thermodynamics1 chemical thermodynamics (chapter 19)

Thermodynamics 1

Chemical ThermodynamicsChemical Thermodynamics

(Chapter 19)(Chapter 19)

Thermodynamics 2

General ReviewGeneral Review• Thermodynamics is the study of energy transformations• We often study transformations that involve the exchange

of energy between a system and its surroundings– In chemical thermodynamics the system is generally defined as the

chemicals under study– The surroundings are the rest of the universe– We ordinarily try to set boundaries to study a limited portion of the

the universe, such as the contents of a calorimeter

• One key quantity that is frequently studied is the Internal Energy (E) of a system

• The Internal Energy of a system includes all forms of energy, and is much too complex to measure directly

• We can, however, study the transfer of such energy between a system and its surroundings

Thermodynamics 3

General ReviewGeneral Review• We can say:

Ef - Ei E = q + wWhere q and w are the heat and work exchanged with the surroundings, respectively

• q can be measured experimentally, generally by studying the heat transferred to / from the surroundings

• In chemical systems, w is generally observed as pressure-volume work (PV) as the system expands to do work on its surroundings, or contracts as the surroundings do work on the system

• Not that, though we don’t know the actual values of Ef and Ei, it is still possible to experimentally determine the exchange between the system and its surroundings

Thermodynamics 4

General ReviewGeneral Review

• To simplify our study of energy transformations, we generally study quantities called state functions

• State functions are defined as thermodynamic quantities whose values do not depend on their history

• Examples of state functions include temperature (T), potential energy (PE), and one studied previously, change in enthalpy (H)

• State functions, as opposed to process functions, like work, are given capital-letter symbols

Thermodynamics 5


• Normally q, the heat transfer, is not a state function. However, when q is measured under constant pressure, designated a qp, it becomes a state function:

qp = H

• Thus enthalpy is defined as the heat change in a system measured under constant pressure

• The constant pressure requirement is not particularly stringent since most reactions are studied in open containers under relatively constant atmospheric pressure

Thermodynamics 6


• Hess’s Law, which states the change in enthalpy observed for two states in a system is independent of the pathway taken during the change

• This law based on the First Law of Thermodynamics and on the fact that enthalpy changes are state functions.

• The First Law of Thermodynamics can be stated in many ways, but it is an expression of the principle of conservation of energy - the total amount of energy in the universe is constant - energy can neither be created nor destroyed.

Thermodynamics 7


• Some other considerations when using Hess’s Law:– Enthalpic quantities are extensive quantities, meaning

that the amount of heat exchanged depends on the amount of material undergoing change

– The physical state of substances in a reaction, gas, liquid, or solid, affects the enthalpy calculation

– When a chemical equation is written backwards, the sign of the enthalpy change for the reaction is reversed

• The methods used in solving enthalpy problems using Hess’s law will, with a few modifications, also be used for other quantities that are also state functions.

Thermodynamics 8

General ReviewGeneral Review• Lets review the application of Hess’s Law by determining the

enthalpy change for the combustion of propane, as shown below:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

• To determine the enthalpy of combustion, we will need to use enthalpy of formation values from tables in the back of the text.

• It is important to remember that the values in the tables represent the enthalpies of formation for a substance from its elements, under standard conditions (1 M concentrations, 1 atm pressure and 25o C)

Thermodynamics 9

General ReviewGeneral Review• We remember that elements, in their normal

elemental state, have no formation values• For other substances in the equation, we write

formation equations, including their enthalpy values from the table:

3 C(s) + 4 H2(g) C3H8(g) H = -103.85 kJ/mol

C(s) + O2(g) CO2(g)H = -393.5 kJ/mol

H2(g) + O2(g) H2O(g) H = -241.82 kJ/mol

• These equations, and their accompanying enthalpies, can now be adjusted and combined to give the target equation, and its enthalpy

€

12

Thermodynamics 10

General ReviewGeneral Review• We will make the following adjustments to the

equations and enthalpies– The first equation will be written backward because C3H8 is

a reactant, not a product. The sign of the enthalpy will also be reversed.

