Course Outline

General Chemistry--Unit 2

04/25/09

�Before You Begin:

To be successful at this material you need to have a basic understanding of heat and energy and be able to perform calculations involving them. You can find an introduction to heat and energy in the first part of the thermodynamics outline.

The Zeroth Law of Thermodynamics

The zeroth law of thermodynamics states that, if system A is in thermodynamic equilibrium with system B, and if system B is in thermodynamic equilibrium with system C, system A is also in thermodynamic equilibrium with system C. It may seem obvious because the form of the zeroth law is similar to the transitive property of equality from elementary algebra. However, the zeroth law was formalized long after the other three laws of thermodynamics. It was given the name “zeroth” because it is more fundamental, yet numbers one through three were already taken. This law gives thermometers meaning. If you place an oral thermometer under your tongue and leave it long enough to reach thermal equilibrium, we can assume that your tongue and the thermometer are in thermal equilibrium with a third system, an imaginary and ideal system on which the Celsius temperature scale is based.

The First Law of Thermodynamics

There are many ways to state the first law, but the easiest to remember is probably: during any chemical change energy is neither created nor destroyed. Energy may be transferred or changed in form, however. Another way of stating the first law of thermodynamics is the internal energy lost or gained by a system is equal to the sum of the heat absorbed or given off and the work done on or by the system:

ΔE = q + w

If we use a chemical reaction to accomplish work, it is at the expense of heat, and vice versa.

Enthalpy, H

Enthalpy, H, is a thermodynamic property equal to the internal energy added to the product of the pressure and volume of the system:

H = E + PV

The enthalpy is a state function (this means that the value of the enthalpy change is independent of path so it doesn’t matter how we accomplish the change). It is pretty sloppy thinking, but enthalpy can be thought of as the portion of the internal energy that is most like heat, q, while the product of the pressure and volume is the part of the internal energy most like work, w.

Like the internal energy, we cannot measure the enthalpy of a system; instead we measure the enthalpy change that accompanies a chemical or physical process:

ΔH = H_final – H_initial

The enthalpy change, ΔH, is the heat absorbed or given off during a process that occurs at constant pressure:

ΔH = q_p (subscript refers to constant pressure)

We can use calorimetry to measure q for reactions. If the calorimeter is at constant pressure (like a “coffee cup” calorimeter), the value measured for q is equal to the enthalpy change.

Endothermic Vs Exothermic Reactions

Chemical and physical processes can give off or absorb heat from the surroundings. When a system gives off heat, the change is said to be exothermic. When a system absorbs heat, the change is said to be endothermic. An exothermic process has an enthalpy change less than zero (ΔH has a negative sign). An endothermic process has an enthalpy change greater than zero (ΔH has a positive sign). For an exothermic process, the temperature of the surroundings increases (the thermometer goers up). For an endothermic process, the temperature of the surroundings decrease (the thermometer goes down).

If a chemical reaction is exothermic, the reverse reaction is endothermic. In fact, for any reaction, the reverse reaction has the same enthalpy change but with the opposite sign.

Similar terms can be used to describe internal energy changes. Exogonic processes are chemical or physical changes which give off energy to the surroundings. Endogonic processes absorb energy.

Hess’s Law

Remember that enthalpy is a state function and that state functions are independent of path. Hess’s Law states that, because enthalpy does not depend on path, the enthalpy change for a single step reaction is the same as the enthalpy change for a multi-step reaction, even if those steps are only theoretical, as long as the initial and final states are the same. This means that we can use the enthalpy changes of certain reactions with known values to deduce the values for enthalpy changes of reactions we don’t know.

Born-Haber Cycle

A Born-Haber Cycle is a graphical method for using Hess’s Law to determine enthalpy changes for reactions that would be difficult to measure directly. A Born-Haber cycle is a chart in which vertical lines represent enthalpy change data. The length of the line represents the magnitude of the enthalpy change. The direction of the line represents the sign of the enthalpy change (up is positive or endothermic). The following figure is a simple but imaginary Born-Haber cycle.

In this example, the initial and final states are the same for the single step and for the multi-step reaction. In the multi-step reaction, intermediates, (D and E) are formed then used as reactants in the next step of the sequence. The single step reaction has an enthalpy change of 160 kJ. The sum of the enthalpy changes of the multi-step reaction is also 160 kJ.

