use this value as a conversion factor:
-285.8 kJ
1 mole H2
-285.8 kJ
0.5 mole O2
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Concept Test:
Given the following equation, what is the enthalpy change associated with the
formation of 5.67 moles of HCl?
H2 (g) + Cl2 (g)  2HCl H = -184.6 kJ
What is the enthalpy change associated with the formation of 1 mole of HCl?
½ (H2 + Cl2  2HCl H = -184.6 kJ)
½ H2 + ½ Cl2  HCl H =
Now what is the enthalpy change associated with the formation of 5.67 moles of HCl?
5.67( ½ H2 + ½ Cl2  HCl H =
)
2.84 H2 + 2.84 Cl2  5.67 HCl H =
neat huh??? but longer/more steps 
Concept Test:
What is the enthalpy change when 12.8 grams of H2 reacts with excess Cl2 to form HCl?
(hint: convert grams to moles, then use conversion factor!)
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Ultimately we are talking about potential energy and kinetic energy of the system. The energy
of motion (translational, rotational, vibrational, and the motion of the electrons) combined with
the potential (location of atoms, forces between the nucleus and the electrons, the forces
between the electrons, the forces between the protons and the neutrons, and the forces between
the atoms as the bond). What changes significantly in a chemical reaction? The bond. Bonds
are broken in chemical reactions and bonds are formed. Thus, this type of potential energy
plays a large role in the H values for reactions.
A system must absorb energy in order to break its bonds and conversely, it therefore must give
off or release energy when it forms its bonds. Whether or not a reaction is endothermic or
exothermic comes from carefully examining the values associated with bonds breaking and
forming. When 1 mole of H-H bonds and 1 mole of F-F bonds break, they absorb a quantity of
energy. The breaking of these bonds results in the formation of 2 moles of H-F bonds which
releases energy. The overall H value for this reaction is (-) signifying an exothermic reaction.
This means that the energy associated with the bond formation is more than the sum of the
energies of the H2 and F2 bonds breaking. Thus, H-F is a stronger, more stable, bond than H-H
or F-F is independently. H-F forms because the bonds in the reactants, H2 and F2 are weaker
and less stable than the product bond H-F that is formed.
Thus, the energy released or absorbed during the chemical reaction is due to the difference
between the strengths of the bonds in the reactants compared to the bonds in the products (the
bonds that are broken compared to the bonds that are formed).
The calorimeter is an instrument which is used to measure the heats of a reaction. To measure
the heat at a constant pressure (say the pressure in lab) we must construct surroundings that
retain the heats and then measure the T value that results using a thermometer or temperature
probe. Then we can relate that quantity of heat released or absorbed to that temperature
change.
The more heat an object absorbs, the hotter that object gets such that the amount of heat that an
object absorbs is proportional to its T. Every object has its own capacity for absorbing heat.
And this is called an object’s specific heat capacity (c). Using this information, q can be
determined. And q is directly related to H in reactions at constant pressures.
q = mCT
CH2O= 4.184 J/goC
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In the calorimeter, heat is transferred from the system to the surroundings. Typically we are
therefore measuring the temperature of the surroundings (example, the water temperature
change that results from the reactions of Mg with HCl in lab).
Constant-Pressure Calorimeters: the coffee cup calorimeter in lab. In lab you can measure the
heat associated with the reaction of Mg with dilute HCl (HCl in water) and the heat associated
with the reaction of MgO with dilute HCl. The water gained the heat that was given off during
the chemical reaction. Thus we can say that:
qH2O = -qsolid
Thus:
-(csolid x masssolid x Tsolid) = cH2O x massH2O x TH2O
and all the quantities but one will be given!
Constant-Volume Calorimeters: using a bomb calorimeter: This calorimeter can precisely
measure the change in temperature during a chemical reaction. Since the volume is constant,
the heat measured does NOT equal H, but instead equals E (no work is done by the system
so w = 0, therefore q = E). Constant volume reactions are evident by looking at the chemical
reactions if the number of moles does NOT change. Constant volume reactions are also
reactions that occur with solids and liquids (as solids and liquids do not have changes in
volume! – cannot really be compressed!) If a gas reaction is observed, the number of moles of
gas that you start with must be the same number of moles of gas t hat you end up with.
