AP Chemistry - Introduction to Chemical Thermodynamics

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.
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
 

Energy Sources
Chemical Potential Energy - Energy that is bound up in chemicals.  This energy is usually thought of as the kinetic molecular energy of the moecules and the energy associated in the bonds.

Electrical Potential Energy - Energy that is associated with the movement of electricity, usually through a resistor that produces a release of heat because of resistance.

Nuclear Potential Energy - Energy associated with the strong and weak forces in a nucleus.  Fission and fusion reactions create atoms that have less energy than  the atoms from which they were made.  This loss of energy is usually released in differnet forms, heat being only one of them.

Mechanical Energy - Energy that is released due to friction and compression.

Geothermal Energy - Energy associated with the Earth.  Usually caused by the Earth's fluidic magma coming in contact with ground water.

Thermometers and Liquids:
Thermometer - from the Latin Thermos or heat.  A device used to measure the kinetic energy (thermos) of something else.  Thermometers work because at a microscopic level liquids expand when they gain energy.  As a liquid gains energy the molecules of the liquid move faster.  Since the molecules move faster they collide more often.  More collisions mean that the molecules push their neighbours further apart.  At a macroscopic level we see the liquid expand up the bore of the thermomter tube.

The corollary of this is the cooling of a liquid.  Cooling is  aloss of heat energy.  A loss of heat means a loss of kinetic energy and a slowing down of the molecules.  A slowing of the molecules means fewer collisions and the molecules take up less space.  They can pack in closer together and we see the liquids contract or shrink when cooled.

Thermometers cannot measure extreme temperatures because they reach the point where they solidify when cooled or boil when heated.  Temperature by definition is a measure of the average kinetic energy available in an environment. i.e. At 0 K there is 0 J of energy and 0 kinetic energy.  At 0 K all motion has stopped, even the vibrational movement of molecules.  As we increase the temperature substances first begin to vibrate, then melt into lquids, then expand, then boil, then expand again as gases.

Ancient Theories of Heat
Four Element Theory - All matter is made up of combinatiosn of Earth, Air, Fire and Water.  ie. Add fire, matter warms up.  Cooling is the removal of fire from the matter.

Caloric Theory - An attempt at quantifying the fire. There were thought to be particles called "calorics".  An object gained "calorics" as it warmed and lost "calorics" as it cooled.  These "calorics" came from fire as the matter was warmed.  i.e. fire was a source of calorics and calorics where transferred from a hot substance to a cooler substance or lost to the environment as something cooled.  Count Rumford of Bavaria disproved this theory.  If there was such a thing as a caloric then the particle must has a mass.  Therefore when something warms up there must be an increase in mass.  When something cooled there must be a decrease in mass.  Using very careful measurements on brass cannons that were being bored, Count Rumford noted that a tremendous amount of heat was being liberated due to the boring but no change in mass was observed.  i.e. the mass of the block of brass was equal to the mass of the cannon and the brass turnings.  If the caloric theory was true then the cannon and turnings should have weighted much less because the cannons got very hot during boring and this heat was observed to be lost to the air around the boring machine.
Measuring Heat
Definitions:
Joule (J) - named after James Joule. Labelled in an equation as "Q", a unit of energy.   It requires 4.184 J to raise the temperature of 1 gram of water by 1 oC.
Watt (W) - after James Watt.  A derived unit:   1 Watt  = 1 J/second
Sample Problem #1    An electric kettle delivers 144,00 J of energy to the water in it in 2 minutes.
What is the power of the kettle?

P = Q/t =  144,000 J / 2 min * 60 sec/ min = 144,000 J / 120 sec = 1200 J/s = 1200 W
 

Sample Problem #2     A 100 watt immersion heater used in a tropical fish tank is used to heat the water in a beaker for 3 minutes.  How much energy was transferred to the water?

P = Q/t  therefore   Q = P * t = 100 W * 3 min = 100 W * 180 sec = 18,000 J = 18 kJ
 