Born-Haber cycles are used to find the lattice energy of ionic compounds. The lattice energy is the enthalpy change that accompanies the process in which gaseous ions combine to form an ionic crystal. This is an interesting piece of information because it represents the inherent stability of the crystal structure. Unfortunately, it is not possible to measure the lattice energy directly. For those of you who are just dying to know how to use Born-Haber cycles to find the lattice energy, your Chem Professor will post this calculation in the FAQ as soon as someone asks!

Bond Enthalpies and Hess’s Law

The bond enthalpy is the amount of enthalpy that must be supplied to a molecule in the gas phase to break a particular bond. For example, 435.990 kJ of heat must be added to the system to break the hydrogen-hydrogen single bond, i.e. to convert a mole of diatomic hydrogen into two moles of hydrogen atoms.

We can use bond enthalpies and Hess’s law to find the enthalpy change for a reaction. A single step reaction is broken into theoretical steps: each bond in the reactant molecules is broken then the atoms are assembled to form the products. The enthalpy change for the bond breaking steps is bond enthalpy. The enthalpy for the bond making steps is the negative of the bond enthalpy, because it is the reverse of a bond breaking reaction. Although the reaction probably does not happen this way, Hess’s law states that the sum of the enthalpy changes for a multi-step reaction are the same as the enthalpy change for a single step even if those steps are only theoretical.

4Concept Check: use bond enthalpies to find the enthalpy change for this reaction:

2HBr_(g) + Cl_2(g) à 2HCl_(g) + Br_2(g)

Bond enthalpies can be found on the internet at the WebElements™ periodic table site. You can access bond enthalpy values from a tool bar at the left of the window on the professional edition pages for a given element. These tables list gas phase diatomic bonds by periodic group.

Most chemistry text books have bond enthalpy tables organized by bond order, i.e. single, double, and triple bonds. To use those tables, you must first draw dots structures to determine the number and kinds of bonds.

Answer: the steps and their enthalpies (Winter, M., 2005, Bromine: bond enthalpies and Chlorine: bond enthalpies) are

Break two moles of H-Br                   ΔH = 2(336.35 kJ/mol)

Break one mole of Cl-Cl                     ΔH = 242.58 kJ/mol

Make two moles of H-Cl                    ΔH = -2(431.62 kJ/mol)

Make one mole of Br-Br                     ΔH = -192.81 kJ/mol

Total                                                    ΔH = -140.77 kJ/mol

!Warning! A word of warning—students usually like these problems. The multi-step reactions are straightforward to deduce and the calculations are pretty easy. Unfortunately, this method is not very accurate. Using bond enthalpies to find enthalpy changes for reactions only works if all of the reactants and products are gases. Another problem is that bond enthalpies are measured for diatomic molecules. If we use bond enthalpy to calculate reactions involving molecules with more than one bond, we have to assume all the bonds are equivalent. The energy needed to break the bond in the high energy N-H molecule is not really the same as the energy needed to break each of the three bonds in ammonia.

Enthalpies of Formation

The formation reaction is defined as the reaction that creates one mole of a compound from its elements in their standard states. For example, let’s look at the formation reaction for liquid water. Water contains the elements hydrogen and oxygen both of which are diatomic gases in their standard (lowest energy) elemental form. The formation reaction is H_2(g) + ½O_2(g) à H₂O_(l) which has an enthalpy change of -285.83 kJ/mol. Values for enthalpies of formation can be found on the National Institute of Standards & Technology, NIST web site under the Chemistry Web Book databases. Another source is the thermodynamic databases at the  Physical Science Information Gateway site, PSIGate. We prefer the NIST site because it has so many more values, however,  the PSIGate site is easier to use.

The entropy change for a reaction varies slightly with temperature. The enthalpy change for two moles of a product is twice the enthalpy change for one mole of a product, so values also depend on quantity. For gases, the pressure is analogous to concentration, so values of entropy also depend on pressure. For clarity, chemists have defined a thermodynamic standard state as being one mole product, temperature at 289 K, and pressure at 1 atm. The symbol for a state function at thermodynamic standard state has a degree mark, ΔH°, for example. Values of enthalpies of formation are usually listed at standard state.

