Thermochemistry and Enthalpy

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Thermochemistry and Enthalpy Powered By Docstoc
					Introduction to Thermochemistry
What is Energy?
Work, as understood by scientists, means more than towing barges and
lifting bales. The concept goes beyond physical labour to embrace a number
of actions, but they all come down to the pushing or pulling of something
against an opposing force. For example, when hot gases expand in the
cylinder of a gasoline engine, they push back a piston and ultimately move
the car. This is of course mechanical work.

Everyone agrees that a battery has energy. It also has the ability to do
work, and it delivers this energy by pushing electrons through a wire. This
pushing of electrons is referred to as electrical work. The current may run
a small motor (and so be changed into mechanical work); or it can be
passed through a bulb and converted into both heat and light, two other
forms of energy.

There is kinetic energy, the energy of motion which is derived from the
formula: KE = ½mv2 where m is the mass of the object in kilograms and
velocity is in meters/sec. The units derived from this equation is kg m2/s2.
The other kind is potential energy and it can be derived one way using PE =
mgh

where m is again the mass in kilograms, g is the gravitational constant of
9.8 m/s2 and h is the height that the object is to fall. Again the units derived
are kg m2/s2.

Origin of Chemical Energy
The term chemical energy is the special name often given to the form of
potential energy that arises from the forces of attraction that bind atoms
together in compounds. These forces of attraction are called chemical
bonds. What is important now is the idea that when chemicals react to form
new substances, atoms are exchanged as old bonds break and new bonds
form. This process changes the potential energies of the atoms. Sometimes a
reaction's products have more potential energy than it's reactants; in other
reactions the products have less potential energy potential energy than the
reactants.

In general all chemical reactions either liberate (exothermic reactions) or
absorb (endothermic reactions) heat. The origin of chemical energy lies in
the position and motion of atoms, molecules and subatomic particles. The
total energy possessed by a molecule is the sum of all the forms of potential
and kinetic energy associated with it. Kinetic energy actually contributes
very little to the energy of a substance. The bonding energy is what is really
important.
Thermochemistry: Units of Energy
To measure the amount of energy involved in chemical reactions, we must
have a unit of energy. Over the decades many units have emerged, some
more useful than others. The SI energy unit is derived from the SI base
units. The unit is the joule, J, and it is based upon 1 J = 1 kg m2/s2.

If you dropped 2 kg of butter on your foot from a height of about 10 cm,
you would deliver about 1 J of energy to your foot. This is actually a very
small amount of energy, especially when we consider chemicals reacting on
a mole scale. A more common unit is the kilojoule, kJ, which is equal to
1000 J.

The Special Place of Heat
One of the important facts about our world is that all forms of energy can
be converted quantitatively into heat. For example, the mechanical energy
of a moving car is converted entirely into heat by the frictional action of the
brakes, and the brake shoes and drums become very hot indeed. When a
current of electrons is directed into a poor conductor, something with high
resistance such as the heating element in a toaster, the electrical energy
changes into mostly heat with some light. The full convertibility of energy in
other forms into heat gives us a way to measure the other kinds of energy,
and the measurement of heat is relatively simple.

The traditional unit of heat energy was the calorie, abbreviated cal, and was
originally defined as the amount of heat needed to raise the temperature of
1 gram of water by 1 degree Celsius. The equivalent amount in Joules is
4.184 J.

ie. 1 cal = 4.184 J
Specific Heat, Molar Heat and Heat Capacity
The physical properties of a substance that concern its ability to absorb
heat without changing chemically are called its thermal properties. Three
examples are heat capacity, molar heat capacity, specific heat capacity, which
is usually just called specific heat.

Specific Heat and Molar Heat Capacity
The specific heat of any substance is the energy needed to raise the
temperature of one gram of it by one degree Celsius. It can be calculated by
the equation

                           (energy absorbed)
specific heat = (mass of sample in g) X (temperature change in oC)
mathematically this is expressed as C =      E
                                          g ΔoC

You should be able to rearrange the above equation into the other 3
possible equations.

The easier equation to remember is E = mΔt. The energy involved in joules
is equivalent to the mass in grams times the specific heat capacity of the
substance times the change in temperature in degrees Celsius.

Table of Specific Heats
     Substance       Specific Heat
                        (25oC)
                         J/goC
   Carbon
                         0.711
  (graphite)
   Copper                 0.387
 Ethyl alcohol             2.45
    Gold                  0.129
   Granite                0.803
     Iron                0.4498
     Lead                 0.128
   Olive Oil                2.0
    Silver                0.235
 Water, (liquid)         4.1796
The molar heat capacity is a more useful quantity to work with in
chemistry. It is the energy required to raise one mole of any substance by
one degree Celsius.

molar heat capacity = _ J_ _
                    mole oC

A useful relationship is   J X g          =       J
                         o
                       g C    mole            mol oC

or specific heat X molar mass = molar heat capacity
Thermochemistry Worksheet on Specific Heats and Heat Units
1.   If a gold ring with a mass of 5.5 grams changes temperature from
     25.0oC to 28.0oC, how much energy (in joules) has it absorbed?
2.   The temperature of a sample of 250 g of water is changed from 25.0oC
     to 30.0oC. How much energy was transferred into the water to cause
     this change? Calculate your answer in joules and in calories.
3.   How many kilojoules can 250 g of water absorb for each degree Celsius
     change in the temperature range near room temperature? In other
     words, what is the heat capacity of 250 g of water?
4.   Suppose the 250 g sample of water above underwent a change in
     temperature from 25.0oC to 26.4oC. How much heat caused this
     change?
5.   The temperature of 335 g of water changed from 24.5oC to 26.4oC. How
     much heat did this sample absorb? Calculate the answer in kilojoules
     and kilocalories.