The Capacity to Hold Heat
When heat is added to a substance it's temperature increases.  There is a direct relationship between the temperature reached and the heat added.
There is another direct relationship between the mass of a substance and the heta needed to make the substance undergo a given change in temperature.
Water is the principle substance used because it was water that was used to define the energy unit and water is readily accessible and relatively safe to use.
Imperial and Metric Heat Unit
The calorie (cal) was defined to be the amount of heat required to raise 1 gram of water by 1 degree Celsius.
If you had 100 grams of water it would take 100 calories of heat energy to raise it's temperature by 1oC.
This is became a definition and was called the Specific Heat Capacity. The symbol used in equations to describe the Specific Heat Capcaity is 'c' and the units are J/goC
For example the Specific Heat Capacity of water is 1 cal/goC = 4.184 J/goC (Metric definition)
Heat Capacity is either the amount of heat required to increase the temperature of a substance by 1.0 oC or the amount of heat released to the environment to cool a substance by 1.0 oC
Specific Heat Capacity Table
Substance Specific Heat Capacity 
at 25oC in J/goC
H2 gas 14.267
He gas 5.300
H2O(l) 4.184
lithium
3.56
ethyl alcohol
2.460
ethylene glycol
2.200
ice @ 0oC
2.010
steam @ 100oC
2.010
vegetable oil
2.000
sodium
1.23
air
1.020
magnesium
1.020
aluminum
0.900
Concrete
0.880
glass
0.840
potassium
0.75
sulphur
0.73
calcium
0.650
iron
0.444
nickel
0.440
zinc
0.39
copper
0.385
brass
0.380
sand
0.290
silver
0.240
tin
0.21
lead
0.160
mercury
0.14
gold 0.129
To calcualte the Specific Heat Capacity you need the following equation:  Q = mcΔt
c = Q / m Δt       where  c is the specific heat capacity
                                              Q is the energy
                                              m is the mass of the substance
                                          Δt is the temperature change
Sample Problem #3
An immersion heater isused to warm 500 grams of a liquid from 35oC to 55oC.   If 20 kJ of neergy are given to the liquid, determine the specific heat capacity and use it to suggest an identity for the liquid.
c = Q/m Δt = 20 kJ / 500 g * (55oC - 35oC) = 20,000 J / 500 g * 20oC =  2.00 J/goC
Based on the calculated value of 'c' the substance may be vegetable oil.
Principle of Heat Transfer
The KMT theory can be used to explain the process of heat conduction. When one end of a rod of metal is warmed the particles a tthat end gain energy and vibrate more rapidly.  They collide with the cooler particles next to them.  the cooler particles in turn speed up due to the collision.  They in turn collide with their neighbours.  This action continues in a chain reaction along the metal.

Solids conduct heat, liquids and gases display heat convection. Heat also can be radiated.  Heat is actually part of the electromagnetic spectrum.  Remember that infrared heat from the electromagnetic spectrum chart.  It radiates just like any other form of  light radiation.

Since heat is a form of radiant energy it can be transmitted (passed through an object), it can be absorbed (taken into an object and stored) or it can be reflected (bounced off an object).
 

This comes from an understanding of the Law of Conservation of Mass and Energy.  You know that hotter objects transfer heat energy to colder objects.  Hotter objects get colder, colder objects get hotter.
             QHeat Released by the Hot Object = QHeat Gained by the Cold Object

              since Q = mc Δt

              then       mhch Δth  =   mccc Δtc
 

and if both substances are the same then ch = cc can be elimnated.  (where ch is the specific heat capacity of   the hot substance and cc is the specific heat capacity of the cold object.)
mh Δth  =   mc Δtc   
Sample Problem #4
If 80 grams of water at 70oC is mixed with a caertain mass of cool water at 20oC, the final temperature of the mixture is 60oC.  What is the mass of the cool water.
Qh = Qc
mhch Δt h = mcccΔt c
since we are using water for both the hot and cold sides of the equation we can cancel out the ch = cc
therefore
mh Δth = mcΔtc
mcmh Δth   = 80 g X (Ti  -  Tf)    =  80 grams X (70oC - 60oC)
           Δtc             (T- Ti)                       (60oC - 20oC) 

      = 80 g X 10oC   =  20 grams
                40oC

The initial mass of the cold water used was 20 grams.
Sample Problem #5
If 400 grams of water at 60oC are mixed with 100 grams of water at 10oC, what is the final temperature of the mixture?
            Qh = Qc
            mh Δth = mcΔtc
            400 g X (60oC - Tf)  =  100 g X (Tf - 10oC)
            24000 - 400Tf oC = 100Tf oC - 1000
            25000 = 500Tf oC
            Tf  = 25000/500 oC = 50oC
Sample Problem #6
A piece of metal with a mass of 500 grams and unknonwn specific heat capacity is placed in boiling water at 100oC.  The hot metal is then transferred quickly into a 200 grams sample of water at 20oC.  If the final temperature of the water-metal mixture is 30oC what is the specific heat capacity of the metal and identify the possible identity of the metal.
      Qh = Qc
      mhch Δth = mccc Δtc

      ch = mccc Δtc   =  200 g X 4.184 J/goC X 10oC   =  0.239 J/goC
                mh Δth               500 g X 70oC

      The metal is probably silver.
 

Molar 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.
The molar heat capacity is defines as:
 
molar heat capacity = _     J_  _
                                      mole oC


It is amount energy lost or gained when 1 mole of a subsstance changes its temperature by 1 degree Celcius.