A chemical reaction can occur directly in one step, or indirectly in two or more consecutive
steps. For example, carbon dioxide can be produced in one step by burning graphite in the
presence of excess oxygen:
C(graphite) + O2(g) → CO2 (g) H= -393.5 kJ
OR, it can be produced by first burning graphite in the presence of a limited supply of oxygen to
obtain carbon monoxide
C(graphite) + ½ O2 (g) →
CO(g)
H = -110.5 kJ
And then burning the carbon monoxide in additional oxygen
CO(g) + ½ O2 (g) → CO2 (g) H = -283.0 kJ
-110.5 kJ + -283.0 kJ = -393.5 kJ (same as above)
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The net overall change is the same, either way. The q and w values are different for the two
pathways, but the overall H value remains the same. 1 mole of carbon dioxide formed from 1
mole of C and 1 mole of O2. This is an example of Hess’s Law.
Hess’s Law: the total enthalpy change for a reaction is the same whether the reaction occurs in
one, or in several step.
This is based on the fact that enthalpy is a state function. And therefore the H value is
independent of the path. In going from the initial state of C (graphite) to the final state of CO 2,
the enthalpy change will be the same regardless of the intermediate steps that might be used to
carry out the reaction.
Hess’s law is often used to help determine enthalpy values for reactions that might be very
difficult to carry out. But other reactions can be, and those H values can be used to determine
the H for the reaction of interest.
When performing Hess’s Law calculations is it important to know where you are going. You
will be adding reactions together, multiplying reactions and H values by factors, or dividing
reactions and H values by factors, you might be reversing chemical reactions and changing the
signs of H values. It will be easiest to examine WHERE you are going. Which species are the
reactants? Which species are the products. How can I rearrange my given chemical reactions
such that the reactants are on the reactant side and the products are on the product side. When
I add my reactions together, does my overall chemical reaction equal what I am looking for? If
it does, then you can simply sum the H values for the individual reactions together. KEEP in
mind, if you manipulate the equations in ANY way do not forget to manipulate the H value!!
Concept Test:
Use the two equations below to calculate the enthalpy change for the third reaction
(1) C (graphite) → C (g)
(2) C (graphite) + O2 (g) → CO2 (g)
H = 716.7 kJ
H = -393.5 kJ
(3) C (g) + O2 (g) → CO2 (g) H = ???
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Let’s try something a little bit more complicated!
Concept Test:
Use the following chemical reactions to calculate the H value for the
following chemical reaction:
2H2S (g) + SO2 (g) → 3S(s) + 2H2O(l)
H = ?????
(1) H2 (g) + ½ O2 (g) → H2O (l) Hf = -285.8 kJ
(2) H2 (g) + S (s) → H2S (g) Hf = -20.6 kJ
(3) S(s) + O2 (g) → SO2 (g) Hf = -296.8 kJ
In the above reaction, we used the heats of formation values for a variety of compounds to
determine the Hrxn for an independent chemical reaction. Hess’s Law and the fact that H is a
state function allows us to do this. This type of calculation shows an imaginary pathway that is
equivalent to performing the decomposition reaction.
When you examine what was just
rearranged in order to do the Hess’s Law calculation you will see that you summed the heats of
formation of the products and then subtracted the sums of the heats of formations of the
reactants.
oHrxn
=  [np(Hf)p] -  [nr (Hf)r]
Hrxn = sum of the number of moles of the products times the Hf of the products minus the
sum of the moles of the reactants times the Hf of the reactants
Some important notes: An element in its standard state is assigned an Hf value of 0.
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And most compounds have negative Hf values which indicates an exothermic reaction for
their formation AND that in most cases, the product is more stable than its component
elements.
Concept Test:
6CO2 (g) + 6H2O(l) → C6H12O6 (s) + 6O2 (g)
Calculate the Hrxn given the following Hf values:
Hf CO2 (g) = -393.5 kJ/mole
Hf H2O(l) = -285.8 kJ/mole
Hf C6H12O6 (s) = -1260 kJ/mole
Hf O2 (g) = 0 kJ/mole
Some final things to be careful of when performing heats of formation reactions: be CAREFUL
to choose the correct Hf value for the PHASE of the substance!! Notice in the table in the back
of your book that the Hf value for mercury liquid (its natural state) is 0 – BUT the Hf value
for mercury as a gas is NOT 0 – in fact, it is 61.317 kJ/mole.
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