Calculating Reaction Enthalpies Using Hess’s Law and Formation Enthalpies

Using Hess’s law and the Enthalpy of formation is an extremely handy and accurate way to find the enthalpy change of a reaction. Hess’s law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for a multi-step reaction, as long as the initial and final states are the same. To use formation enthalpies to calculate a reaction enthalpy, we break a single step reaction into the following multi-step reaction: break each reactant apart into its elements then use the elements to form the products. The enthalpy changes for the product making reactions are the enthalpies of formation. The enthalpy changes for the reactant breaking reactions are the negatives of the enthalpies of formation, since they are the reverse reactions. The enthalpies are tabulated for one mole of product. To use these values for other quantities, multiply the enthalpy change by the number of moles of product.

4Concept Check: find the standard free energy for this reaction using the tabulated values from PSIgate: CCl_4(l) + H₂O_(l) à COCl_2(g) + 2HCl_(g)

Answer: By Hess’s law, we are breaking this single step reaction into the following multiple steps with these enthalpies (RDN, “Molar thermodynamic properties,” 2002):

CCl_4(l) à C_(s) + 2Cl_2(g)                                     ΔH = -(-135.44kJ/mol)

H₂O_(l) à H_2(g)+ ½O_2(g)                         ΔH = -(-285.83 kJ/mol)

C(s) + ½O_2(g)  + Cl_2(g)à COCl_2(g)                           ΔH = (-218.8 kJ/mol)

H_2(g) + Cl_2(g) à 2HCl_(g)                                    ΔH = 2(-92.307 kJ/mol)

 CCl_4(l) + H₂O_(l) à COCl_2(g) + 2HCl_(g)             ΔH = 17.9 kJ/mol

The first two reactions are the reverse of the formation, so the sign of the formation enthalpy is reversed. The last reaction forms two moles of product, so the formation enthalpy is doubled.

Note that the intermediates that are produces during some of the steps (carbon, hydrogen, oxygen and chlorine) are used as reactants in subsequent steps so that they ‘cancel out.’

4Concept Check: What does this result mean?

Answer: When we combine carbon tetrachloride with water and add about 18 kJ of enthalpy, we make phosgene and hydrogen chloride gases. This reaction is slightly endothermic. But don’t try this at home! All of these are toxic except the water.

Once we recognize the connection between this calculation and Hess’s law, we can reduce the logic to a formula. Algebraically, we get the formula

ΔH° = ΣnΔH°_f(products) – ΣmΔH°_f(reactants)

where ΔH° is the enthalpy change for a reaction under thermodynamic standard conditions, ΣnΔH°_f(products) is the sum of the enthalpies of formation for the products each multiplied by its stoichiometric coefficient, and ΣmΔH°_f(reactants) is the sum of the enthalpies of formation for the reactants each multiplied by its stoichiometric coefficient. In other words, we look up tabulated values for the enthalpies of formations, multiply each by its coefficient from the balanced reaction, add together the products and subtract the reactants.

4Concept Check: Repeat the previous calculation using the formula ΔH° = ΣnΔH°_f(products) – ΣmΔH°_f(reactants)

Answer: Using the tabulated values from PSIGate (RDN, “Molar thermodynamic properties,” 2002) in the previous example, multiplying each compound by its stoichiometric coefficient and subtracting the reactants’ values from the products’ values, the calculation looks like this:

CCl_4(l) + H₂O_(l) à COCl_2(g) + 2HCl_(g)             

ΔH = [1(-218.8 kJ/mol) + 2(-92.307 kJ/mol)] – [1(-135.44 kJ/mol) + 1(-285.830 kJ/mol)]

= 17.9 kJ/mol

The Second Law of Thermodynamics

Entropy, S

Entropy, S, is the degree of randomness or disorganization of the system. Natural processes tend to have increases in entropy. A real world analogy is how disorganized the ChemProfessor desk gets as time passes. Papers, tools, coffee cups, and miscellaneous junk get all mixed together and stacked in random piles. It is possible to make the ChemProfessor desk more organized; we just have to devote work to making that change. However, the desk never cleans itself.

Many chemical and physical changes result in a higher level of randomness in the system. Gases are more random than liquids which are more random than solids. Imagine an ice cube at the bottom of the otherwise empty glass sitting on the ChemProfessor desk. In ice, the water molecules that are held in a regular repeating crystal structure by hydrogen bonds. When an ice cube melts, the molecules move freely throughout the liquid. The situation in which the individual molecules are changing position with respect to each other is much more random than the situation in which the individual water molecules are locked into a three dimensional pattern. As some of the liquid water molecules vaporize, they become even more randomly distributed throughout the room.

