Strand 2  Energy and Rates
Unit 3 Thermodynamics Title


Forms of EnergyThere are numerous forms of energy we could investigate.  All forms of energy can be converted to other forms, but within a closed system no energy is every created or lost.  The size of this system depends upon one's relative view.  A beaker could be the system in which case the classroom would be outside the system and so heat could be lost from the system making it seem as energy was lost, but we know that in reality it was just lost to the surroundings.  If we consider the entire classroom the system, then no energy was in fact lost, but only transferred.  Thus the law of conservation of energy holds true even when it does not appear to.  In fact if we consider the system to be to the entire universe, then the law is never broken.  We we consider reactions we must consider both the system and the surroundings.  In chemical reactions only two forms of energy are consistently prevalent although others appear from time to time.  These ubiquitous forms are kinetic and potential energy.

    Kinetic energy is the energy of motion. An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy.  Recall from quantum chemistry that electrons are constantly in motion in a variety of ways (orbiting the nucleus, moving to new levels, spinning and etcetera) and thus represent kinetic energy.  If more kinetic energy is given to an object, some of this energy will be transferred to increased electron motion.  This form of kinetic energy is therefore always present and always affected by energy changes.  However, its impact is largely limited to the sub-subatomic level (aren't you glad).

    Other forms of kinetic energy include: vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another).  In every case, all four forms of motion are present.

Vibration of A Carbon Dioxide Molecule

Rotational MotionRotational Motion

    To keep matters simple, we will focus upon translational kinetic energy since it has the largest impact on chemical reactions.  The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) which an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.

Kinetic Energy

  This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four; for a threefold increase in speed, the kinetic energy will increase by a factor of nine; and for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem-solving, but also a guide to thinking about the relationship between quantities.

    An object can store energy as the result of its position. For example, the heavy ram of a pile driver is storing energy when it is held at an elevated position. This stored energy of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the result of its position. When assuming its usual position (i.e., when not drawn), there is no energy stored in the bow. Yet when its position is altered from its usual equilibrium position, the bow is able to store energy by virtue of its position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy of position possessed by an object.

Gravitational Potential EnergyElastic Potential Energy

    The two examples above illustrate the two different forms of potential energy - gravitational potential energy and elastic potential energy. Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height). The energy is stored as the result of the gravitational attraction of the Earth for the object. The gravitational potential energy of the heavy ram of a pile driver is dependent on two variables - the mass of the ram and the height to which it is raised. There is a direct relation between gravitational potential energy and the mass of an object; more massive objects have greater gravitational potential energy. There is also a direct relation between gravitational potential energy and the height of an object; the higher that an object is elevated, the greater the gravitational potential energy.

    The second form of potential energy, elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Elastic potential energy can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow, etc. The amount of elastic potential energy stored in such a device is related to the amount of stretch of the device - the more stretch, the more stored energy.    Springs are a special instance of a device which can store elastic potential energy due to either compression or stretching. A force is required to compress a spring; the more compression there is, the more force which is required to compress it further. 

    It is the second form of potential energy that is of interest to chemists since it influences how far apart to interacting particles will be. For example there is potential energy created due to the distance between nuclear particles (protons and neutrons).  There is also potential energy stored in intermolecular and intramolecular bonds.  Hence changing potential energy can have a huge impact on the state of a substance and its reactivity.

    When we look at the heating of something we can investigate how kinetic and potential energy relate.  The combined kinetic and potential energy of something is called its thermal energy.  Thermal energy curves tend to look like this:

General Heating Curve

    Consider the heating of water:

Water Heating Curve

    The heating of ice at -25 °C to +125 °C at constant pressure (1 atm) will exhibit the following characteristics:

  • Initially, the heat input is used to increase the temperature (KE) of the ice, but the ice does not change phase (remains a solid).
  • As the temperature approaches some critical point (i.e. the melting temperature of ice), the kinetic energy of the molecules of water is sufficient to allow the molecules to begin sliding past one another. 
  • As the ice begins to melt, additional input of heat energy does not raise the temperature of the water, rather it is used to overcome the intermolecular attraction (PE) during the phase change from solid to liquid.
  • Once the water is in a liquid phase, increasing the amount of heat input raises the temperature of the liquid water.
    As the temperature approaches another critical point (the vaporization, or boiling, temperature of water) the kinetic energy of the molecules is sufficient to allow the separation of molecules into the gas phase.  As the liquid begins to boil. Additional input of heat energy does not raise the temperature of the water, rather it is used to overcome the intermolecular attractions during the phase change from liquid to gas. Once the water is in the gas phase, additional heat input raises the temperature of the water vapor.  (Note: greater energy is needed to vaporize water than to melt it).

