Chapter 19
Chemical Thermodynamics
Lecture Presentation
Dr. Subhash C. GoelSouth GA State College
Douglas, GA© 2012 Pearson Education, Inc.
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ReviewUnits of Energy
• The SI unit of energy is the joule (J):
• An older, non-SI unit is still in widespread use: the calorie (cal):
1 cal = 4.184 J
1 J = 1 kg m2
s2
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Heat• Energy can also
be transferred as heat.
• Heat flows from warmer objects to cooler objects.
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First Law of Thermodynamics• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.
• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.
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Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.
By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial
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WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston:
w = −PV
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Enthalpy
• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV
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Enthalpy
• When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
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Enthalpy
• Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:
H = E + PV
H = (q + w) − w
H = q
• So, at constant pressure, the change in enthalpy is the heat gained or lost.
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Enthalpy of Reaction
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
H = enthaly of reaction or heat of reaction
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Hess’s Law
Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
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Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.
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Calculation of H
We can use Hess’s law in this way:
H = nHf,products – mHf°,reactants
where n and m are the stoichiometric coefficients.
ChemicalThermodynamics
Spontaneous Processes
• Spontaneous processes are those that can proceed without any outside intervention.
• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.
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ChemicalThermodynamics
Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
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ChemicalThermodynamics
Spontaneous Processes• Processes that are spontaneous at one temperature may
be nonspontaneous at other temperatures.• Above 0 C, it is spontaneous for ice to melt.• Below 0 C, the reverse process is spontaneous.
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ChemicalThermodynamics
Exercise 1 Identifying Spontaneous Processes
Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or in equilibrium:
(a)Water at 40 °C gets hotter when a piece of metal heated to 150 °C is added.
(b)Water at room temperature decomposes into H2(g) and O2(g).
(c)Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C.
ChemicalThermodynamics
Reversible Processes
In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.
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ChemicalThermodynamics
Irreversible Processes
• Irreversible processes cannot be undone by exactly reversing the change to the system.
• Spontaneous processes are irreversible.
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ChemicalThermodynamics
Entropy
• Entropy (S) is a term coined by Rudolph Clausius in the nineteenth century.
• Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, .q
T
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ChemicalThermodynamics
Entropy• Entropy can be described as a
measure of how spread out or dispersed the energy is among the different possible ways that system can contain energy.
• The greater is the dispersal, greater is the entropy.
• It is related to the various modes of motion in molecules.
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ChemicalThermodynamics
Entropy
• Like total energy, E, and enthalpy, H, entropy is a state function.
• Therefore,
S = Sfinal Sinitial
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ChemicalThermodynamics
Entropy
For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:
S =qrev
T
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ChemicalThermodynamics
Exercise 2 Calculating ΔS for a Phase Change
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is –38.9 °C, and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?
ChemicalThermodynamics
Second Law of Thermodynamics
The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.
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ChemicalThermodynamics
Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
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ChemicalThermodynamics
Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases.
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ChemicalThermodynamics
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of entropy on the molecular level.
• Temperature is a measure of the average kinetic energy of the molecules in a sample.
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ChemicalThermodynamics
Entropy on the Molecular Scale• Molecules exhibit several types of motion:
– Translational: Movement of the entire molecule from one place to another.
– Vibrational: Periodic motion of atoms within a molecule.– Rotational: Rotation of the molecule about an axis or
rotation about bonds.
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ChemicalThermodynamics
Entropy on the Molecular Scale• Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.– This would be akin to taking a snapshot of all the
molecules.
• He referred to this sampling as a microstate of the thermodynamic system.
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ChemicalThermodynamics
Entropy on the Molecular Scale• Each thermodynamic state has a specific number
of microstates, W, associated with it.• Entropy is
S = k ln W
where k is the Boltzmann constant, 1.38 1023 J/K.
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ChemicalThermodynamics
Entropy on the Molecular Scale• The change in entropy for a process,
then, is
S = k ln Wfinal k ln Winitial
Wfinal
Winitial
S = k ln
• Entropy increases with the number of microstates in the system.
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ChemicalThermodynamics
Entropy on the Molecular Scale
• The number of microstates and, therefore, the entropy, tends to increase with increases in– Temperature– Volume– The number of independently moving
molecules.
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ChemicalThermodynamics
Entropy and Physical States
• Entropy increases with the freedom of motion of molecules.
