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ChemicalThermodynamics
Chemical ThermodynamicsThermo Part 2
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ChemicalThermodynamics
First Law of Thermodynamics
• You will recall from Chapter 5 that energy cannot be created nor destroyed.
• Therefore, the total energy of the universe is a constant.
• 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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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 go back to one container 3
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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 0C it is spontaneous for ice to melt.
• Below 0C the reverse process is spontaneous.
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ChemicalThermodynamics
SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes
Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150°C is added to water at 40°C, the water gets hotter. (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.
PRACTICE EXERCISEUnder 1 atm pressure CO2(s) sublimes at –78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100ºC and 1 atm pressure?
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ChemicalThermodynamics
Reversible ProcessesIn 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.8
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ChemicalThermodynamics
Entropy• Entropy (S) is a term coined by Rudolph Clausius
in the 19th century.
• Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered,
qT
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• Entropy can be thought of as a measure of the randomness of a system.
• 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
T11
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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 ThermodynamicsIn 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 on 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 lnW
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 lnWfinal k lnWinitial
lnWfinal
lnWinitial
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
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. 23
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ChemicalThermodynamics
SAMPLE EXERCISE 19.3 Predicting the Sign of S
Predict whether S is positive or negative for each of the following processes, assuming each occurs at constant temperature:
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ChemicalThermodynamics
PRACTICE EXERCISEIndicate whether each of the following processes produces an increase or decrease in the entropy of the system:
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ChemicalThermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy
Choose the sample of matter that has greater entropy in each pair, 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.
PRACTICE EXERCISEChoose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.
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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
SAMPLE EXERCISE 19.5 Calculating S from Tabulated Entropies
Calculate S° for the synthesis of ammonia from N2(g) and H2(g) at 298 K:
PRACTICE EXERCISEUsing the standard entropies in Appendix C, calculate the standard entropy change, S°, for the following reaction at 298 K:
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Step 1 Formula with substitutions:
Step 2 Gather data from chart or question provided. We would fine these in Appendix C of your book:
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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
• This becomes:
Suniverse = Ssystem +
Multiplying both sides by T,
TSuniverse = Hsystem TSsystem
Hsystem
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
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. 37
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ChemicalThermodynamics
SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change from Free Energies of Formation
(b) What is G° for the reverse of the above reaction?
(a) By using data from Appendix C, calculate the standard free-energy change for the following reaction at 298 K:
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ChemicalThermodynamics
PRACTICE EXERCISEBy using data from Appendix C, calculate G° at 298 K for the combustion of methane:
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ChemicalThermodynamics
SAMPLE EXERCISE 19.7 Estimating and Calculating G°
(a) Without using data from Appendix C, predict whether G° for this reaction is more negative or less negative than H°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?
In Section 5.7 we used Hess’s law to calculate H° for the combustion of propane gas at 298 K:
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ChemicalThermodynamics
PRACTICE EXERCISEConsider the combustion of propane to form CO2(g) and H2O(g) at 298 K:
Would you expect G° to be more negative or less negative
than H°?
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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 the entropy term.
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ChemicalThermodynamics
Free Energy and Temperature
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ChemicalThermodynamics
SAMPLE EXERCISE 19.8 Determining the Effect of Temperature on Spontaneity
Assume that H° and S° for this reaction do not change with temperature. (a) Predict the direction in which G° for this reaction changes with increasing temperature. (b) Calculate the values of G° for the reaction at 25°C and 500°C.
The Haber process for the production of ammonia involves the equilibrium
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ChemicalThermodynamics
SAMPLE EXERCISE 19.8 continued
PRACTICE EXERCISE(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate H° and S° at
298 K for the following reaction: (b) Using the values obtained in part (a), estimate G° at 400 K.
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ChemicalThermodynamics
Free Energy and Equilibrium
Under any conditions, standard or nonstandard, the free energy change can be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)
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ChemicalThermodynamics
SAMPLE EXERCISE 19.9 Relating G to a Phase Change at Equilibrium
As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of G° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.20 to estimate the normal boiling point of CCl 4.
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ChemicalThermodynamics
PRACTICE EXERCISEUse data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.)
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ChemicalThermodynamics
SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions
We will continue to explore the Haber process for the synthesis of ammonia:
Calculate G at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.
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ChemicalThermodynamics
PRACTICE EXERCISECalculate G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.
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ChemicalThermodynamics
Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT lnK
• Rearranging, this becomes
G = RT lnK
or,
K = eG/RT
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ChemicalThermodynamics
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from G°
Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process:
The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: G° = –33.3 kJ/mol = –33,300 J/mol.
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ChemicalThermodynamics
PRACTICE EXERCISEUse data from Appendix C to calculate the standard free-energy change, G°, and the equilibrium constant, K, at 298 K for the reaction
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ChemicalThermodynamics
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
(a) Calculate the value of G° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of G° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will G° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?
Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions:
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