Reversible and Irreversible Processes

The concept of entropy was first proposed as a thermodynamic potential such that, during a reversible change in a system that occurs at constant temperature, the entropy change is equal to the heat absorbed or given off by the system divided by the temperature:

ΔS = q_reversible/T

A change is reversible if it can occur in a single step, work is not required for the change, and the reverse of the change can also occur in a single step without work. Physics and chemistry textbooks always describe pistons and gas cylinders with partitions and other complicated arrangements that can or can’t be reversed in a single step. Instead we will use a chemical example, the auto-ionization of water. Imagine a water molecule. One of the covalent bonds between a hydrogen atom and an oxygen atom can break forming a hydrogen positive ion and a hydroxide ion:

H₂O à H⁺¹ + OH^-1

This reaction occurs in a single step without work, though it requires a little heat. The hydrogen and hydroxide ions that are produced can react with each other to form a water molecule. This is also a single step reaction that doesn’t require work, though it gives off a little heat. This pair of chemical reactions exactly counter-acts each other. Both are possible (spontaneous). Neither one requires work. In any beaker of pure water at room temperature at any given instant 1 x 10^-5 % of the water molecules are busily breaking apart and re-assembling. This is a reversible process.

Now imagine you are in the kitchen heating some water to make a cup of instant coffee. As you apply thermal energy from the surroundings, some of the water molecules gain enough kinetic energy to break free from the intermolecular attractions of their neighbors and escape into the air (the water boils). We know that this change can be used to perform useful work—that is how steam engines operate. The water vapor spreads throughout the room. In order to take it back to the initial state of liquid water in a pan on the stove, we would have to collect the air, separate the water from the other gases, and condense it back into a liquid. It is possible to change the system back to its initial state, but we can’t do it in one step without applying work. This change is irreversible. Note that chemical changes are not called ‘irreversible’ because they can never return to their original state. However, re-establishing the original state requires work from the surroundings.

Spontaneity

A reaction is spontaneous if it will occur as devised. A reaction is non-spontaneous if it will not occur without work being added to the system from the surroundings. Spontaneous processes reflect the natural direction of change. We can tell we are watching film in reverse because we are accustomed to changes having a spontaneous direction.  We do not expect to see the ice cream leap back into the cone. Spontaneous processes tend to increase randomness and/or give off heat.

The Second Law of Thermodynamics

The Second Law of Thermodynamics states that it is not possible for a cyclic process to take heat from a reservoir and convert it into work with no other result.  This is the same as saying that perpetual motion machines are impossible. No matter how clever we are, we cannot build a machine that can run itself forever.

Another way of stating the second law: The entropy of an isolated system increases during any natural process. During any spontaneous process, the entropy of the universe increases. If the entropy of the system decreases, the entropy of the surroundings must have an even larger increase. These two statements seem very different. One deals with converting heat to work and the other deals with entropy. What is the connection? The entropy is the part of the thermal energy that cannot be used to perform work. It is off limits (as far as useful work is concerned) because it is the part of the molecular motion that leads to or results from its disorganization.

Standard State Entropy of Reaction

Unlike enthalpy and internal energy, the entropy of a pure substance or simple mixture can be calculated. We can use calorimetry to measure the heat capacity of the system at constant volume and pressure. Then we use calculus to convert the heat capacity to entropy. Rather, ‘we’ don’t actually do the calculus; people with well equipped laboratories, great math skills, and lots of time on their hands have calculated values for entropy and collected these in data bases. We at Chem Professor like the NIST data bases, but they can be a little difficult to navigate. The PSIGate thermodynamic values page is a lot easier to use.

Because entropy is a state function, we can calculate entropy changes using techniques similar to those for calculating the enthalpy changes of a reaction:

ΔS° = ΣnS°_f(products) – ΣmS°_f(reactants)

where ΔS° is the entropy change for a reaction at standard conditions, ΣnS°_f(products) is the sum of the entropies of the products each multiplied by its stoichiometric coefficient, and ΣmS°_f(reactants) is the sum of the entropies of the reactants each multiplied by its stoichiometric coefficient. In other words, we look up tabulated values for the entropies, multiply each by its coefficient from the balanced reaction equation, add together the products and subtract the reactants. In this calculation we are using the idea that entropy is a state function, so ΔS° = S°_final - S°_initial. The change in entropy is the difference between the total entropy of the final state, ΣnS°_f(products), and the total entropy of the initial state, ΣmS°_f(reactants).