Thermochemistry: Energy Changes in Chemical Reactions

Thermochemistry deals with transfers of energy between reacting
chemicals and the world around them.

Systems, Surroundings, and Boundaries
The word system refers to that particular part of the universe we wish to
study. The system might be the chemicals reacting in a beaker or the
chemicals in a battery cell reacting to give electricity, or the system of a
living cell.

The word surroundings refers to whatever is entirely outside the defined
system, everything in the universe except the system itself.

A boundary, real or imaginary, separates the system from its surroundings.
When the system is in a beaker, the boundary exists wherever the solution
contacts the beaker or the air above it. If the boundary can prevent any
transfer of heat between the system and the surroundings, we say that the
system is insulated from its surroundings. Styrofoam makes a good
insulating boundary for keeping a cup of coffee hot, but no material is a
perfect insulator.

Another term that is used frequently is the state of a system. Each system
has a state defined by listing its temperature, pressure, volume, and
composition (including concentration terms). We say that a system
undergoes a change of state whenever any change occurs in one or more of
the variables that define the system.

Heats of Reaction
In thermochemistry we are concerned with the exchange of energy between
a chemical system and its surroundings. Sometimes chemical changes are
able to bring energy into the system. These are endothermic changes. An
example is the charging of a battery, in which energy from an external
source becomes stored in the battery in the form of chemical potential
energy. Photosynthesis is also endothermic as far as the plant is concerned,
and the needed energy (only 0.04%) is imported from the sun. When
endothermic changes occur by themselves within an uninsulated system, we
often notice a cooling effect in the surroundings. This is what happens to
cool your drink with ice or when you use an "instant cold" compress from
the drugstore.

Many chemical reaction are able to release energy to the surroundings.
Such changes are described as exothermic. A typical example is the
combustion of gasoline. Heat transfer away from the system (if uninsulated)
and into the surroundings, where the temperature increases.

The form of the energy absorbed or released during a change can vary. It
sometimes appears as light, or electrical work, but most often occurs only
as heat. When the entire energy change of a reaction involves heat, the
amount of heat is called the heat of reaction and is usually represented by
the symbol 'q'.

We show exothermic reactions by q = -ve meaning that energy has been lost
from the system. Endothermic reactions are documented by q = +ve
meaning that energy was absorbed by the system.

The actual amount of heat of reaction for a given change in a system
depends to some extent on the conditions under which we carry out the
reaction. It depends on the physical states of the reactants and products; it
depends on the initial temperature of the system; and it depends somewhat
on whether the volume of the system or its pressure is held constant or is
permitted to change.

To simplify matters a great deal, chemists noticed a long time ago that most
reactions are carried out in open beakers or vats under atmospheric
pressure. So we will also limit ourselves to these conditions and some new
terms to explain these conditions.


Enthalpy
The first term, enthalpy, refers to the total value of energy of a system when
it is at constant pressure. It is symbolized by the letter 'H'. When a system
reacts at constant pressure it will either gain or lose energy and we say that
the enthalpy of the system has gone through a change or an enthalpy
change, which is symbolized by ΔH. Δ means "change in".
ΔH is defined by the equation:        ΔH = Hfinal - Hinital
Hfinal is the enthalpy of the system in its final state and Hinitial is the enthalpy
of the system in its initial state.

For a chemical reaction the above equation can be expressed much more
nicely as

       ΔH = Hproducts - Hreactants

The above equation simply put means "the total heat content of the
products minus the total heat content of all the reactants".

After having gone to all this effort to give a formal definition of H, it is
perhaps a bit disappointing to learn that we cannot actually calculate it
from measured values of Hfinal and Hinitial. This is because the total enthalpy
of the system depends on its total kinetic energy plus its total potential
energy, and these values can never be determined. The good news is that we
don't need to know it. We care only about what our system could do for us
(or to us!) right here at a particular place on this planet. For example, when
we want to know the yield of energy from burning gasoline, we really do not
care what its total enthalpy is in either its initial or final state. All we care
about is by how much the enthalpy changes, because it is only this enthalpy
change that is available to us. In other words, we don't need Hinital and
Hfinal , but we can calculate H by direct measurement.

Enthalpy Change and Heat of Reaction at Constant Pressure
The enthalpy change for a reaction can make itself known in the
surroundings either as work or as heat energy or as some of each. When all
of the enthalpy change appears as heat, we have a way to measure the
enthalpy change for a reaction, because in this circumstance H is equal to
the heat of reaction at constant pressure.

ΔH = q (at constant pressure)

The sign of q, defined earlier, is actually determined by the sign of ΔH. If
ΔH is negative so q is negative for exothermic changes. ΔH is positive and
so q is positive for endothermic changes.