    Heating ice, water and water vapor:

  • In the region of the curve where we are not undergoing a phase transition, we are simply changing the temperature of one particular phase of water (either solid, liquid or gas) as a function of heat input. (KE)
    The slope of the lines relates temperature to heat input .  The greater the slope, the greater the temperature change for a given unit of heat input .  When water finally changes to a gas any further heat serves to both increase the motion of the molecules (KE) and the distance between them (PE).

    The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. The relationship between heat and temperature change is usually expressed in the form shown below where c is the specific heat. The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature. 

Calculating Quantities of Heat

    All three factors must be considered when looking at how much heat something will absorb.  The greater the mass, the more heat that can be absorbed without significantly changing the temperature.  Similarly, the greater the temperature difference for any mass of a substance, the greater the heat difference between those two temperatures.  The last variable, c (specific heat capacity) is dependent upon the type of material present.  Each substance has a unique specific heat capacity (although two substances can have the same value).  Specific heat capacity is influenced by the density and arrangement of particles within the substance which is determined by its atomic and quantum structure.  Consider some examples:

1. Calculate the amount of heat needed to increase the temperature of 250 g of water from 20oC to 56oC.

Q = m c(Tf - Ti)

m = 250 g
c = 4.18 JoC-1 g-1 (from table above)
Tf = 56oC
Ti = 20oC

Q = (250)(4.18)(56 - 20) = 37 620 J 

2. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15 g of copper from 25oC to 60oC.

Q = m c(Tf - Ti)

q = 204.75 J
m = 15g
Ti = 25oC
Tf = 60oC

204.75 = (15)c(60 - 25)
204.75 = 525c
c = 204.75 ÷ 525 = 0.39 JoC-1 g-1

3. The initial temperature of 150 g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol?  (Specific heat capacity of ethanol is 2.44 JoC-1g-1).

Q = m c(Tf - Ti)

q = 3240 J
m = 150 g
c = 2.44 JoC-1g-1
Ti = 22oC

3240 = (150)(2.44)(Tf - 22)
3240 = 366 (Tf - 22)
3240÷366 = (Tf - 22)
8.85 = Tf - 22

Tf = 30.9oC

    These values are actually determined using a calorimeter which can precisely measure changes in temperature associated with heating changes.  It is is the same device used to estimate how much energy (calories) is released through the burning (cellular digestion) of food.



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

    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 meaning that energy has been lost from the system. Endothermic reactions are documented by +Q 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, 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  means "change in".  As a result we look at:

= Hfinal - Hinital

where 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:

= 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 (released or added) by direct measurement. 

    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. 

    Consider the following series of energy changes (don't worry about what the entire reaction is doing):

Energy Changes

  • step 1 is endothermic since the products are less stable and possess more energy than the reactants (energy was added and thus temperature drops)
  • steps 2 and 3 are both exothermic since in each case the products possess less energy (more stable) than the respective reactants (energy was released and thus temperature increases)
  • step 4 is endothermic since the products possess more energy than the reactants (energy was added and thus temperature drops)
Hess's Law

    Chemical scientists have agreed upon a standard 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.  Together they are referred to STAP.

    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 STAP, the reaction releases 92.38 kJ. Hence for the reaction as written Ho = -92.38 kJ. 

    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. 

    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 perform, 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. 

    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 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   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.

  This is the Law of Heat Summation (Hess's Law) which states that 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 Ho for a reaction whose Ho is unknown or cannot be measured. Hess's law says that we can add thermochemical equations, including their values of Ho, to obtain some desired thermochemical equation and its Ho

    Consider the following example:  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)           Ho = -26.74 kJ 

Use this equation and the equation for the combustion of CO 

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

to calculate the value of Ho for the reaction 

          2 Fe(s) + 1½O2(g) Fe2O

     Combine the equations in such a way that we can add them to the final target equation. Then we add the corresponding Ho's to obtain the Ho of the target equation. 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. 

          2 Fe(s) + 3 CO2(g) Fe2O3(s) + 3 CO(g)           Ho = +26.74 kJ 

    There must be 1½O2 of 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. 

