Definitions and Purpose of H, S and G

Enthalpy H

What Enthalpy is:

Enthalpy is defined by the function dH = dU + PdV where dU = dQ + dW.  

(H - enthalpy, U - internal energy, P - pressure, V - volume, Q - heat, W - work)

What the purpose of enthalpy is:

Enthalpy takes advantage of the 1st law of thermodynamics to simplify energy exchanges between a system and surroundings.  Because we can often control the surroundings to transfer all energy as heat (such as by using a calorimeter), we can equate the energy gained or lost by the surroundings to the enthalpy gained or lost by the system (ΔHsys = Qsurr).  When energy transfer occurs during a chemical or physical process, the system will often undergo multiple transformations.  This complexity is simplified by using enthalpy which includes expansion/contraction as well as thermal energy gains.  

It is very difficult to assign a single energy value to a system.  It is much easier to describe the system by its changes in energy.  Enthalpy is a mathematical tool to simplify this analysis down into a single step instead of having to concern kinetic (thermal) and potential (phase/expansion/contraction) energy considerations.  

What the sign of ΔH means:

When ΔH is +, this means that the system is gaining energy from the surroundings.  This is referred to as endothermic, but only for the system.  When ΔH is -, this means that the system is losing energy to the surroundings.  For both of these it is important to note that the total energy change is 0 per the 1st law of thermodynamics.  So when a system has ΔH as a - value, this means that the change was exothermic with respect to the system and endothermic with respect to the surroundings.  All changes that involve energy exchange have an endothermic and exothermic component, it depends on what the system is defined as.  For example, students are often confused by the fact that ice melting is endothermic while a chemical reaction in solution that heats up is exothermic.  To a student, both of these show signs of increase in energy yet meltings is assigned endothermic status and the reaction is exothermic.  The reason for this discrepancy is that ice is considered to be the system but the water is considered to be the surroundings to the chemical mixture.  When a system is not defined in the problem, it is nearly always assumed to be the chemical mixture involved in the physical or chemical change occurring.  

How to calculate ΔH:

There are four typical methods of calculating ΔH.  The first is called calorimetry and involves a chemical or physical change happening in a calorimeter.  

1. Calorimetry

The heat gained or lost by the calorimeter is equivalent to the enthalpy change of the process being analyzed per unit mole of the reaction.  To calculate this we use ΔH = Q/mol or             ΔHrxn = Qcal/molrxn.  The Q is calculated using Q = mCΔT or Q = H.C.ΔT or both where C is the specific heat capacity and H.C. is the heat capacity.  In a coffee cup calorimeter, the heat going into the cup itself is calculated using Qcup = H.C.cup*ΔT and the heat going into the water in the coffee cup is calculated using QH2O = mCΔT.  The total heat is the sum of those two values.  This can be divided by the number of moles of chemical used to find the enthalpy change of the process.  

2.  Hess’s Law

Hess’s Law is used when you are not able to determine the enthalpy of a reaction, but do know several other reactions that can be combined into the one you do not know.  The sum of the enthalpies for the known reaction will add up to that of the unknown reaction as long as the known reactions will sum to the exact unknown reaction.  For example, converting diamond into graphite is extremely difficult to carry out and would be very expensive and difficult to accurately measure the enthalpy change of.  

C(diamond) → C(graphite) ΔHrxn = ???

But burning diamond and burning graphite is much easier to accomplish and measure.  

a. C(diamond) + O2(g) → CO2(g) ΔHrxn = -395.4 kJ/mol

b.    C(graphite) +  O2(g) → CO2(g) ΔHrxn = -393.5 kJ/mol

We can produce the unknown reaction by adding reaction a to the reverse of reaction b.  The CO2 and O2 will be present as reactants and products and thus cancel when the two reactions are added.  By reversing equation b, we do change the sign from - to +.  

1. C(diamond) + O2(g) → CO2(g) ΔHrxn = -395.4 kJ/mol

+    b.  CO2(g) → C(graphite) +  O2(g)        + ΔHrxn = +393.5 kJ/mol

  unk.  C(diamond) → C(graphite) ΔHrxn = -1.9 kJ/mol

3.  Standard enthalpy of formation ΔHf°

A standard enthalpy of formation value is the enthalpy change that accompanies the formation of an element or compound from its elements under standard conditions.  These can be determined experimentally using calorimetry and this has been done for many compounds and can be found in appendices of textbooks.  The standard enthalpy of formation for CH4(g) is the enthalpy change that accompanies the following transformation:

C(s, graphite) + 2H2(g) → CH4(g) ΔHf°=-74.9 kJ/mol

The standard enthalpy of formation for any element is 0 as long as the element is in its standard state under normal conditions.  For example, bromine is typically Br2 and liquid under standard conditions.  The enthalpy of formation of liquid bromine is 0 but gaseous bromine is not.  

