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Chapter 16Chapter 16• Thermodynamics: Entropy,Free Energy,and Equilibrium
• Thermodynamics: Entropy,Free Energy,and Equilibrium
spontaneous
nonspontaneous
In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.
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Spontaneous ProcessesSpontaneous Processes
Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.
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Universe: System + SurroundingsUniverse: System + Surroundings
The system is what you observe; surroundings are everything else.
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Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
energy, pressure, volume, temperature
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Laws of Thermodynamics Laws of Thermodynamics
1) Energy is neither created nor destroyed.
2) In any spontaneous process the total Entropy of a system and its surrounding always increases. We will prove later: ∆G = ∆Hsys – T∆Ssys
∆G < 0 Reaction is spontaneous in forward direction3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)
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Enthalpy, Entropy, and Spontaneous ProcessesEnthalpy, Entropy, and Spontaneous Processes
State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.
Enthalpy Change (H): The heat change in a reaction or process at constant pressure.
Entropy (S): The amount of Molecular randomness change in a reaction or process at constant pressure.
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Enthalpy ChangeEnthalpy Change
= +40.7 kJHvap
CO2(g) + 2H2O(l)CH4(g) + 2O2(g)
H2O(l)H2O(s)
H2O(g)H2O(l)
2NO2(g)N2O4(g)
= +3.88 kJH°
= +6.01 kJHfusion
= -890.3 kJH°
= +57.1 kJH°
Endothermic:
Exothermic:
Na1+(aq) + Cl1-(aq)NaCl(s)H2O
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Entropy ChangeEntropy Change
S = Sfinal - Sinitial
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EntropyEntropy
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EntropyEntropy
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The sign of entropy change, S, associated with the boiling of water is _______.
1. Positive
2. Negative
3. Zero
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Correct Answer:
Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; S must be positive.
1. Positive
2. Negative
3. Zero
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Entropy and Temperature 02Entropy and Temperature 02
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Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)
Ssolid < Sliquid < Sgas
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Entropy Changes in the System (Ssys)
When gases are produced (or consumed)
• If a reaction produces more gas molecules than it consumes, S0 > 0.
• If the total number of gas molecules diminishes, S0 < 0.
• If there is no net change in the total number of gas molecules, then S0 may be positive or negative .
What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s)
The total number of gas molecules goes down, S is negative.
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Standard Molar Entropies and Standard Entropies of ReactionStandard Molar Entropies and Standard Entropies of Reaction
Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.
Calculated by using S = k ln W , W = Accessible microstates
of translational, vibrational and rotational motions
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Review of Ch 8: Review of Ch 8:
• Calculating H° for a reaction:
H° = H°f (Products) – H°f (Reactants)
• For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.
aA + bB cC + dD
H° = [cH°f (C) + dH°f (D)] – [aH°f (A) + bH°f (B)]
See Appendix B, end of your book
•∆G = ∆H – T∆S see page 301 of your book
We will show the proof of this formula later in this chapter
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Entropy Changes in the System (Ssys)
aA + bB cC + dD
S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
S0rxn nS0
(products)= mS0 (reactants)-
The standard entropy change of reaction (S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)
S0(CO) = 197.6 J/K•molS0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
S0rxn = 427.2 – [395.2 + 205.0] = -173.0J/K•mol
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Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)
∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]
∆∆SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + [2 mol (130.6 J/K•mol) +
1 mol (205.0 J/K•mol)]1 mol (205.0 J/K•mol)]∆∆SSoo = -326.4 J/K = -326.4 J/KNote that there is a Note that there is a decrease in Entropy decrease in Entropy because 3 because 3
mol of gas give 2 mol of liquid.mol of gas give 2 mol of liquid.
Calculating ∆S for a ReactionCalculating ∆S for a ReactionCalculating ∆S for a ReactionCalculating ∆S for a Reaction
∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)
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Calculate S° for the equation below using the standard entropy data given:
2 NO(g) + O2(g) 2 NO2(g)
S° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205.
1. +176 J/mol
2. +147 J/mol
3. 147 J/mol
4. 76 J/mol
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Correct Answer:
S° = 2(240) [2(211) + 205]S° = 480 [627]
S° = 147
reactantsproducts SSS
1. +176 J/K
2. +147 J/K
3. 147 J/K
4. 76 J/K
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Thermite ReactionThermite Reaction
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Spontaneous Processesand Entropy 07Spontaneous Processesand Entropy 07
• Consider the gas phase reaction of A2 molecules (red) with B atoms (blue).
(a) Write a balanced equation for the reaction.(b) Predict the sign of ∆S for the reaction.
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Entropy Changes in the Surroundings (Ssurr)
Exothermic ProcessSsurr > 0
Endothermic ProcessSsurr < 0
In any spontaneous process the total Entropy of a system and its surrounding always increases.