Energy Conservation
One important truth has been strongly implied so far, but never stated in so
many words - the amount of energy that leaves a system is exactly the same
as the amount that goes into the surroundings. No energy is lost. It just
transfers from one place to another, and some of it might change from one
form to another. The formal statement of this truth is called the law of
conservation of energy.

Thermochemistry: Conservation of Mass and Energy
                               Nuclear chemistry forces us to modify the Law
                               of Conservation of Mass to include an energy
                               term as well. The energy term is derived from
                               Albert Einstein's famous E=mc2 equation.
                               Actually this equation was rewritten for the
                               layman, in its true form it is E = moc2, where E
                               is the change in energy that takes place, mo is
                               the change in rest mass, and c is the speed of
                               light.

                               Because the speed of light is very large and it's
                               square even larger, even a very small change
                               in mass would translate into an enormous
                               change in energy.

                               For example lets take a look at an ordinary
                               chemical reaction.

The combustion of the gas methane.

CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(g)        Ho = -890 kJ/mol

since E = moc2 then upon rearrangement we get mo = E
                                                  c2

Fact: 1 J = 1 kg m2 This a fact that should be memorized.
               s2

therefore mo = ___890 kJ X 1000 J X 1 kg m2
                (3.0 x 108 m)2 1 kJ      s2

             = 9.89 x 10-12 kg

             = 9.89 x 100 ng

The 890 kJ of energy released by the combustion of one mole of methane
thus originates from the conversion of 9.89 ng of mass into energy. Such a
small change, about 10 ng out of 80 g cannot be detected by balances. It
amounts to the loss of 1.0 x 10-7% of the mass. So we ignore Einstein's
equation when doing regular chemical stoichiometric calculations. However
this equation is very useful in nuclear chemistry.
Nuclear Binding Energy
The sum of the rest masses of the nucleons of an atom does not equal the
measured mass of any nucleus. The actual mass of an atomic nucleus is
always a trifle smaller then the sum of the rest masses of all it's nucleons (p+
+ no). This mass difference is changed into energy as the nucleus formed,
and was emitted as high energy electromagnetic radiation. It would cost
this much energy to break the nucleus apart into its nucleons again, so the
energy is called the 'binding energy of the nucleus'.

Lets take a look at the manufacture of a Helium nucleus.

Binding energy of 42He - actual rest mass is 4.001506 amu.

The rest masses of the nucleons are:

p+ = 1.007277 amu no = 1.008665 amu You should include these numbers in
your data book.

For 42He then 2 p+ = 2 x 1.007277 amu = 2.014554 amu
              2 no = 2 x 1.008665 amu = 2.017330 amu
                                       4.031884 amu

The difference in mass = calculated mass - actual mass
                      = 4.031884 amu - 4.001506 amu
                      = 0.030378 amu

1 kg = 1000 g       1 amu = 1.6606 x 10-24 g {Include these facts as well}

so

E = moc2

= (0.030378 amu X 1.6606 x 1024 g x 1 kg )(3.0 x 108 m)2
                       amu         1000 g            s

= 4.54 x 10-12 kg m2
                s2

= 4.54 x 10-12 J This is the energy release for 1 atom of 42He

A mole of He would be 6.02 x 1023 nuclei more, therefore,
6.02 x 1023 nuclei/mole * 4.54 x 10-12 J/nuclei
= 2.73 x 1012 J/mole (enough energy to power a 100 watt light bulb for 900
years)

How do I know this?

P=E/t from grade 11 chemistry class.

therefore t = E/P

= 2.73 x 1012 J
    100 W

= 2.73 x 1012 J
     100 J/s

= 2.73 x 1010 s (X 1 min/60 s)

= 4.55 x 108 min (X 1 hr/60 min)

= 7.583 x 106 h (X 1 day/24 h)

= 315972.2 days (X 1 y/365.25 days)

= 865.09 years
Nuclear Conservation of Energy Worksheet
1.   Why isn't the sum of the masses of all nucleons in one nucleus equal to
     the mass of the actual nucleus?
2.   Calculate the binding energy in joules per nucleon of the nucleus of an
     atom of iron-56. The observed mass of one atom is 55.9349 amu.
3.   Calculate the binding energy in joules per nucleon of the nucleus of an
     atom of uranium-235. The observed mass of one atom is 234.98835
     amu. Calculate the amount of energy evolved from the fission of 1
     mole of U-235.
4.   Calculate the binding energy in joules per nucleon of the nucleus of an
     atom of plutonium-244. The observed mass of one atom is 234.8991
     amu. Calculate the amount of energy evolved from the fission of 1
     mole of Pu-244.
5.   Suppose that the total mass of reactants in a chemical reaction was
     100.000000 g. How many kilojoules of energy would have to evolve
     from this reaction if the total mass of the products could be no greater
     than 99.99900 g?
The First Law of Thermodynamics
Law of Conservation of Energy: The energy of the universe is constant;
it can be neither created or destroyed, but only transferred and
transformed.