To calculate the ΔHrxn using ΔHf° values you would add the total of all of the products ΔHf° values, add the total of all of the reactant ΔHf° values and subtract products - reactants.  

Σproducts ΔHf° - Σreactants ΔHf° = ΔH°rxn

4.  Bond enthalpies

Bond enthalpies or bond energies are defined as the energy required to break 1 mol of a particular bond.  These have to be averaged over multiple molecules because the exact bond energy varies depending on the polarization caused by neighboring atoms.  These values can also be found in most appendices of chemistry textbooks.

To calculate ΔH°rxn using bond enthalpies just add up all of the bond enthalpies for every bond broken and every bond formed.  Bond breaking is endothermic and so these values are positive.  Bond forming is exothermic so these values are negative.  If the bonds broken are stronger (larger energy) than the bonds formed, the overall reaction is endothermic.  If the bonds formed are stronger than the bonds broken, the overall reaction is exothermic.  

Entropy S

What Entropy is:

Entropy is defined by the functions dS = dQ/T and S = k ln W.  W is a quantitative description of how many microstates there are and k is the Boltzmann constant.  

What the purpose of entropy is:

It is important to show the definition of entropy, either from the historical definition proposed by Rudolph Clausius and applied be Willard Gibbs, or through the statistical analysis of Ludwig Boltzmann.  It should be stressed that an understanding of what entropy is lies beyond that of a high school student.  You can state that entropy correlates with disorder, but it should not be defined as this and if it is, it should be articulated clearly that this is a placeholder until higher levels of math, statistics and physics are available.  

The purpose of entropy is that the rules of physics make it so that entropy can be used to make predictions and also explain whether or not something will happen based upon all energy considerations.  A precise definition of entropy is not needed in order to use entropy to predict whether or not a process will occur without work being applied.  

What the sign of ΔS means:

If ΔS = +, this means that entropy has increased and this can be stated as the energy in a system has increased its dispersal.  This generally happens via two pathways.  Either more energy was added to the system or the system has expanded in volume so that the energy in that system has a wider distribution in space.  ΔS being + is associated with the following (with a few exceptions):

• increasing temperature
• Changing from solid to liquid, liquid to gas or solid to gas (melting, boiling, sublimating, vaporizing)
• dissolving
• expansion of a gas
• chemical reactions where the # of gas molecules increases

ΔS being - is associated with the following (with a few exceptions):

• decreasing temperature
• Changing from gas to liquid, liquid to solid, gas to solid (condensing, freezing, deposition)
• crystallization
• contraction of a gas
• chemical reactions where the # of gas molecules decreases

Chemicals also have entropy values calculated for them.  In order to do this, it is assumed that the entropy of the chemical is 0 at absolute zero (0 K) and the specific heat capacity of the substance is used to determine the entropy under different conditions.  Larger molecules and gases tend to have higher entropies.  

How to calculate ΔS:

Standard entropy values have previously been determined and can be found in the appendices of textbooks.  This will be the only method used to calculate entropy and often we will use signs to make a qualitative assignment of entropy changes.

Σproducts ΔS° - Σreactants ΔS° = ΔS°rxn

Gibbs Free Energy G

What Gibbs free energy is:

dG = dU + PdV - TdS        or       ΔG = ΔH - TΔS

When Clausius introduced the entropy term, many did not understand what it was.  It was Gibbs that developed the physical interpretations of this term and combined it into the first equation shown above.  Later Walther Nernst combined the internal energy and pressure-volume work functions into a single enthalpy term which gives us the equation most common to introductory chemistry today.  

What the purpose of Gibbs free energy is:

The purpose of Gibbs free energy depends on whether you are talking about ΔG or ΔG°.  ΔG° is used to compare the energy stability of a chemical reaction mixture at standard conditions.  This can be used to make determinations about a reaction or physical change but is limited to standard conditions and while at equilibrium.  ΔG° can tell you information about the equilibrium constant of the reaction and the electromotive force (EMF) for a redox reaction.  ΔG on the other hand is not limited to being at equilibrium.  ΔG gives you information about a specific mixture of chemicals and how they will change based on their amounts as well as all of the other information about them.  ΔG° will tell you how the amounts of products and reactants compare at equilibrium and ΔG will tell you how a specific mixture under specific conditions will proceed to achieve equilibrium.  

ΔG° can be related to the equilibrium constant using ΔG° = -RT ln K.  ΔG° can be related to E°cell by the equation ΔG° = -nFE°cell.  ΔG can be used with the equation ΔG = ΔG°+RT ln Q.  This allows us to use the energy comparisons of products/reactants (ΔG°) along with the reaction conditions (RT ln Q) to determine the maximum amount of work (free energy) capable of being produced as the reaction proceeds to equilibrium.  