– The second equation, and its accompanying enthalpy, will be tripled because our target equation contains three CO2

– The third equation, and its accompanying enthalpy, will be quadrupled because our target equation contains four H2O

• Now, if the contributing equations can be combined to give the target equation, the combination of their adjusted enthalpies will represent the target equation

Thermodynamics 11

General ReviewGeneral ReviewC3H8(g) 3 C(s) + 4 H2(g) H = -103.85 kJ/mol

3[C(s) + O2(g) CO2(g)] H = 3(-393.5 kJ/mol)

4[H2(g) + O2(g) H2O(g)] H = 4(-241.82 kJ/mol)

or

C3H8(g) 3 C(s) + 4 H2(g) H = 103.85 kJ/mol

3C(s) + 3 O2(g) 3 CO2(g) H = -1180.5 kJ/mol)

4H2(g) + 2 O2(g) 4H2O(g)H = -967.28 kJ/mol)

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12

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) H = -2043.93 kJ/mol

Since the adjusted contributing equations can be combined to give the correct target equation, Hess’s Law states that the combination of their adjusted enthalpies will give the correct overall enthalpy

Thermodynamics 12

General ReviewGeneral Review• Having demonstrated, by combining equations, how Hess’s

Law works, be aware that a simpler method, defined below, is usually employed by chemists

• That is, taking into consideration the coefficients in the balanced equation, the summation of the enthalpies of formation of the products minus the summation of the enthalpies of formation of the reactants can be combined to give the enthalpy of the reaction

• Verify this shortcut approach with the problem just done

€

H reaction = ΔH fo∑ reactants - ΔH f

o∑ products

Thermodynamics 13

Beyond EnthalpyBeyond Enthalpy

• Thermodynamics is often concerned with the question: can a reaction occur?

• First Law of Thermodynamics: energy is conserved.• Any process that occurs without outside intervention

is spontaneous.• When two eggs are dropped they spontaneously break.• The reverse reaction (two eggs leaping into your hand

with their shells back intact) is not spontaneous.• We can conclude that a spontaneous process has a

direction.

Thermodynamics 14

Spontaneous ProcessesSpontaneous Processes• A process that is spontaneous in one direction is not spontaneous

in the opposite direction.

• The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0C. Water turning to ice is spontaneous at T < 0C.

Thermodynamics 15

Spontaneous ProcessesSpontaneous ProcessesReversible and Irreversible ProcessesReversible and Irreversible Processes• A reversible process is one that can go back and forth

between states along the same path.– When 1 mol of water is frozen at 1 atm at 0C to form 1 mol

of ice, q = Hfus of heat is removed.

– To reverse the process, q = Hfus must be added to the 1 mol of ice at 0C and 1 atm to form 1 mol of water at 0C.

– Therefore, converting between 1 mol of ice and 1 mol of water at 0C is a reversible process.

• Allowing 1 mol of ice to warm is an irreversible process. To get the reverse process to occur, the water temperature must be lowered to 0C.

Thermodynamics 16

Spontaneous ProcessesSpontaneous Processes• Chemical systems in equilibrium are reversible.• In any spontaneous process, the path between

reactants and products is irreversible.• Thermodynamics gives us the direction of a process.

It cannot, however, predict the speed at which the process will occur.

• Generally, nature favors lower energy states, that is, spontaneous processes are usually exothermic.

• Some processes, such as the spontaneous melting of ice at room temperature, are endothermic.

• Why are some endothermic reactions spontaneous?

Thermodynamics 17

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsSpontaneous Expansion of a GasSpontaneous Expansion of a Gas• Why do spontaneous processes occur?• Consider an initial state: two flasks connected by a

closed stopcock. One flask is evacuated and the other contains 1 atm of gas.

• The final state: two flasks connected by an open stopcock. Each flask contains gas at 0.5 atm.

• The expansion of the gas is isothermal (i.e. constant temperature). Therefore the gas does no work and heat is not transferred.

• Why does the gas expand?

Thermodynamics 18

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof Thermodynamics

Spontaneous Spontaneous Expansion of Expansion of an Ideal Gasan Ideal Gas

Thermodynamics 19

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsEntropyEntropy• Entropy, S, is a measure of the disorder of a system.• Spontaneous reactions proceed to lower energy or

higher entropy (nature favors disorder).• In ice, the molecules are very well ordered because of

the H-bonds.• Therefore, ice has a low entropy.• As ice melts, the intermolecular forces are broken

(requires energy), but the order is interrupted (so entropy increases). Water is more random than ice, so ice spontaneously melts at room temperature.