State functions
A state function is any physical property whose value does not depend on
the system's history. Some examples are pressure, volume, and
temperature. For example, a system's temperature at any particular
moment does not depend on what its temperature was the day before, nor
does it depend on how the system reached its current temperature. If the
system's temperature is now 25oC, this is all we have to know about its
temperature. We do not have to say how it got to be that temperature. Also,
if the temperature was to rise to 35oC, the change in temperature, t, is
simply the difference between the final and the initial temperature.

             Δt = tfinal - tinitial

We do not have to know what caused the temperature to change to
calculate this difference. All that we need are the initial and final values.
This independence from the method by which a change occurs is an
especially important property of state functions, and being able to recognize
when some function or property is a state function simplifies many useful
calculations.

Enthalpy is a particularly important state function. The enthalpy of a
system in a given state cannot depend on how the system arrived in that
state. This is useful to know, because when we measure the heat of a
reaction we do not have to worry about how the reaction is occurring, but
only that it is. To determine H, we only have to be sure of our initial and
final states and then measure the total amount of heat absorbed or evolved
as the system changes between these states.
Measuring Heats of Reaction: Calorimetry
The changes in temperature caused by a reaction, combined with the values
of the specific heat and the mass of the reacting system, makes it possible to
determine the heat of reaction.

Heat energy can be measured by observing how the temperature of a
known mass of water (or other substance) changes when heat is added or
removed. This is basically how most heats of reaction are determined. The
reaction is carried out in some insulated container, where the heat absorbed
or evolved by the reaction causes the temperature of the contents to change.
This temperature change is measured and the amount of heat that caused
the change is calculated by multiplying the temperature change by the heat
capacity of the system.

The apparatus used to measure the temperature change for a reacting
system is called a calorimeter (that is, a calorie meter). The science of using
such a device and the data obtained with it is called calorimetry. The design
of a calorimeter is not standard and different calorimeters are used for the
amount of precision required. One very simple design used in many general
chemistry labs is the styrofoam "coffee cup" calorimeter, which usually
consists of two nested styrofoam cups.

When a reaction occurs at constant pressure inside a Styrofoam coffee-cup
calorimeter, the enthalpy change involves heat, and little heat is lost to the
lab (or gained from it). If the reaction evolves heat, for example, very
nearly all of it stays inside the calorimeter, the amount of heat absorbed or
evolved by the reaction is calculated.

Example Problem
The reaction of an acid such as HCl with a base such as NaOH in water
involves the exothermic reaction

                 HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O


In one experiment, a student placed 50.0 mL of 1.00 M HCl in a coffee-cup
calorimeter and carefully measured its temperature to be 25.5oC. To this
was added 50.0 mL of 1.00 M NaOH solution whose temperature was also
25.5oC. The mixture was quickly stirred, and the student noticed that the
temperature of the mixture rose to 32.4oC. What was the heat of reaction?

Assumptions
These are solutions, not pure water. The specific heat of water is 4.184
J/goC. Assume that these solutions are close enough to being like water that
their specific heats are also 4.1984 J/goC.

The density of water is 1.00 g/mL and even though these are solutions we
can assume that they are close enough to water to have the same density.

Solution
Calculate the heat actually evolved.

                          q = mcΔt

Fill in the missing info. We have mL's and we need grams.
Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1     g       = 100 grams of solution. (m = V X D)
              mL

Find the temperature change.

    Δt =tfinal - tinitial = 32.4oC - 25.5oC = 6.9oC

  q = mcΔt
    = 100 grams X 4.184      J X 6.9oC
                             o
                            gC

    = 2.9 X 103 J

This is the heat gained by the water, but in fact it is the heat lost by the
reacting HCl and NaOH, therefore q = -2.9 x 103 J.

i.e. it is an exothermic reaction, heat was lost to the water and it got
warmer.

This only gets us part way. This is the heat evolved for those specific
amounts used. (Notice we used identical amounts to keep these solutions
simple). We need to find the amount of heat released per mole.

How much HCl did we actually use anyways?

50.0 mL of HCl X 1.00 mol HCl = 0.0500 mol HCl
                1000 mL HCl

The same quantity of base, 0.0500 mole NaOH, was used.

To calculate the energy per mole of acid or base, divide the number of
joules by the number of moles.

i.e. molar enthalpy = J/mol = -2.9 x 103 J / 0.0500 mol
                        = -5.8 x 104 J/mol
                        = -58000 J/mol
                        = -58 kJ/mol

Therefore, for the neutralization of HCl and NaOH, the enthalpy change,
often called the enthalpy of reaction is ΔH = -58 kJ/mol
The Bomb Calorimeter
A type of calorimeter used in very precise measurements of heats of
reaction is called the bomb calorimeter. It is used to measure energy
changes for reactions that will not happen until they are deliberately
initiated, for example, combustions which must be ignited. The reactants
are put into the "bomb", which is then sealed and immersed in a large,
well-insulated vat of water. When the reaction is set off, any heat that is
liberated is absorbed by the bomb, the water, and any piece of the
equipment sticking into the water, and the temperature of the entire
contents of the vat rises. The stirrer ensures that any heat released becomes
uniformly distributed before the final temperature is read. From the
temperature change and the heat capacity of the calorimeter (water plus
everything in the water), the heat liberated is calculated.