• Therefore,
S(g) > S(l) > S(s)
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ChemicalThermodynamics
Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
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ChemicalThermodynamics
Predict whether is positive or negative for each process, assuming each occurs at constant ΔS temperature:(a) H2O(l) H2O(g)(b) Ag+(aq) + Cl–(aq) AgCl(s)(c) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)(d) N2(g) + O2(g) 2 NO(g)
Exercise 3 Predicting the Sign of ΔS
ChemicalThermodynamics
In each pair, choose the system that has greater entropy and explain your choice: (a)1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b)2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c)1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.
Exercise 4 Predicting Relative Entropies
ChemicalThermodynamics
Entropy Changes
• In general, entropy increases when
– Gases are formed from liquids and solids;
– Liquids or solutions are formed from solids;
– The number of gas molecules increases;
– The number of moles increases.
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ChemicalThermodynamics
Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.
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ChemicalThermodynamics
Standard Entropies
• These are molar entropy values of substances in their standard states.
• Standard entropies tend to increase with increasing molar mass.
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ChemicalThermodynamics
Standard Entropies
Larger and more complex molecules have greater entropies.
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ChemicalThermodynamics
Entropy Changes
Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:
S = nS(products) — mS(reactants)
where n and m are the coefficients in the balanced chemical equation.
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ChemicalThermodynamics
Exercise 5 Calculating ΔS° from Tabulated Entropies
Calculate the change in the standard entropy of the system, ΔS° , for the synthesis of ammonia from N2(g)and H2(g) at 298 K:
N2(g) + 3 H2(g) 2 NH3(g)
ChemicalThermodynamics
Entropy Changes in Surroundings
• Heat that flows into or out of the system changes the entropy of the surroundings.
• For an isothermal process:
Ssurr =qsys
T
• At constant pressure, qsys is simply H for the system.
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ChemicalThermodynamics
Entropy Change in the Universe
• The universe is composed of the system and the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
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ChemicalThermodynamics
Entropy Change in the Universe• Since
Ssurroundings =
and
qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T, we get
TSuniverse = Hsystem TSsystem
Hsystem
T
qsystem
T
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ChemicalThermodynamics
Gibbs Free Energy
TSuniverse is defined as the Gibbs free energy, G.
• When Suniverse is positive, G is negative.
• Therefore, when G is negative, a process is spontaneous.
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ChemicalThermodynamics
Gibbs Free Energy
1. If G is negative, the forward reaction is spontaneous.
2. If G is 0, the system is at equilibrium.
3. If G is positive, the reaction is spontaneous in the reverse direction.
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ChemicalThermodynamics
Exercise 6 Calculating Free–Energy Change from ΔH°, T, and ΔS°
Calculate the standard free–energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K:
N2(g) + O2(g) 2 NO(g)given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these conditions?
ChemicalThermodynamics
Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G:
f
G = nG (products) mG (reactants)f f
where n and m are the stoichiometric coefficients.
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ChemicalThermodynamics
Exercise 7 Calculating Standard Free–Energy Change from Free Energies of Formation
(a) Use data from Appendix C to calculate the standard free–energy change for the reaction P4(g) + 6 Cl2 (g) 4 PCl3 (g) run at 298 K.(b) What is ΔG° for the reverse of this reaction?
ChemicalThermodynamics
Free Energy Changes
At temperatures other than 25 C,
G = H TS
How does G change with temperature?
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ChemicalThermodynamics
Free Energy and Temperature
• There are two parts to the free energy equation:H— the enthalpy term– TS — the entropy term
• The temperature dependence of free energy then comes from theentropy term.
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ChemicalThermodynamics
Exercise 8 Determining the Effect of Temperature on Spontaneity
The Haber process for the production of ammonia involves the equilibrium
Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for the reaction changes with increasing temperature. (b) Calculate ΔG° at 25 °C and 500 °C.
ChemicalThermodynamics
Free Energy and Equilibrium
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT ln Q
(Under standard conditions, all concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.)
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ChemicalThermodynamics
Exercise 9 Calculating the Free–Energy Change under Nonstandard Conditions
Calculate at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process:
ChemicalThermodynamics
Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT ln K
• Rearranging, this becomes
G = RT ln K
or
K = e
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G/RT
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