Thermodynamics 20

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsOrder in ice crystalsOrder in ice crystals

Thermodynamics 21

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsEntropyEntropy• There are often multiple entropic effects occurring

simultaneously. • For example, when an ionic solid is placed in water

two things happen:– the water organizes into hydrates about the ions (so the

entropy decreases), and

– the ions in the crystal dissociate (the hydrated ions are less ordered than the crystal, so the entropy increases).

Thermodynamics 22

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsEntropyEntropy

Thermodynamics 23

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsEntropyEntropy• Generally, when an increase in entropy in one process

is associated with a decrease in entropy in another, the increase in entropy dominates.

• Entropy is a state function.

• For a system, S = Sf - Si

• If S > 0 the randomness increases, if S < 0 the order increases.

Thermodynamics 24

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof Thermodynamics• Disorder can also be considered from a probability

perspective. • For example, consider a simple system containing

two chambers, separated by a stopcock, where two molecules are located in one of the chambers, and the other is evacuated. – When the stopcock is opened and the gas molecules are

free to move independently, several new states become possible

– The number of new states possible is given as:Number of states = 2n, where n represents the number of gas molecules in the system in this case n = 2.

Thermodynamics 25


• The experiment described previously is shown below. Note that the initial condition (a) with two molecules gives rise to 22 = 4 equally possible states (b).

• Now consider a more complex system containing a much larger number of molecules, such as a mole (6.02 x 1023) molecules.– The number of possible states would be– The vast majority of these states would put equal numbers of molecules randomly in

each chamber. Only two states would put all molecules in one chamber or the other.

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26.02 x 1023

Thermodynamics 26


• Ludwig Boltzmann formalized the statistical approach to entropy with his famous equation (engraved on his headstone!)

S = k ln W

Where k = Boltzmann’s constant: k = 1.381 x 10-23 J K-1

ln W represents the natural log of the number of ways a system can be arranged

€

k = Ideal gas constant

Avogadro' s number =

8.314 J/mol - K

6.02 x 1023 mol

⎛

⎝ ⎜

⎞

⎠ ⎟

Thermodynamics 27


• While the statistical approach is interesting, it will be more useful for us to consider entropy in terms of disorder and other thermodynamic quantities. Specifically, it can be shown that:

• If a system changes reversibly between state 1 and state 2. Then, the change in entropy is given by:

– at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.)

– Note that reversible processes are generally associated with equilibrium systems

T

qS rev=

Thermodynamics 28


• The relationship suggests that changes in

entropy are related to changes in heat. Does this seem

reasonable?

• Remember, in our modern understanding, heat is

defined in terms of molecular motion.

• Intuitively we understand that a system containing

more heat will certainly exhibit greater disorder,

especially in terms of degrees of freedom of

translational, rotational and vibrational motion.

T

qS rev=

Thermodynamics 29

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsAdditional Entropy ConsiderationsAdditional Entropy Considerations• Since motion is associated with disorder, entropy values for

the physical states of a substance are not the same. In general: Sgases > Sliquids > Ssolids

• Entropy generally increases with increasing molar mass• Entropy generally increases with increasing numbers of atoms

in molecules• Entropy values for elements are NOT zero• As with enthalpy, the entropy change associated with a process

can be calculated by subtracting the total entropy table values of the reactants from the total entropy table values of the products

• Also note that entropy values are reported in joules, not kilojoules

Thermodynamics 30

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsThe Second Law of ThermodynamicsThe Second Law of Thermodynamics• Let’s now take our understanding of entropy and derive, in an

informal manner, the Second Law of Thermodynamics.

• Let’s begin by looking a little more closely and the formula we just discussed. For a reversible process at constant temperature:

• Now, since qp = H at constant pressure, we can say at constant temperature and pressure for a reversible process:

€

S =ΔH

T€

S =qrev

T

Thermodynamics 31

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsThe Second Law of ThermodynamicsThe Second Law of Thermodynamics• Now, if we multiply through by T and gather Now, if we multiply through by T and gather

terms we get:terms we get:H - TH - TS = 0S = 0

• The quantity, The quantity, H - TH - TS, is symbolized as S, is symbolized as G G and is called the Free Energy (formerly Gibb’s and is called the Free Energy (formerly Gibb’s Free Energy) of the process.Free Energy) of the process.