Example Problem:
A sample of sucrose (table sugar) with a mass of 1.32 g is burned in a bomb
calorimeter. The heat capacity of this calorimeter had been previously
found to be 9.43 kJ/oC. The temperature changed from 25.00oC to 27.31oC.
Calculate the heat of combustion of sucrose in kilojoules per mole. The
formula of sucrose is C12H22O11.

Solution
The Δt is 2.31OC. For each degree increase, the reaction has evolved 9.43
kJ, as we know from the heat capacity. Therefore the total heat evolved is

               E = 2.31oC X 9.43 kJ = 21.8 kJ
                            o
                              C
This heat was produced by the combustion of 1.32 g of sucrose.

         moles = g/molecular mass
              = 1.32 g / 342.3 grams/mole
              = 3.86 x 10-3 mol of sucrose.

Therefore, the heat evolved per mole of sucrose is

         21.8 kJ       = 5.65 x 103 kJ/mole
     3.86 x 10-3 mol

Since the combustion is exothermic, this should be given a minus sign and
reported as -5.65 x 103 kJ/mol for the heat of combustion for sucrose.

Calorimetry Worksheet
1.   When sulphuric acid dissolves in water, a great deal of heat is given off.
     The enthalpy change for this process is called the enthalpy of solution.
     To measure it, 175 g of water was placed in a coffee-cup calorimeter
     and chilled to 10oC. Then 49.0 g of pure sulphuric acid, also at 10.0oC
     was added, and the mixture was quickly stirred with a thermometer.
     The temperature rose rapidly to 14.9oC. Assume that the value of the
     specific heat of solution is 4.184 J/goC. You may assume that the
     specific heat of the resulting sulphuric acid solution will also be 4.184
     J/goC. Calculate q for the formation of this solution, and calculate the
     enthalpy of solution in kilojoules per mole of H2SO4.
2.   Gram for gram, fats in food have much more chemical energy than
     sugar. One component of fat is stearic acid, C18H36O2. When a sample
     of 1.02 g of stearic acid was burned completely in a bomb calorimeter,
     the temperature of the calorimeter rose by 4.26oC. The heat capacity of
     the calorimeter was 9.43 kJ/oC. Calculate the molar heat of combustion
     of stearic acid in kilojoules per mole.
3.   The reaction of 2.000 mol of gaseous hydrogen with 1.000 moles of
     gaseous oxygen to form 2.00 mol of liquid water releases 517.8 kJ,
     provided that all reactants and products are brought to 25oC and 1 at.
     Write a thermochemical equation for the formation of 1.00 mol of
     liquid water.
4.   Ethanol, C2H5OH, is made industrially by the reaction of water with
     ethylene, C2H44. Calculate the value of Ho for the reaction

     C2H4(g) + H2O(l) ----> C2H5OH(l)

     given the following thermochemical equations:
    C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O(l) Ho = -1411.1 kJ

    C2H5OH(l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O(l) Ho = -1367.1 kJ
Standard Heats of Reaction
Chemical scientists have agreed upon a standard reference set of conditions
for temperature and pressure. These conditions have of course been chosen
so that experiments can be done easily. Thus the standard reference
temperature is 25oC, which is just slightly above normal room temperature.
This is easily controlled in a water bath. The reference pressure is 1
atmosphere or 101.325 kPa.

When the enthalpy change of a reaction is determined with all reactants
and products at 1 atm and some given temperature, and when the scale of
the reaction is in the moles specified by the coefficients of the equation, then
H is the standard enthalpy change or the standard heat of reaction. To
show that a pressure of 1 atm is used, the symbol H is given a superscript, o,
to make the symbol Ho. Values of Ho usually correspond to an initial and
final temperature of 25oC, unless otherwise specified.

The units of Ho are normally kilojoules. For example, a reaction between
gaseous nitrogen and hydrogen produces gaseous ammonia according to
the equation

                  N2(g) + H2(g) ----> 2 NH3(g)

When 1.00 mol of N2 react with 3.00 mol of H2 to form 2.00 mol of NH3 at
25o and 1 atm, the reaction releases 92.38 kJ. Hence for the reaction as
written Ho = -92.38 kJ.

Thermochemical Equations
Often it is useful to make the enthalpy change of a reaction part if its
equation. When we do this we have to be very careful about the coefficients,
and we must indicate the physical states of all the reactants and products.
The reaction between gaseous nitrogen and hydrogen to form gaseous
ammonia, for example, releases 92.38 kJ is 2.00 mol of NH3 forms.

But if we were to make twice as much, or 4.00 mol, of NH3 from 2.00 mol of
N2 and 6.00 mol of H2, then twice as much heat (184.8 kJ) would be
released. On the other hand, if only 0.50 mol of N2 and 1.50 mol of H2 were
to react to form only 1.00 mol of NH3, then only half as much heat (46.19
kJ) would be released.

An equation that includes its value of Ho is called a thermochemical
equation. The following three thermochemical equations for the formation
of ammonia, for example, give the quantitative data describe in the
preceding paragraph and correctly specify the physical states of all
substances.