There are tabulated values for standard entropies just as there are for standard enthalpies of formation reactions. The thermodynamic standard state is one molar quantity at 1 atm pressure and 298 K temperature. If you scan the PSIGate table of thermodynamic properties, you will see that the first entry is silver solid. The standard molar entropy of silver solid is 42.55 J/mol K (RDN, “Molar thermodynamic properties,” 2002) rather than zero. Formation enthalpies, ΔH_f, for elements in their standard states are zero because the formation reaction of an element from an element does not involve a change. Tabulated values of entropies, on the other hand, are not formation reaction entropy changes. These values represent the amount of entropy, S, of the system rather than entropy change, ΔS, of a theoretical reaction. One mole of silver solid at room temperature has about 43 joules of randomness.

4Concept Check: find the standard entropy change for this reaction: CCl_4(l) + H₂O_(l) à COCl_2(g) + 2HCl_(g)       

Answer: Using the tabulated values from PSIGate (RDN, “Molar thermodynamic properties,” 2002), multiplying each compound by its stoichiometric coefficient and subtracting the reactants’ values from the products’ values, the calculation looks like this:

ΔS° = [1(283.53 J/mol K) + 2(186.908 J/mol K)] – [1(216.40 J/mol K) + 1(69.91 J/mol K)] = 371.04 J/mol K

4Concept Check: Does the sign of the entropy make sense?

Answer: A positive sign for entropy change means that the final state was more random than the initial state. Under standard conditions, this reaction has an increase in entropy, so the system becomes more disordered. The net change is that two moles of liquid (fairly random) is converted into three moles of gas (very random). This is an increase in entropy.

The Third Law

Gibb’s Free Energy, G

Natural processes tend to give off heat (be exothermic) and/or have an increase in entropy. The Gibb’s free energy function was devised to combine these two trends into one handy thermodynamic state function. The Gibb’s free energy, G, is defined as the difference between the enthalpy, H, and the product of the temperature, T, and the entropy, S:

G = H – TS

For a chemical change occurring at constant temperature, the free energy change is 

ΔG = ΔH – TΔS

What is ‘free’ about the free energy? The free energy change is the amount of energy available to do work; it is the maximum amount of work that can be accomplished by a chemical reaction. Of course, work doesn’t have to be done. We have to design a machine that can use the release of chemical energy to move a mass in the surroundings. For a great many situations, no work is done even if great amounts of energy are given off by the system. However, if we want to apply that energy, it is nice to know what maximum is possible.

For reactions that are not spontaneous, the Gibbs free energy is the amount of work that must be applied to the system to force the reaction to occur. Our human experience makes us used to change as being irreversible. Children never turn into their parents. However, chemical and physical processes are often reversible. For every reaction equation that we can write there is a reverse reaction in which the substances we have thought of as products change into what we have thought of as reactants. An analogy for Gibbs free energy is the potential energy of a rock on a hill. The rock has potential energy due to its position. The natural direction of change is for the rock to roll down the hill until it reaches a point of minimum potential energy in the valley. The rock can be pushed back up the hill, but it would take work. It takes a certain minimum amount of work to push the rock back to its starting point, so the rock would never spontaneously appear at the top of the hill on its own. That doesn’t mean we can’t push it up there, though.

The Gibbs free energy has another handy quality. We can find the free energy change rather easily, either by using Hess’s Law or by using values for temperature, enthalpy and entropy change. The Gibbs free energy function combines the two thermodynamic potentials that are involved in the nature of spontaneity. The sign of the Gibbs free energy change for a reaction indicates whether or not a reaction is spontaneous.

• If ΔG is negative, the forward reaction is spontaneous.
• If ΔG is zero, the forward and reverse reactions are both spontaneous. The system is at chemical equilibrium.
• If ΔG is positive, the reverse reaction is spontaneous.