• This quantity for a reversible process, one at This quantity for a reversible process, one at equilibrium, has a value of equilibrium, has a value of G = 0.G = 0.

• For processes not at equilibrium, For processes not at equilibrium, G has a G has a nonzero value.nonzero value.

Thermodynamics 32

Entropy and the Second Law Entropy and the Second Law of Thermodynamicsof ThermodynamicsThe Second Law of ThermodynamicsThe Second Law of Thermodynamics• When all of the thermodynamic quantities are combined we When all of the thermodynamic quantities are combined we

get an important master thermodynamic equation:get an important master thermodynamic equation:G = G = H - TH - TSS

• Spontaneous processes involve either exothermic reactions (Spontaneous processes involve either exothermic reactions (H H is negative) or increases in entropy (is negative) or increases in entropy (S is positive). When S is positive). When those conditions occur, those conditions occur, G will have a negative sign. In reality, G will have a negative sign. In reality, when the effects of when the effects of H and H and S combine so that S combine so that G has a G has a negative sign, we can state that the process will be spontaneous. negative sign, we can state that the process will be spontaneous. Indeed, Indeed, G is the key indicator of reaction spontaneity.G is the key indicator of reaction spontaneity.

• We can summarize with three important generalizations:We can summarize with three important generalizations:– For reversible, equilibrium processes:For reversible, equilibrium processes: G = 0G = 0– For spontaneous processesFor spontaneous processes G < 0G < 0– For nonspontaneous processesFor nonspontaneous processes G > 0G > 0

Thermodynamics 33


• When we consider the thermodynamic equation just discussed, we can make four important generalizations, based on H and S values:

Enthalpy sign

Entropy sign

Spontaneity

1 - + Spontaneous at all temperatures

2 - - Spontaneous only at lower temperatures

3 + + Spontaneous only at higher temperatures

4 + - Nonspontaneous at all temperatures

Thermodynamics 34


• Useful as the generalizations in the previous slide are, even more important conclusions can be drawn from the Second Law of Thermodynamics, which we will derive and define now.

• We will not do a rigorous derivation, but the one we use should be complete enough to justify the validity of the law. First, lets make some necessary assumptions:

• For a process, the entropy for the universe is:

Suniv = Ssys + Ssurr • For a reversible process at equilibrium the entropy change

for the universe is zero, as disorder is exchanged only between the system and its surroundings, and we can say:

Ssys = -Ssurr

Thermodynamics 35


Gsys for a reversible process can be defined as:

Gsys = Hsys - TSsys

• Dividing through by T we get:

• Now, since and since Ssys = -Ssurr with substitution we get:

• Multiplying through by -1, we get:

€

G sys

T =

ΔH sys

T - ΔSsys

€

Ssys = ΔH sys

T

€

G sys

T = - ΔSsurr - ΔSsys

€

- ΔG sys

T = ΔSsurr + ΔSsys

Thermodynamics 36


• Finally, since we know: Suniv = Ssys + Ssurr Substituting we get:

• This equations states mathematically that a negative value for G will be accompanied by an increase in entropy for the universe.

• This is a statement of the Second Law of Thermodynamics:A spontaneous process will lead to an increase in entropy in the universe.

€

- ΔG sys

T = ΔSuniv

Thermodynamics 37


• The consequences of the Second Law of Thermodynamics are profound.– All spontaneous processes in the universe, and there are

many, will lead to greater disorder in the universe

– Energy transformations will always be less than 100% efficient, with entropy changes accompanying all of them

– The Second Law states that energy sources will eventually run down as energy becomes more evenly dispersed

Thermodynamics 38

Third Law of ThermodynamicsThird Law of Thermodynamics

• This would be an appropriate time to define the Third Law of Thermodynamics. It states:

• A state of zero entropy can only be found in a perfectly pure crystalline solid at absolute zero.

• Since perfect purity is impossible, and since achieving absolute zero is not possible, the Third Law refers to a theoretical state - approachable but not actually attainable.