             N2(g) + 3 H2(g) ----> 2 NH3(g) Ho = -92.38 kJ

            2 N2(g) + 6 H2(g) ----> 4 NH3(g) Ho = -184.8 kJ

         ½ N2(g) + 1½ H2(g) ----> NH3(g) Ho = -46.19 kJ

When you read a thermochemical equation, always interpret the coefficient
as moles. This is why we must use fractional coefficients in such an
equation, where normally we try to avoid them (because we cannot have
fractions of molecules). In thermochemical equations, however, fractions
are allowed, because we can have fractions of moles.

Thermochemical Equations for Experimentally Difficult Reactions
Once we have the thermochemical equation for a particular reaction, we
automatically have all the information we need for the reverse reaction.
The thermochemical equation for the combustion of carbon to give carbon
dioxide is:

                 C(s) + O2(g) -------> CO2(g) Ho = -393.5 kJ

The reverse reaction would be experimentally impossible to preform, for
reasons that will become clear later. But from the above equation we can
write the reverse reaction.

                     CO2 (g) ------> C(s) + O2 (g) Ho = +393.5 kJ

If we have the Ho for a given reaction, the Ho for the reverse reaction has
the same numerical value, but its algebraic sign is reversed.
Thermochemistry: Standard Heats of Formation Worksheet
1. Calculate Ho in kilojoules for the following reactions.
   a) 2 NO(g) + O2(g) ---> 2 NO2(g)
   b) NaOH(s) + HCl(g) ----> NaCl(s) + H2O(g)
2. Use a standard enthalpies of formation table to determine the change in
   enthalpy for each of these reactions.
   a) 2 CO(g) + O2(g) ---> 2 CO2(g)
   b) CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l)
   c) 2 H2S(g) + 3 O2(g) ---> 2 H2O(l) + 2 SO2(g)
3. Calculate Ho for these reactions. In each case, state whether the reaction
   is exothermic or endothermic, rewrite the equation as a thermochemical
   equation to include the heat term, and indicate whether the products
   have a greater or smaller enthalpy than the reactants. Hfo of NH4Cl =
   314.4 kJ/mol.
   a) SO2(g) + ½ O2(g) ---> SO3(g)
   b) CaO(s) + H2O(l) ---> Ca(OH)2(s)
   c) N2(g) + 3 H2(g) ---> 2 NH3(g)
   d) C6H6(l) + 1½ O2(g) ---> 6 C(s) + 3 H2O(l)
   e) NH3(g) + HCl(g) ---> NH4Cl(s)
Hess's Law of Heat Summation
When thermochemical equations are added to give some new equation,
their values of Ho are also added to give the Ho of the new equation.

The enthalpy change for a reaction is a state function. Its value is
determined only by the enthalpies of the initial and final states of the
chemical system, and not by the path taken by the reactant as they form the
products. To appreciate the significance of this, let us consider again the
combustion of carbon.

      C(s) + O2(g) -------> CO2(g) Ho = -393.5 kJ

This is only one possible way to make CO2.

The second pathway to CO2 involves two steps. The first is the combination
of carbon with just enough oxygen to form carbon monoxide. Then, in the
second step, this CO is burned in additional oxygen to produce CO2. Both
steps are exothermic, and their thermochemical equations are:

      C(s) + ½O2(g) ------> CO(g) Ho = -110.5 kJ
     CO(g) + ½O2(g) ------> CO2(g) Ho = -283.0 kJ

Note please that if we add the amount of heat liberated in the first step to
the amount released in the second, the total is the same as the heat given off
by the one-step reaction that was described first.

                 (-110.5 kJ) + (-283.0) = -393.5 kJ
Enthalpy Diagrams
The enthalpy relationships involved in adding thermochemical equations
are most easily visualized by means of an enthalpy diagram, such as the one
below. Horizontal lines in such a diagram correspond to different absolute
values of enthalpy, H. A horizontal line drawn higher in the diagram
represents a larger value of H. Changes in enthalpy, H, are represented by
the vertical distances between the lines. Take another look at the diagram
below. It shows all three of the CO2 reactions already discussed. The total
decrease in energy, however, is the same regardless of which path is taken,
so the total energy evolved in the two-step path has to be the same as in the
one-step reaction.




Standard Heats of Formation and Hess's Law
The preceding sections have lead to this point.

    ΔHo = [Ho products] - [Ho reactants]

What goes into the products and reactants brackets are the standard heats
of formation. We'll begin by the defining the term standard state. Any
substance in its most stable physical form (gas, liquid, or solid) at 25oC and
under a pressure of 1 atm is said to be in its standard state. The element
oxygen for example, is in its standard state when its exists as a molecule of
O2 -not as atoms and not as molecules of O3. The element carbon is in its
standard state when it exits as graphite - not as diamond - at 25oC.
Diamond is also a form of carbon, but it is actually slightly less stable than
graphite.