A useful relationship is       J       g         =        J
                                        g oC        mole            mol oC

or the                    specific heat X molar mass = molar heat capacity

Go to the Specific Heat Worksheet
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 ΔtΔ 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

           Δ H = 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
Δ H = q (at constant pressure)
The sign of q, defined earlier, is actually determined by the sign of Δ H.   If Δ H is negative so q is negative for exothermic changes.   Δ H 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.

The First Law of Thermodynamics:  The 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.
  Δ t = 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.

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 = mcΔ t
 

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.
       Δ t =tfinal - tinitial = 32.4oC - 25.5oC = 6.9oC
    q = mcΔ t
       = 100 grams X 4.184    J   X 6.9oC
                                          goC
       = 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 Δ H = -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
                                                                            oC
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.

Go to the Enthalpy Calorimetry Worksheet
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.

Go to the Standard Heats of Formation Worksheet

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) ΔHo = -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) ΔHo = -110.5 kJ
         CO(g) + ½O2(g) ------> CO2(g) ΔHo = -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.
Law of Heat Summation (Hess's Law) For any reaction that can be written in steps, the standard heat of reaction is the same as the sum of the standard heats of reactions for the steps.
One of the most useful applications of Hess's law is the calculation of the value of ΔHo for a reaction whose ΔHo is unknown or cannot be measured. Hess's law says that we can add thermochemical equations, including their values of ΔHo, to obtain some desired thermochemical equation and its ΔHo.
Example Problem:
Consider the following thermochemical equations:

                    C(s) + ½O2(g) ------> CO(g)      ΔHo = -10.5 kJ
                CO(g) + ½O2(g) ------> CO2(g)     ΔHo = -283.0 kJ
 

Use them to find the Ho in kilojoules for the reaction.

                                 C(s) + O2(g) ------> CO2(g)
 

Solution
This is a particularly simple problem, but it illustrates a few important points about all such problems. Let's add the two given thermochemical equations:
             C(s) + ½O2(g) ------> CO(g)           ΔHo = -10.5 kJ
         CO(g) + ½O2(g) ------> CO2(g)          ΔHo = -283.0 kJ

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

The resulting equations can be simplified. Cancel the CO(g) because it appears on both sides. You can do this as long as you have the same chemical in the same physical state. Add the two oxygen terms together. This gives us the target equation.
                   C(s) + ½O2(g) ------> CO(g)         ΔHo = -10.5 kJ
                CO(g) + ½O2(g) ------> CO2(g)       ΔHo = -283.0 kJ
                     C(s) + O2 (g) ------> CO2(g)        ΔHo = -393.5 kJ
Example Problem #2
Carbon monoxide is often used in metallurgy to remove oxygen from metal oxides and thereby give the free metal. The thermochemical equation for the reaction of CO with iron(III) oxide, Fe2O3, is
              Fe2O3(s) + 3 CO(g) ------> 2 Fe(s) + 3 CO2 (g)        ΔHo = -26.74 kJ
Use this equation and the equation for the combustion of CO

                          CO(g) + ½O2(g) ------> CO2(g)     ΔHo = -283.0 kJ
 

to calculate the value of Ho for the reaction

                              2 Fe(s) + 1½O2 (g) -----> Fe2O3
 

Solution
Combine the equations in such a way that we can add them to the final target equation. Then we add the corresponding ΔHo's to obtain the ΔHo of the target equation.
Step   1  The target equation must have 2 Fe on the left, but the first equation above has 2 Fe on the right. To move it left, reverse the entire equation and remember to reverse the sign of Ho. When the equation is flipped over the Fe2O3 also falls into the correct position.
Step 2  There must be 1½O2 on the left, and we must be able to cancel three CO and three CO2 when the equations are added. Multiply the second equation by 3 and we get the necessary coefficients. Multiply the Ho values for the second equation by 3 as well. The adjusted equations are:

2 Fe(s) + 3 CO2(g) ----> Fe2O3(s) + 3 CO(g)          ΔHo = +26.74 kJ
3 CO(g) + 1½O2(g) ----> 3 CO2(g)                        ΔHo = 3(-283.0 kJ) = -849.0 kJ
 2 Fe(s) + 1½O2 (g) -----> Fe2O3                                     ΔHo = -822.26 kJ
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.
                     Ho = [nHfo(N) + mHfo(M) + ...] - [aHfo(A) + bHfo(B) + ...]
Putting this into a more useful form we get:
                     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 Ho 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, we 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.
Bond Energies
Bond energy is a property that as a profound influence on the chemical properties of a compound.   Bond energy is the amount of energy required to break a chemical bond to give electriclally neutral framnets.  Bond energies can tell a lot about a molecule, like wether it has single, double, or triple bonds.  Bond energy tends to increase as the number of bonds beween atoms increase. i.e. a double bond has more energy than a single bond.