What the sign of ΔG means:

If ΔG° = -, the reaction under standard conditions is spontaneous and will achieve an equilibrium where K is greater than 1.  The voltage under standard conditions (including activities) will be positive.  

If ΔG° = +, the reaction under standard conditions is not spontaneous and will achieve an equilibrium where K is smaller than 1.  The voltage under standard conditions (including activities) is negative.  

If ΔG = -, the particular reaction mixture will form more products to achieve equilibrium.  This will occur spontaneously.

If ΔG = +, the particular reaction mixture will form more reactants to achieve equilibrium.  In order to form more products, work must be done on the system.

If ΔG° = 0, the reaction under standard conditions will have an equilibrium constant of 1.  The voltage under standard conditions is 0.  If ΔG = 0, the system is at equilibrium and ΔG° can be used to determine what the composition of the reaction mixture is.  

How to calculate ΔG:

Calculate ΔG° with the equation ΔG° = ΔH° - TΔS°.  Calculate ΔG with the equation ΔG = ΔH - TΔS.  It is important to note that the units of ΔG and ΔH are often presented in kJ/mol while ΔS are in units of J/(mol*K).  A conversion is usually needed.  ΔGf° values can also be used in the same manner as ΔHf° using the equation:

Σproducts ΔGf° - Σreactants ΔGf° = ΔG°rxn

The equation ΔG = ΔG° + RT ln Q can be expanded to include redox chemistry calculations as well.  Because ΔG = ΔG° + RT ln Q and ΔG° = -nFE°cell, we can combine these two equations to solve for the voltage supplied by a reaction not under standard conditions to be

Ecell = E°cell  -  (RT/nF) ln Q or Ecell = E°cell  -  (0.0257/n) ln Q or Ecell = E°cell  -  (0.0591/n) log Q.  These equations are referred to as the Nernst equation and allow us to simplify a very complicated analysis of ions and electrons into a much easier calculation.  

Calculations with ΔG are often done qualitatively using only the signs of ΔS and ΔH.  Because a negative value of ΔG is often desirable, we can see that an exothermic reaction       (ΔH = -) will cause ΔG to decrease.  ΔS on the other hand will cause ΔG to decrease when it is positive because it is subtracted and temperature in Kelvins is always positive.  

ΔH	ΔS	ΔG	spontaneous?
-	+	-	always
+	-	+	never
-	-	depends on temperature	when |ΔH| > |TΔS|
+	+	depends on temperature	When |TΔS |> |ΔH|

Final Note

The definitions of enthalpy, entropy and Gibb’s free energy are highly technical and not of significant value for a student beginning in chemistry.  Their purposes as predictive tools are much more relevant and these should be the focus for new students.  Many educators struggle to define these terms adequately or consistently but still strive for this to be a focus.  Momentum does not have good definition because it is a mathematical construct that is useful for making predictions about collisions.  Knowing the equation in physical terms is meaningless in nearly all introductory situations in physics.  But knowing that you can use this “momentum” tool to determine motion before or after a collision is a powerful tool and this should be the focus.  The predictive measures of ΔH, ΔS and ΔG should be the focus of introductory physical chemistry.  A wonderful summary of the history of these developments can be found in the thermodynamics section of Great Physicists by William H. Cropper.

Demonstration Comparing Endpoint and Equivalence Point

 Oops, I Used the Wrong Indicator

This demonstration shows four acid-base titrations using two different indicators.  Two titrations are performed using 0.1 M HC2H3O2 and 0.1 M NaOH.  The first titration uses methyl red that goes from pink to yellow near a pH of 5.  The second titration uses phenolphthalein that changes from colorless to pink near a pH of 9.  The demonstration can be viewed below:

Background

To effectively highlight the distinction between endpoint and equivalence point we must start with some background.  Equivalence point is defined as the point in a titration where stoichiometrical equivalent moles of acid and bases have been added.  For these reactions where acid and base react in a 1:1 fashion, this is where equal moles of acid and base have been added.  At the equivalence point the products of the neutralization establish an equilibrium with the water present.  If a strong acid and strong base are mixed, the products do not effect the equilibrium significantly and the solution will be neutral with a pH near seven.  But when weak acids or weak bases are used, the conjugates of these will react with the water and establish an equilibrium with unequal amounts of acid and base.  The pH of these mixtures will typically not be seven.  If a weak acid is neutralized by a strong base, the conjugate base of the weak acid will react with water to increase the hydroxide concentration and lower the hydronium concentration.  