          3 CO(g) + 1½O2(g) 3 CO2(g)           Ho = 3(-283.0 kJ) = -849.0 kJ 

    If we then sum our equations we get the equation we want.  If we sum the Ho we get the value we were looking for.

          2 Fe(s) + 3 CO2(g) Fe2O3(s) + 3 CO(g)           Ho = +26.74 kJ 
          3 CO(g) + 1½O2(g) 3 CO2(g)                          Ho = -849.0 kJ 
          2 Fe(s) + 1½O2(g) Fe2O3                                      Ho = +26.74 kJ + -849.0 kJ  = -822.26 kJ 

    Notice that some species cancel out of the reactions when you sum them.

    If you look at the reactions when you are summing them, it becomes apparent the the summation of the heat terms always follow a predictable pattern which is o = [Hoproducts] - [Horeactants

    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 Hfo is 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.  Standard heats of formation have been calculated and tabulated.  Note that 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." 

    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. 

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

Putting this into a more useful form we get: 

o = [sum of the Hfo of products]-[sum of the Hfo of reactants] 

     Consider the following example:  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 o for this reaction in kilojoules.

o = 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. 

    Note that any Ho can be calculated using o using either the reactions or the summation equation.  You would just leave one of the values in the o equation as unknown and solve the equation.

Nuclear Energy

    Nuclear fusion is the energy-producing process which takes place continuously in the sun and stars. In the core of the sun at temperatures of 10-15 million degrees Celsius, Hydrogen is converted to Helium providing enough energy to sustain life on earth.  It is the process by which two or more nuclei are joined together upon collision to form new products and energy.  It is this energy that is of interest. 


    For energy production on earth different fusion reactions are involved. The most suitable reaction occurs between the nuclei of the two heavy forms (isotopes) of Hydrogen - Deuterium (D) and Tritium (T); eventually reactions involving just Deuterium or Deuterium and Helium (3He) may be used. 


    At the temperatures required for the D-T fusion reaction - over 100 Million deg. C - the fuel has changed its state from gas to PLASMA. In a plasma, the electrons have been separated from the atomic nuclei (usually called the "ions"). Understanding plasma required major developments in physics.  Plasmas are now used widely in industry, especially for semi-conductor manufacture. 

    There are some definite advantages of fusion.  Aside from the vast, new source of energy due to the fact that the fuels are plentiful, it is inherently safe since any malfunction results in a rapid shutdown of the reactor (safety protocols are very stringent).  Furthermore, no atmospheric pollution leading to acid rain or "greenhouse" effect is created and the radioactivity of the reactor structure, caused by the neutrons, decays rapidly and can be minimized by careful selection of low activation materials; therefore provision for geological time span disposal is not needed. 

    Deuterium is abundant as it can be extracted from all forms of water. If all the world's electricity were to be provided by fusion power stations, Deuterium supplies would last for millions of years. Tritium does not occur naturally and will be manufactured from Lithium within the machine. Lithium, the lightest metal, is plentiful in the earth's crust. If all the world's electricity were to be provided by fusion, known reserves would last for at least 1000 years.  Once the reaction is established, even though it occurs between Deuterium and Tritium, the consumables are Deuterium and Lithium.  10 grams of Deuterium which can be extracted from 500 litters of water and 15g of Tritium produced from 30g of Lithium would produce enough fuel for the lifetime electricity needs of an average person in an industrialized country. 

    Nuclear fission is the process of splitting atoms, or fissioning them.  This can happen naturally or in a controlled nuclear reaction.  The challenge is to split the nuclei of an atom.  To do this something substantial needs to impact the nucleus with great energy and speed.  This something also needs to be small to actually hit just the nucleus and in particular the components of the nucleus.  Often neutrons are used to bombard nuclei to cause them to break because they fit all the criteria.  Protons can be used but they take considerably more energy since they must move fast enough to overcome the great positive repulsion created the nucleus being bombarded.


    Fissile isotopes are isotopes of an element that can be readily or easily split through fission. Only certain isotopes of certain elements are fissile. For example, one isotope of uranium, 235U, is fissile, while another isotope, 238U, is not. Other examples of fissile elements are 239Pu and 232Th. An important factor affecting whether or not an atom will fission is the speed at which the bombarding neutron is moving. If the neutron is highly energetic (and thus moving very quickly), it can cause fission in some elements that a slower neutron would not. For example, thorium 232 requires a very fast neutron to induce fission. However, uranium 235 needs slower neutrons. If a neutron is too fast, it will pass right through a 235U atom without affecting it at all. 