Second Law Of Thermodynamic:
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Entropy and the Second Law of ThermodynamicsEntropy and the Second Law of Thermodynamics
ssurr - H)
ssurr T
1
ssurr = T
- H
See example of tossing a rock into a calm waters vs. rough watersPage 661 of your book
)
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2nd Law of Thermodynamics 012nd Law of Thermodynamics 01
• The total entropy increases in a spontaneous
process and remains unchanged in an equilibrium
process.
Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0
Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0
• The system is what you observe; surroundings are
everything else.
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2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq) O(liq) ∆H ∆H °° = -571.7 kJ = -571.7 kJ
Sosurroundings =
- (-571.7 kJ)(1000 J/kJ)
298.15 KSo
surroundings = - (-571.7 kJ)(1000 J/kJ)
298.15 K
2nd Law of 2nd Law of ThermodynamicsThermodynamics
∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K
Ssurr Hsys
T
Calculate change of entropy of surrounding in the following reaction:
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Spontaneous Reactions 01Spontaneous Reactions 01
• The 2nd law tells us a process will be spontaneous
if ∆Stotal > 0 which requires a knowledge of ∆Ssurr.
spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0
S sur Hsys
T
•∆Stotal = ∆Ssys + ( ) > 0 H sys
T
-T (∆Stotal = ∆Ssys + ( ) ) < 0 H sys
T
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Gibbs Free Energy02Gibbs Free Energy02• The expression –T∆Stotal is equated as Gibbs free
energy change (∆G)*, or simply free energy change:
•
•
• ∆G = ∆Hsys – T∆Ssys
• ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction.
-T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0
–T∆Stotal = ΔG
* Driving force of a reaction, Maximum work you can get from a system.
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Calculate ∆GCalculate ∆Go o rxn for the following:rxn for the following:Calculate ∆GCalculate ∆Go o rxn for the following:rxn for the following:
CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)
Use enthalpies of formation to calculateUse enthalpies of formation to calculate ∆∆HHoo
rxnrxn = -1256 kJ = -1256 kJ
Use standard molar entropies to calculate ∆SUse standard molar entropies to calculate ∆Soorxnrxn ( see page ( see page
658, appendix A-10)658, appendix A-10) ∆∆SSoo
rxnrxn = -97.4 J/K or -0.0974 kJ/K = -97.4 J/K or -0.0974 kJ/K
∆ ∆GGoorxnrxn = -1256 kJ - (298 K)(-0.0974 kJ/K) = -1256 kJ - (298 K)(-0.0974 kJ/K)
= -1227 kJ= -1227 kJ
Reaction is Reaction is product-favoredproduct-favored in spite of negative ∆S in spite of negative ∆Soorxnrxn. .
Reaction is Reaction is “enthalpy driven”“enthalpy driven”
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1. +33 kJ
2. +66 kJ
3. 66 kJ
4. 33 kJ
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Correct Answer:
H° = 92 kJS° = 198.3J/mol . K
G° = H° TS°G° = (92) (298)(0.1983)
G° = (92) + (59.2) = 32.9 KJ
1. +33 kJ
2. +66 kJ
3. 66 kJ
4. 33 kJ
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Gibbs Free Energy03Gibbs Free Energy03
Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S.
1) If both ∆H and ∆S are positive,
∆G will be negative only when the temperature value is large.
Therefore, the reaction is spontaneous only at high temperature.
2) If ∆H is positive and ∆S is negative,
∆G will always be positive.
Therefore, the reaction is not spontaneous
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3) If ∆H is negative and ∆S is positive,
∆G will always be negative.
Therefore, the reaction is spontaneous
4) If both ∆H and ∆S are negative,
∆G will be negative only when the temperature value is small.
Therefore, the reaction is spontaneous only at Low temperatures.
Gibbs Free Energy: 04Gibbs Free Energy: 04∆G = ∆H – T∆S
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Gibbs Free Energy04Gibbs Free Energy04
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Gibbs Free Energy06Gibbs Free Energy06
• What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?
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Gibbs Free Energy02Gibbs Free Energy02• The expression –T∆Stotal is equated as Gibbs free
energy change (∆G)*, or simply free energy change:
•
•
• ∆G = ∆Hsys – T∆Ssys
• ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction.
-T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0
–T∆Stotal = ΔG
* Driving force of a reaction, Maximum work you can get from a system.
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Standard Free-Energy Changes for ReactionsStandard Free-Energy Changes for Reactions
Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:
S° = -198.7 J/K
2NH3(g)N2(g) + 3H2(g)
= -33.0 kJ
H° = -92.2 kJ
G° = H° - TS°
1000 J
1 kJ
K
-198.7 Jx x=
-92.2 kJ - 298 K
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Gibbs Free Energy05Gibbs Free Energy05
• Iron metal can be produced by reducing iron(III)
oxide with hydrogen:
Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(g)
∆H° = +98.8 kJ; ∆S° = +141.5 J/K
1. Is this reaction spontaneous at 25°C?
2. At what temperature will the reaction become
spontaneous?