The quantities that we'll use from now on to compute values for Ho are
called the standard enthalpies of formation of standard heats of formation.
The standard heats of formation of a compound Hfois the amount of heat
absorbed or evolved when one mole of the compound is formed from its
elements in their standard states. Thus, the thermochemical equation for the
formation of one mole of liquid water from oxygen and hydrogen in their
standard states is

          H2(g) + ½O2(g) ------> H2O(l)           Hfo = -285.8 kJ/mol
The standard enthalpy change for this reaction, that is, the enthalpy change
at 25oC and 1 atm, is called the standard heat of formation of liquid water.
This is a point that often causes confusion among students. Each of the
following equations, for example involves the formation of CO(g)

                   C(s) + O2(g) ----> CO2(g)
                CO(g) + ½O2(g) ----> CO2(g)
                2 C(s) + 2 O2(g) ----> 2 CO2(g)

However, only the first involves just the elements as reactants and the
formation of just one mole of CO2. In the second equation, one of the
reactants is a compound, carbon monoxide, and in the third, two moles of
CO2 are formed. Only the first equation, therefore, has a standard enthalpy
change that we identify as Hfo.

Your databook has tables of standard heats of formation for a variety of
substances. They are listed alphabetically and in some cases not logically.
There are also standard heats of formation values for the organic
chemicals, too.

***************** IMPORTANT **********************
The Hfo for any element in its standard state is zero. (0 kJ/mol)

This makes sense if you think about it. There would be no enthalpy change
if you "form an element in its standard state from itself."

Using Hess's Law with Standard Heats of Formation
The Ho for any reaction must be the difference between the total enthalpies
of formation of the products and those of the reactants. Any reaction can be
generalized by the equation

                aA + bB + ..... -----> nN + mM + .....

where a, b, n, m, etc, are the coefficients of substances A, B, N, M, etc. The
value of Ho can be found using Hess's law equation.

            Δ Ho = [nHfo(N) + mHfo(M) + ...] - [aHfo(A) + bHfo(B) + ...]

Putting this into a more useful form we get:

            Δ Ho = [sum of the Hfo of products]-[sum of the Hfo of reactants]
Example Problem
Some chefs keep baking soda, NaHCO3, handy to put out grease fires.
When thrown on the fire, baking soda partly smothers the fire, and the heat
decomposes it to give CO2, which further smothers the flame. The equation
for the decomposition of NaHCO3 is

                 2 NaHCO3(s) -------> Na2CO3(s) + H2O(g) + CO2(g)

Calculate the ΔHo for this reaction in kilojoules and kilocalories.

Solution
        ΔHo = sum of products - sum of reactants
          = [ Na2CO3(s) + H2O(g) + CO2(g) ] - [ (2)NaHCO3(s)]

Look up the values in the databook tables for each substance.
Make sure the physical states are identical.

           = [-1130.7 -241.8 -393.5 kJ/mol ]-[(2)(-950.8 kJ/mol)]
           = -1766 kJ/mol - (-1901.6 kJ/mol)
           = +135.6 kJ/mol

Under standard conditions, the reaction is endothermic by 135.6 kJ/mol.
To calculate kilocalories, remember the conversion factor of

              1 cal = 4.184 J or 1 kcal = 4.184 kJ

In kilocalories the reaction is endothermic by 135.6 kJ/mol
                                              4.184 kJ/cal

which is 32.41 kcal/mole.

Example Problem #2
What is the ΔHo in kilojoules for the combustion of 1 mol of ethanol,
C2H5OH(l), to form gaseous carbon dioxide and gaseous water?

Solution
First write and balance the combustion equation.

              C2H5OH(l) + 3 O2(g) -----> 2 CO2(g) + 3 H2O(g)

Hess's law for this equation is:

         ΔHo = [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l) + (3)O2(g)]
            = [ (2)-393.5 + (3)-241.8 kJ/mol] - [ -277.1 + (3)0 kJ/mol]
            = [-787 -725.4 kJ/mol ] - [ -277.1 kJ/mol ]
            = -1512.4 + 277.1 kJ/mol
            = -1235.3 kJ/mol

The reaction for the combustion of ethanol is exothermic by 1235.3 kJ/mol.
Values of Hfo from Standard Heats of Combustion
To measure directly the heat of formation of sucrose, C12H22O11, you would
have to carry out the following reaction:

               12 C(s) + 11 H2(g) + 5½ O2(g) ------> C12H22O11


But no one has ever been able to figure out how to make this reaction occur
directly under any conditions, so there is no direct way to measure the Hfo
of sucrose. How, then, can we obtain values of Hfo for compounds such as
sucrose?

If the compound in question can be burned - which is usually far easier to
do than make it from its elements - then we have a source of energy data
that we can use to calculate its Hfo. This is because the products of
combustion are nearly always compounds whose values of Hfo are known or
can be measured by direct means. The combustion of sucrose in an
atmosphere of pure oxygen proceeds by the following equation:

            C12H22O11(s) + 12 O2(g) -----> 12 CO2(g) + 11 H2O(l)


If the standard enthalpy change for this reaction can be measured, and if
we can look up the values of Hfo for three of the four chemicals in the
equation, then we can use the Hess law equation to find the Hfo of the
remaining substance, sucrose.

The above equation is a combustion equation and your data tables have Hco
values (standard heat of combustion). Putting these into a Hess's Law
equation you should get:

        Hco = [(12)CO2(g) + (11)H2O(l)] - [C12H22O11(s) + (12)O2(g)]


All values are in kJ/moles.