An acid-base indicator is a weak acid/weak base that has different colors when it is in the acidic or basic form.  An indicator is often represented by the formula HIn where HIn is the acidic form and In- is the conjugate base of the indicator.  When the indicator is put into an acidic solution, it will be of the form HIn and in a basic solution it will be mostly in the In- form.  The indicator has a pKa value that gives us guidance into when this color change will occur.  When the pH is equal to the pKa of the indicator, both forms will be present and the color will be a mixture of the two.  When pH is 1 unit lower than the pKa value, the acidic form will be present in a ratio of 10:1.  When pH is 1 unit higher than the pKa value, the basic form will be present in a ratio of 10:1.  So we expect the indicator to change color over a range of 2 pH units.  In this experiment, methyl red (pKa = 5) is used which will change from pink to yellow over a pH range of 4-6.  Phenolphthalein (pKa = 9) is the second indicator used which changes from colorless to pink over the pH range of 8-10.  These color changes are called the endpoint of the titration.  If you were without a pH sensor, you would rely on these to determine when equivalent amounts of acid and base are present.  

Analysis

Figure 1:  Titration of acetic acid with sodium hydroxide

In Figure 1, the methyl red is in the pink form as the pH is below the pKa.  As the pink color fades and the yellow color increases we will reach our endpoint of the titration.

Figure 2:  Titration of acetic acid with sodium hydroxide

In Figure 2, we now see the pink color changing to a mixture of yellow and pink.  This is our endpoint of the titration.  The arrow marks on the LabQuest 2 unit where this moment is.  Note that this is not occuring during an inflection point where the slope rises rapidly.  This endpoint is poor because it is not happening at the same time or near the same time as the equivalence point.  If he had not had a pH sensor available we would have received an incorrect volume required to complete the reaction.  

Figure 3:  Titration of acetic acid with sodium hydroxide

In Figure 3 we continue to see the color change from pink to yellow.  We have still not seen the inflection point indicating that the acid and base are present in equivalent amounts.  

Figure 4:  Titration of acetic acid with sodium hydroxide

In Figure 4 above we now finally can see where the equivalence point is.  Note that the pH at equivalence is above 7, as we would expect for a weak acid being titrated with a strong base.  At the point of neutralization the acetate ion establishes an equilibrium where hydroxide is present at larger concentration than hydronium.  We can also now compare our equivalence point (green arrows) and endpoint (red arrows).  It is difficult to pinpoint exactly where the endpoint is because the color change is not sharp as the color change occurs during a buffering region.  We have given a rather generous endpoint that the reader can choose to examine with a more critical eye if desired.  Many will note that the two are not at the same point on the curve and that this is going to cause error.  To determine how much error, we must then determine the volume of base added at the time of the endpoint where the color change occurred and the volume of base added at the time of equivalence.  The smaller arrows at the bottom of the display can be used for this.  The small red arrow in Figure 4 should be compared to the small green arrow in Figure 4.  The gap between these should then be compared to the total time of the titration to that point to determine the approximate error assuming time can be correlated with volume of base added.  In these figures a buret is constantly emptying into the flask, so as time progresses the volume of base added is increasing.  It is the volume that we use to find out answer.  We are typically doing a titration to use a known concentration to react with an unknown concentration so that we can determine the unknown concentration.  This is done using the volume and concentration of the known to calculate the unknown concentration.  In the above example, using the wrong indicator appears to be giving us about 20% error in volume before any other errors are considered.  This is why choosing an appropriate indicator is critical for a titration.  

Now we shall run through the same titration done with phenolphthalein.  Phenolphthalein is colorless in the protonated form and pink in the deprotonated form.  The transition from colorless to pink occurs in the pH range of 8-10.  

Figure 5:  Titration of acetic acid with sodium hydroxide

In Figure 5 above the solution is colorless.  The splash of pink is from the base not being evenly distributed as anyone who has run a titration before is familiar with.  This buret emptied faster than the previous one, but this point for the previous one would have been near the endpoint.   

Figure 6:  Titration of acetic acid with sodium hydroxide

In Figure 6 we are now at the endpoint of the titration.  The pH is rising rapidly as well which would indicate that we are probably near the equivalence point as well.

Figure 7:  Titration of acetic acid with sodium hydroxide

In Figure 7 our titration is now complete and excess base is in the system. The pink color is not vibrant and we have surpassed the equivalence point. The green arrow is meant to indicate the location of the equivalence point and the red arrow the endpoint. However, both of those are difficult to differentiate because of the rapid flow of sodium hydroxide into the beaker. They may be even closer than indicated above. Nonetheless the error from endpoint and equivalence point is nearly erase. The pH of a 0.05 M solution of sodium acetate (what is present at equivalence) projects to be a pH of 8.7 and the half-equivalence point should have a pH of 4.7