The Third Law of Thermodynamics

The third law of thermodynamics states that, in an ideal crystalline substance, entropy approaches zero as the temperature approaches absolute zero. This implies that, if we could cool something to absolute zero, the atoms would become perfectly ordered. The converse is that, since nothing can be perfectly ordered, we can never reach absolute zero. Nothing can be perfectly ordered because, even if the atoms and molecules ceased moving in the bulk and froze into perfect alignment, any real sample of matter will have impurities and interfaces. Where these different atoms meet, their interactions can’t be the same as the interactions between pure substances and are bound to introduce randomness. For substances that are not ideal crystalline solids, the entropy approaches some constant as the temperature approaches absolute zero. For both (ideal and non-ideal) as the temperature gets close to zero, molecular motion slows and chemical and physical processes stop.

Hess’s Law and Free Energy

Because the Gibbs free energy was designed to be a state function, we can calculate free energy changes using techniques like those for calculating the enthalpy changes of a reaction:

ΔG° = ΣnΔG°_f(products) – ΣmΔG°_f(reactants)

where ΔG° is the Gibbs free energy change for a reaction at standard conditions, ΣnΔG°_f(products) is the sum of the Gibbs free energies of formation for the products each multiplied by its stoichiometric coefficient, and ΣmΔG°_f(reactants) is the sum of the Gibbs free energies of formation for the reactants each multiplied by its stoichiometric coefficient. In other words, we look up tabulated values for the free energies of formations, multiply each by its coefficient from the balanced reaction, add together the products and subtract the reactants. This is an application of Hess’s Law. We are breaking the single step reaction into a series of theoretical steps for the purpose of calculating the free energy change. The theoretical reactions are: take every reactant and break it apart into its elements then take all the elements and form the products from them. The first steps have free energy changes equal to the negative of the free energies for the formation reactions because these are the reverse chemical reactions. The last steps have free energy changes equal to the free energies for the formation reactions because they are the formation reactions.

There are tabulated values for standard free energies of formation reactions just as there are for standard enthalpies of formations. The thermodynamic standard state is one molar quantity at 1 atm pressure and 298 K temperature.

4Concept Check:  find the standard free energy for this reaction using the tabulated values from PSIgate: CCl_4(l) + H₂O_(l) à COCl_2(g) + 2HCl_(g)   

Answer: Using the tabulated values from PSIGate (RDN, “Molar thermodynamic properties,” 2002), multiplying each compound by its stoichiometric coefficient and subtracting the reactants’ values from the products’ values, the calculation looks like this:

ΔG° = [1(-204.6 kJ/mol) + 2(-95.299 kJ/mol)] – [1(-65.27 kJ/mol) + 1(-237.129 kJ/mol)]

ΔG° = -92.8 kJ/mol

4Concept Check: What additional information does this answer tell us?

Answer: At standard temperature and pressure, this reaction is spontaneous and can accomplish work. The maximum amount of work that one mole each of carbon tetrachloride and water can accomplish is 92.8 kJ (the negative sign means that the work is done by the system on the surroundings).

Free Energy and Temperature

The Gibbs free energy change for a reaction changes dramatically with temperature, since the function contains temperature: ΔG = ΔH – TΔS. The enthalpy and entropy, on the other hand, don’t change very much with temperature. We can find approximate Gibbs free energy changes for other temperatures if we use standard temperature values for enthalpy and entropy changes and assume that they are close to the values at the temperature of interest.

4Concept Check: what is the Gibbs free energy change for the reaction of carbon tetrachloride and water at 10 K? Assume that the values for the standard enthalpy and entropy changes are similar to those at 10 K. Remember that we found these values in sample problems above. Do not forget to convert the entropy units!

Answer: ΔG = ΔH – TΔS = 17.9 kJ/mol - 10 K(371.04 J/mol K)(1 kJ/1000J) = 14.2 kJ/mol

Note that the reaction is not spontaneous at this temperature.

Natural processes tend to be exothermic and/or increasing entropy. Chemical changes that fit both of these descriptions are spontaneous at all temperature. Chemical changes that have neither of these characteristics are non-spontaneous, no matter what the temperature is. For reactions that are either exothermic or have increasing entropy, the temperature determines whether or not the reaction is spontaneous. Recall that spontaneous reactions have a negative value for the Gibb’s free energy change. The free energy state function has two terms, ΔH and –TΔS.  If the reaction is exothermic, the sign for ΔH is negative. If the reaction also has decreasing entropy (negative) then the –TΔS term is positive. As long as the magnitude of the enthalpy change is greater than the magnitude of the entropy term, the reaction will be spontaneous. At a high enough temperature, the entropy term will be greater in magnitude than the enthalpy term, and the reaction will be non-spontaneous.