 Bond energy is also important in determining the chemical properties of a substance.  During a chemical reaction, bonds within the reactants are broken and new ones are formed as the products appear.  The first step - bond breaking - is one of the factors that controls the reactivity of substances.  Elemental nutrogen, for example has a very low degree of reactivity, which is gneerally attributed to the very strong triple bond in N2.  Any reactions that involve the breaking of this bond in one step simply do not occur.  When N2 does react, it is by a stepwise involvement of its three bonds, one at a time.

Bond energies of simple molecules such as H2, O2 and Cl2 are usually measured spectroscopitally.  That is, the light that is emitted by the moecules when they are energized by a flame or an elecrtic arc is analyzed, and the amount of energy needed to break the bond is computed.

We can use a table of bond energies to calculate the heat of reaction for numerous compounds. 

Bond Bond Energy
kJ/mole
Bond Energy
kcal/mole
Bond Bond Energy
kJ/mole
Bond Energy
kcal/mole
Br-Br 192.9 46.1 O=O 498.3 119.2
Br-H 366.3 87.6 S-S 266 63.6
Cl-Cl 243.4 58.2 S=S 429.2 102.6
F-F 158 37.8 O-S 469 112.2
F-H 568.0 135.8 Si-SI 226 54.0
I-I 151.2 36.2 P=P 485 116
H-I 298.3 71.3 O-Si 466 111.4
H-H 435.9 104.2 O=Si 638 152.6
H-Si 318 76.0 O=Si 805 192.5
H-Ge 285 68.1 P-P 198 47.3
H-N 391 93.5 P=P 485 116
H-P 321 76.8 C-C 347 83
H-As 297 71 C=C 612 146.3
H-O 464 111 C=C 838 200.4
H-S 364 87 C-H 413 98.8
H-Se 313 75 C-F 467 111.7
Na-Na 72 17.2 C-Cl 346 82.7
K-K 49 11.7 C-Br 290 69.3
N-N 158 37.8 C-I 228 54.5
N=N  410 98 C-N 286 68.4
N=N 945.4 226.1 C=N 615 147.1
N-O 214 51.2 C=N 887 212.1
N=O 587 140.4 C-O 336 80.3
N=P 582 139.2 C=O 736 176
O-O 144 34.4 C=O 1077 257.5
O-O in O3 302 72.2



The is also a table of Heats of Formation of Gaseous Atoms
Atom Heat of Formation Heat of Formation Atom Heat of Formation Heat of Formation
C 715.0 170.9 F 78.91 18.86
N 472.7 113.0 Cl 121.0 28.92
O 249.2 59.56 Br 111.9 26.74
H 218.0 52.10 S 274.7 65.65

The convienent thing about covalent bond energies is that they are almost very nearly identical in many differnet compounds.   Since the bond energy does not vary very much we can use them to estimate heats of formation and heats of combustion.

Ex.  Use the bond energy table above to estimate the energy tied up in a molecule of methanol.
Solution
Step 1:  Write out the formula for methanol structurally:
             H
              |
         H-C-O-H
              |
             H

Step 2:  Determine all the bonds types in the molecule.
There are 3 C-H bonds, 1 C-O bond and an O-H bond

Step 3:   Look up the values

          C-H = 413 kJ/mole
          C-O =  336 kJ/mole
          O-H =  464 kJ/mole

Step 4:  Multiply according to the number of bonds present and add the results.

          3 (C-H) = 3 * 413 kJ/mole = 1239 kJ/mole
          1 (C-O) = 1 * 336 kJ/mole =   336 kJ/mole
          1 (O-H) = 1 * 464 kJ/mole =   464 kJ/mole 


                                                          2039 kJ/mole
Step 5:  The Heat of Formation is -2039 kJ/mol and the reaction is exothermic. 

Note:  The formation of CH3OH from gaseous atoms is exothermic - energy is always released when atoms become joined by a covalent bond. 

Ex 2.   Calculate the heat of formation for methanol from its gaseous elements.
Solution:
Step 1:  Write out the equation for the formation of methanol.

               C  +   2 H2   +    ½O2   ---->  CH3OH

Step 2:    Write out the equation structurally to make sure the bonds are correct.

                C   +   2  H-H  + ½ O=O   ---->    CH3OH (see the example above for CH3OH)

Step 3:    Add up the left side.

                 715.0   +     2 (H-H)  + ½ (O=O)   ----->   CH3OH
                 715.0   +     2 (435.9) + ½ (498.3)   ------>  2039 
                 715.0   +     871.8 + 249.15  ----->   2039
                            1835.95         ------->      2039

Step 4: Apply Hproducts - Hreactants

Hreaction = 2039 kJ/mole - 1835.95 kJ/mole 
               =  203.05 kJ/mole 
 

Go to the Bond Energy Worksheet
Go to the Enthalpy Unit Review