  Nuclear fission is easier to do since the nuclear reactors are considerably smaller and the energy is easier to release than in fusion.  However, the vast amounts of radioactive wastes generated create significant environmental challenges.  Also, nuclear reactors can explode expelling their nuclear contents into the atmosphere.  Such a disaster took place on April 26, 1986, in the former Soviet Union at the Chernobyl Nuclear Facility.


    At 1:23 am technicians at the Chernobyl Power Plant allowed the power in the fourth reactor to fall to low levels as part of a controlled experiment which went wrong.  The reactor overheated causing a meltdown of the core. Two explosions blew the top off the reactor building releasing clouds of deadly radioactive material in the atmosphere for over 10 days. The people of Chernobyl were exposed to radioactivity 100 times greater than the Hiroshima bomb. The people of the world and Northern Europe were greeted with clouds of radioactive material being blown northward through the sky. Seventy percent of the radiation is estimated to have fallen on Belarus and 10 years later babies are sill being born with no arms, no eyes, or only stumps for limbs. It is estimated that over 15 million people have been victimized by the disaster in some way and that it will cost over 60 Billion dollars to make these people healthy. More than 600,000 people were involved with the cleanup many who are now dead or sick. It is now estimated that everyone in the world has been exposed to some amounts of Chernobyl radiation.

Chernobyl Radiation

    Today, the fourth reactor is wrapped in thick layers of cement to prevent any further contamination. The other reactors are slated to reopen for energy production.  The cement is bad shape and there are worries of radiation leaks and future explosions at the site.

    So where does the energy from nuclear changes come from.  Recall the pions which hold the protons and neutrons together.  They must possess incredible amounts of energy to prevent the protons from repelling each other over sub-atomic distances.  This is believed to be the source of the energy.  During a nuclear reaction energy is required to break the pions and energy is released due to the formation of new nuclear elements.  If the output exceeds the input, nuclear energy can be harvested. 

    It seems that no mass is converted into energy. However, this is not entirely correct. The mass of an atom is more than the sum of the individual masses of its protons and neutrons, which is what those numbers represent. Extra mass is a result of the binding energy that holds the protons and neutrons of the nucleus together. Thus, when the uranium atom is split, some of the energy that held it together is released as radiation in the form of heat. Because energy and mass are one and the same, the energy released is also mass released. Therefore, the total mass does decrease a tiny bit during the reaction. 

    A resting nucleus always weighs less than the protons and neutrons that made it up. This lost mass is the nuclear binding energy: the energy that holds the nucleus together.  The binding energy can be computed from the mass-energy relationship E = mc2. The amount of missing mass is known as the mass defect.

    To compute the nuclear binding energy, simply total up the masses of the protons and neutrons in a nucleus and compare it to the mass of the nucleus. 

Proton mass: 1.007271605 amu 
Neutron mass: 1.008665 amu 

    Example: What is the binding energy for 1 mole of the very stable 56Fe nucleus? 

Solution: An iron-56 nucleus has 26 protons and (56-26) = 30 neutrons. The total mass is thus (26)(1.007271605) + (30)(1.008665) = 56.44901173 amu.  Looking up the mass of the iron nucleus, we find that it is 55.92066 amu. This means that the mass defect is 56.44901173 - 55.92066 = 0.52835173 amu. Note that we switched m from massafter- massbefore to massbefore-massafter.  This will cause the energy to be a positive value which makes sense since the binding energy is a positive value since it is the amount of energy holding that nucleus together and not the energy released due to the break down of the atom.  To convert to energy, use 

E = mc2
E = (0.52835173)(3 x 108)2
E = 4.76 x 1010 kJ/mole (a huge amount of energy).

    A nuclear equation shows how a nucleus gains or loses subatomic particles. For example:

11H + 94Be 63Li + 42He

    Let's take a look at this part: 94Be . The 9 represents the mass number, or the number of nucleons present (neutrons plus protons) and the number 4 represents the atomic number or number of protons.  In a nuclear equation (assuming no mass is lost to energy), the sum of the mass numbers of the reactants equals the sum of the mass numbers of the products. For example:

A Balanced Nuclear Reaction

    Using this rule, you can figure out any missing particles in the equation.  It is easiest to determine the identify of any unknown elements by first determining the appropriate atomic numbers.  In the example of with 94Be the product He could only be identified by first identifying that this element had 2 protons.  It could only be helium.  Consider the following two examples (the elements in blue were unknown).

More Nuclear Equations