If a chemical reaction is endothermic, the sign for ΔH is positive. If the reaction also has increasing entropy (positive) then the –TΔS term is negative. As long as the magnitude of the entropy term is greater than the magnitude of the enthalpy term, the reaction is spontaneous. At a low enough temperature, the enthalpy term will be greater in magnitude than the entropy term, and the reaction will be non-spontaneous.

4Concept Check: at what temperature does the reaction of carbon tetrachloride and water become non-spontaneous?

Answer: The reaction of carbon tetrachloride and water is spontaneous at standard temperature and pressure BUT NOT AT SLL TEMPERATURES. The reaction is endothermic and increasing entropy, therefore, if the temperature is too low, the reaction will not be spontaneous. To find the temperature at which the reaction becomes non-spontaneous, we can set the Gibbs free energy to zero and solve the formula for temperature:

ΔG = ΔH – TΔS so, if ΔG = 0, T = ΔH/ΔS

T = 17.86 kJ/mol ÷ (371.04 J/mol K)(1kJ/1000J) = roughly 48 K (the answer has four significant figures by significant figure rules but the enthalpy and entropy values vary somewhat with temperature; therefore, the answer is approximate)

4Concept Check: what is the standard molar enthalpy change, the standard molar entropy change, and the Standard molar Gibb’s free energy change for the following reaction? Is it spontaneous? If not, at what temperature does it become spontaneous?

Al₂O_3(s) + H₂O_(l) + 2H¹⁺_(aq) à 2Al³⁺_(aq) + 4OH^1-_(aq)

If you use values from the PSIGate site, aqueous substances are on the “Molar Thermodynamic Properties of Aqueous Solutes” page.

Answer: ΔH = -20.4 kJ/mol; ΔS = -807 J/mol; and ΔG = 220.4 kJ/mol. The reaction is not spontaneous under standard conditions (ΔG > 0). Since it is exothermic and decreasing in entropy, the reaction will become spontaneous at low temperatures. Substituting zero for the free energy change and solving for T gives a temperature of 25K, below which the reaction becomes spontaneous.

The Laws of Thermodynamics, Reprise

The three laws of thermodynamics can be restated as:

You can’t get something for nothing.

The best you can do is break even.

You can only break even when heck freezes over.

1. You can’t get something for nothing. During a chemical change energy is neither created nor destroyed, rather it is transferred into or out of a system as heat and work. Since we can’t create energy, we can’t do work without expending energy.
2. The best you can do is break even. We cannot convert heat to work with perfect efficiency. During any process some of the heat will go toward the entropy increase of the universe.
3. You can only break even when heck freezes over. As the temperature gets lower, the entropy ‘cost’ drops. You could ‘break even’ if you could get the system cold enough. Unfortunately, since all processes stop at absolute zero, you can never get a real system cold enough.

References

Resource Discovery Network. (2002). PSIgate: Physical Sciences Information Gateway. “Molar Thermodynamic Properties of Pure Substances.” Retrieved August 3, 2005, from http://www.psigate.ac.uk/newsite/reference/chemdata/3.html.

Winter, M. (2005). WebElements™ “Bromine—Bond enthalpies in gaseous diatomic species”. Retrieved June 25, 2005, from http://www.webelements.com/webelements/elements/text/Br/enth.html. The author cites Kerr, J. A. CRC Handbook of Chemistry and Physics 1999-2000 : A Ready-Reference Book of Chemical and Physical Data. D.R. Lide, (ed.), CRC Press, Boca Raton, Florida, USA, 81st edition, 2000.

Winter, M. (2005). WebElements™ “Chlorine—Bond enthalpies in gaseous diatomic species”. Retrieved June 25, 2005, from http://www.webelements.com/webelements/elements/text/Cl/enth.html. The author cites Kerr, J. A. CRC Handbook of Chemistry and Physics 1999-2000 : A Ready-Reference Book of Chemical and Physical Data. D.R. Lide, (ed.), CRC Press, Boca Raton, Florida, USA, 81st edition, 2000.

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