      -5639.7 = [(12)-393.5 + (11)-285.8] - [C12H22O11(s) + (12)0]
      -5639.7 = -4722 -3143.8 -[C12H22O11]
       2226.1 = -[C12H22O11]

Therefore the Hfo of sucrose, C12H22O11, is -2226.1 kJ/mole

Sample problem
One of the "building blocks" for proteins such as those in muscles and
sinews is an amino acid called glycine, C2H5NO2. The equation for its
combustion is

         4 C2H5NO2(s) + 9 O2(g) ---> 8 CO2(g) + 10 H2O(l) + 2 N2(g)
The value of Hco for glycine is -973.49 kJ/mole. Using this information and
the values of Hfo calculate the Hfo for glycine.

Solution
For this problem, Hess's law equation becomes

Ho = [(8)CO2(g) + (10)H2O(l) + (2)N2(g)]-[(4)C2H5NO2(s) + (9)O2(g)]

No, I didn't forget the Hco. The first term Ho, is obtained from the standard
heat of combustion of glycine. Since the chemical equation for this reaction
is for the combustion of four moles of glycine, we have to multiple
Hocombustion by four.

Ho = 4 mol x -973.49 kJ/mol = -3894.0 kJ

Now we can substitute into Hess's law equation the correct values.

   -3894.0 kJ = [(8)-393.5 + (10)-285.8 + (2)0] - [(4)C2H5NO2 + (9)0]
     -3894.0 = -3148.0 -2858 -[(4)C2H5NO2]
      2112.0 = -[(4)C2H5NO2]

Therefore by rearranging we get

           Hfo for C2H5NO2 = -2112.0 kJ/mole = -528.0 kJ/mole
                               4 moles

Thus, the standard heat of formation of glycine is -528.0 kJ/mol, and we
have seen how we can determine this quantity without making glycine
directly from its elements.
Thermochemistry Review Worksheet
1. Write thermochemical equations for the combustion of one mole of each
   of the following compounds. The products in each case are gaseous
   carbon dioxide and liquid water. (Standard conditions are assumed.)
     The numbers given in parentheses after each formula are the standard
     heats of combustion.
     (a) Acetylene, C2H2(g) (-1299.6 kJ/mol)
     (b) Methyl alcohol, CH3OH(l) ( -726.51 kJ/mol)
     (c) Diethyl ether, C4H10O(l) (-2751.1 kJ/mol)
     (d) Toluene, C7H8(l) (-3909.0 kJ/mol)
2.   Palmitic acid, C16H32O2, is typical of the materials available from fats
     and oils insofar as chemical energy is concerned.
     (a) Write the thermochemical equation of the complete combustion of
     one mole of this compound, a solid, assuming that gaseous carbon
     dioxide and liquid water form. The standard heat of combustion of
     palmitic acid is -9957.92 kJ/mol.
     (b) Using the molar heat of combustion of palmitic acid and data from
     the appropriate source, estimate the Hfo for palmitic acid.
3.   Construct an enthalpy diagram that shows the enthalpy changes for a
     one-step conversion of germanium, Ge(s), into its dioxide, GeO2(s), and a
     two-step conversion - first to the monoxide, GeO(s), and then the
     oxidation of the monoxide to the dioxide. The relevant thermochemical
     equations are as follows.
     Ge(s) + ½ O2(g) -----> GeO(s)        Ho = -255 kJ
     Ge(s) + O2(g) -----> GeO2(s)       Ho = -534.7 kJ
     Using this diagram, determine the value of Ho for the reaction
     GeO(s) + ½ O2(g) -----> GeO2(s)
4.   Calculate Ho for the following reaction, which describes the preparation
     of an unstable acid, HNO2, nitrous acid.
     HCl(g) + NaNO2(s) ----> HNO2(l) + NaCl(s)
     Use the following thermochemical equations. Calculate the answer in
     kilojoules.
     2 NaCl(s) + H2O(l) --> 2 HCl(g) + Na2O(s)          Ho = +507.3 kJ
     NO(g) + NO2(g) + Na2O(s) ----> 2 NaNO2(s)           Ho = -427.14 kJ
     NO(g) + NO2(g) -----> N2O(g) + O2(g)         Ho = -42.68 kJ
     2 HNO2(l) -----> N2O(g) + O2(g) + H2O(l)        Ho = +34.35 kJ
5.   The standard heat of formation of gaseous hydrogen bromide was first
     evaluated by means of the standard enthalpy values measured for the
     following reactions. The last three are standard heats of solution.
     Cl2(g) + 2 KBr(aq) ----> Br2(aq) + 2 KCl(aq)        Ho = -96.2 kJ
     H2(g) + Cl2(g) ----> 2 HCl(g)                      Ho = -184 kJ
     HCl(aq) + KOH(aq) ----> KCl(aq) + H2O(l)            Ho = -57.3 kJ
     HBr(aq) + KOH(aq) ----> KBr(aq) + H2O(l)            Ho = -57.3 kJ
     HCl(g) --------> HCl(aq)                          Ho = -77.0 kJ
     Br2(g) --------> Br2(aq)                         Ho = -4.2 kJ
HBr(g) --------> HBr(aq)                        Ho = -79.9 kJ
Write the thermochemical equation for the formation of 1 mol of HBr(g)
from its elements, including its value for Hfo.