Thermodynamics 39

Three Laws of ThermodynamicsThree Laws of Thermodynamics

• First Law: The total amount of energy in the universe is constant - energy can be neither created nor destroyed.

• Second Law: Spontaneous processes will lead to and increase in entropy in the universe.

• Third Law: A state of zero entropy can only exist in a perfectly pure crystalline solid at absolute zero

• Someone once said whimsically that the three laws can be summarized as: You can’t win; you can’t break even; and you gotta play.

Thermodynamics 40

Thermodynamics ProblemsThermodynamics Problems• There are many useful types of problems that can be solved

using thermodynamic relationships. For example, if the heat of fusion (melting) is 6.00 kJ/mol , and the heat of vaporization is 40.65 kJ/mol , compare the entropy change observed when a mole of ice melts at 0oC to the entropy change observed when a mole of liquid water vaporizes at 100oC.

Remember, for reversible processes such as these.

Remember also that Kelvin temperatures must be used in these calculations.

Ans: Sfusion = 22.0 J/mol-K Svaporization = 109 J/mol-K€

S = q

T

Thermodynamics 41

Entropy and the Second Law of Entropy and the Second Law of ThermodynamicsThermodynamicsThe Second Law of ThermodynamicsThe Second Law of Thermodynamics• The second law of thermodynamics explains why

spontaneous processes have a direction.• In any spontaneous process, the entropy of the

universe increases. Suniv = Ssys + Ssurr: the change in entropy of the

universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.

• Entropy is not conserved: Suniv is increasing.

Thermodynamics 42

Entropy and the Second Law of Entropy and the Second Law of ThermodynamicsThermodynamics

The Second Law of ThermodynamicsThe Second Law of Thermodynamics• For a reversible process: Suniv = 0.

• For a spontaneous process (i.e. irreversible): Suniv > 0.

• Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases.

• For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process.

Thermodynamics 43

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy• A gas is less ordered than a liquid that is less ordered

than a solid.

• Any process that increases the number of gas molecules leads to an increase in entropy.

• When NO(g) reacts with O2(g) to form NO2(g), the

total number of gas molecules decreases, and the entropy decreases.

Thermodynamics 44

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy

Thermodynamics 45


• There are three atomic modes of motion:– translation (the moving of a molecule

from one point in space to another),– vibration (the shortening and

lengthening of bonds, including the change in bond angles),

– rotation (the spinning of a molecule about some axis).

Thermodynamics 46

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy• Energy is required to get a molecule to translate,

vibrate or rotate.• The more energy stored in translation, vibration and

rotation, the greater the degrees of freedom and the higher the entropy.

• In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order.

• Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

• Entropy changes dramatically at a phase change.

Thermodynamics 47


Thermodynamics 48

The Molecular Interpretation of EntropyThe Molecular Interpretation of Entropy• As we heat a substance from absolute zero, the

entropy must increase.• If there are two different solid state forms of a

substance, then the entropy increases at the solid state phase change.

• Boiling corresponds to a much greater change in entropy than melting.

• Entropy will increase when– liquids or solutions are formed from solids,

– gases are formed from solids or liquids,

– the number of gas molecules increase,

– the temperature is increased.

Thermodynamics 49

Calculations of Entropy ChangesCalculations of Entropy Changes• Absolute entropy can be determined from

complicated measurements.• Standard molar entropy, S: entropy of a substance in

its standard state. Similar in concept to H.• Units: J/mol-K. Note units of H: kJ/mol.• Standard molar entropies of elements are not zero.• For a chemical reaction which produces n products

from m reactants:

( ) ( )∑ ∑ °−°=° reactantsproducts mSnSS

Thermodynamics 50

Gibbs Free EnergyGibbs Free Energy• For a spontaneous reaction the entropy of the

universe must increase.• Reactions with large negative H values are

spontaneous.• How to we balance S and H to predict whether a

reaction is spontaneous?• Gibbs free energy, G, of a state is

G = H - TS• For a process occurring at constant temperature

G = H - TS.

Thermodynamics 51

Gibbs Free EnergyGibbs Free Energy• There are three important conditions:

– If G < 0 then the forward reaction is spontaneous.

– If G = 0 then reaction is at equilibrium and no net reaction will occur.

– If G > 0 then the forward reaction is not spontaneous. (However, the reverse reaction is spontaneous.) If G > 0, work must be supplied from the surroundings to drive the reaction.

• For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.

Thermodynamics 52

Gibbs Free EnergyGibbs Free Energy

Thermodynamics 53

Gibbs Free EnergyGibbs Free EnergyStandard Free-Energy ChangesStandard Free-Energy Changes• We can tabulate standard free-energies of formation,

Gf (c.f. standard enthalpies of formation).

• Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and G = 0 for elements.

G for a process is given by

• The quantity G for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (G > 0) or products (G < 0).

( ) ( )∑ ∑ °−°=° reactantsproducts ff GmGnG

Thermodynamics 54

Free Energy and TemperatureFree Energy and Temperature• Focus on G = H - TS:

– If H < 0 and S > 0, then G is always negative.

– If H > 0 and S < 0, then G is always positive. (That is, the reverse of 1.)

– If H < 0 and S < 0, then G is negative at low temperatures.

– If H > 0 and S > 0, then G is negative at high temperatures.

• Even though a reaction has a negative G it may occur too slowly to be observed.

• Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process.

Example 1Example 1

Thermodynamics 55

Let’s consider a real free energy calculation for a real reaction. Using standard free energies of formation from the tables at the back of the book, calculate the free energy change for the following reaction under standard conditions:

3 O2(g) 2O3(g)

From the tables:

For O2(g):

For O3(g):

Ans: 326.8 kJ/mol€

G f0 = 0 kJ/mol

€

G f0 = 0 kJ/mol

Free Energy Calculations: ConsiderationsFree Energy Calculations: Considerations

Thermodynamics

Under standard temperature conditions (T = 25oC), the calculation of G is simple and direct, and is done as has been shown previously where the G values from the table can be used directly, by subtracting the summation of reactant values from the summation of product values. The example just done for ozone is typical.

Note that when temperatures depart from the standard value, G, which is dramatically affected by temperature, must be calculated by using H and S, which are relatively unaffected by temperature using the standard equation:

G = H - TS

Example 2Example 2

Thermodynamics 57

Substance(kJ/mol) (J/molK) (kJ/mol)

H2(g) 0 130.58 0

N2(g) 0 191.50 0

NH3(g) -46.19 192.5 -16.66

€

H fo

€

So

€

G fo

Calculate G for the Haber Process, shown below, at 25oC and at 500oC. Note that G (like H) for elements, in their most stable elemental state, have a value of zero. Note also that the table values for S have different units (J/molK) that must be made compatible with the other quantities.

3H2(g) + N2(g) 2NH3(g)

From the table:

Ans: (at 25oC) G = -33.32 kJ/mol (at 500oC) G = 60.86 kJ/mol

Would it be possible to determine at what temperature the process ceases to be spontaneous?

Thermodynamics 58

Free Energy and TemperatureFree Energy and Temperature

Thermodynamics 59

Free Energy and the Equilibrium Free Energy and the Equilibrium ConstantConstant• Recall that G and K (equilibrium constant) apply to

standard conditions.• Recall that G and Q (equilibrium quotient) apply to

any conditions.• It is useful to determine whether substances under

any conditions will react:

QRTGG ln+°=

Thermodynamics 60

Free Energy and the Equilibrium Free Energy and the Equilibrium ConstantConstant• At equilibrium, Q = K and G = 0, so

• From the above we can conclude:– If G < 0, then K > 1.

– If G = 0, then K = 1.

– If G > 0, then K < 1.

.ln

.ln0

ln

KRTG

KRTG

QRTGG

−=°∴+°=

+°=

Example 3Example 3

Thermodynamics 61

Using the equation that relates the equilibrium constant to the standard free energy change, estimate DG for the ionization of acetic acid, which has an equilibrium constant of 1.76 x 10-5 M at 25oC.

Ans: 27.1 kJ/mol

Example 4Example 4

Thermodynamics 62

G for the hydrolysis (break down) of a peptide bond, binding two amino acids, represented below, is estimated to be -10 kJ/mol. Estimate the equilibrium constant for the hydrolysis reaction.

AA1-AA2 + H2O AA1 + AA2

Ans: 0.018

thermodynamics1 chemical thermodynamics (chapter 19)

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