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Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol K. At what temperature does this become spontaneous?
1. 121°C2. 395°C3. 848°C4. 1121°C
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Correct Answer:
1. 121°C2. 395°C3. 848°C4. 1121°C
T = H°/S°
T = 178.3 kJ/mol/0.159 kJ/mol K
T = 1121 K
T (°C) = 1121 – 273 = 848
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Standard Free Energies of FormationStandard Free Energies of Formation
G° = G°f (products) - G°f (reactants)
G° = [cG°f (C) + dG°f (D)] - [aG°f (A) + bG°f (B)]
ReactantsProducts
cC + dDaA + bB
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Standard free energy of formation (G0f) is the free-energy
change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atm) states.
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aA + bB cC + dD
G0rxn dG0 (D)fcG0 (C)f= [ + ] - bG0 (B)faG0 (A)f[ + ]
G0rxn nG0 (products)f= mG0 (reactants)f-
The standard free-energy of reaction (G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.
rxn
G0 of any element in its stable form is zero.
f --
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Calculate G° for the equation below using the Gibbs free energy data given:
2 SO2(g) + O2(g) 2 SO3(g)
Gf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4
1. +70 kJ
2. +140 kJ
3. 140 kJ
4. 70 kJ
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Correct Answer:
°° ° reactantsproducts ff GGG
G° = (2 370.4) [(2 300.4) + 0]
G° = 740.8 [600.8] = 140
1. +70 kJ
2. +140 kJ
3. 140 kJ
4. 70 kJ
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Gibbs Free Energy07Gibbs Free Energy07
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Calculation of Nonstandard ∆GCalculation of Nonstandard ∆G
• The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions.
• Under nonstandard( condition where pressure is not 1atm or concentrations of solutions are not 1M) conditions, ∆G˚ becomes ∆G.
∆G = ∆G˚ + RT lnQ
• The reaction quotient is obtained in the same way as an equilibrium expression.
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Free Energy Changes and the Reaction MixtureFree Energy Changes and the Reaction Mixture
C2H4(g)2C(s) + 2H2(g) Qp =C2H4
P
H2P
2
Calculate ln Qp:
Calculate G for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4.
= -11.511002
0.10ln
G = G° + RT ln Q
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Free Energy Changes and the Reaction MixtureFree Energy Changes and the Reaction Mixture
Calculate G:
= 68.1 kJ/mol +
G = 39.6 kJ/mol
(298 K)(-11.51)8.314K mol
J
G = G° + RT ln Q
1000 J
1 kJ
( see page A-13 of your book, page 667)
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Free Energy and ChemicalEquilibrium 05Free Energy and ChemicalEquilibrium 05
• At equilibrium ∆G = 0 and Q = K
•∆G = ∆G˚ + RT lnQ
∆G˚rxn = –RT ln K
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Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium
Calculate Kp at 25 °C for the following reaction:
Calculate G°:
CaO(s) + CO2(g)CaCO3(s)
- [(1 mol)(-1128.8 kJ/mol)]
G° = +130.4 kJ/mol
= [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
G° = [G°f (CaO(s)) + G°f (CO2(g))] - [G°f (CaCO3(s))] Please see appendix B:
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Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium
Calculate K:
=
G° = -RT ln K
-130.4 kJ/mol
ln K = -52.63
(298 K)1000 J
1 kJ8.314
K mol
J
= 1.4 x 10-23
Calculate ln K:
RT
-G°ln K =
-52.63
K = e
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Free Energy and ChemicalEquilibrium 04Free Energy and ChemicalEquilibrium 04
∆G˚rxn = –RT ln K
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Suppose G° is a large, positive value. What then will be the value of the equilibrium constant, K?
1. K = 02. K = 13. 0 < K < 14. K > 1
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Correct Answer:
G° = RTlnK
Thus, large positive values of G° lead to large negative values of lnK. The value of K itself, then, is very small.
1. K = 02. K = 13. 0 < K < 14. K > 1
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More thermo?More thermo? You betcha!You betcha!
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Calculate K for the reactionCalculate K for the reaction
NN22OO44 --->2 NO --->2 NO22 ∆G∆Goorxnrxn = +4.8 kJ/mole = +4.8 kJ/mole ( see page A-11)( see page A-11)
ΔG° = RT lnK
∆∆GGoorxnrxn = +4800 J = - (8.31 J/mol.K)(298 K) ln K = +4800 J = - (8.31 J/mol.K)(298 K) ln K
∆∆GGoorxnrxn = - RT lnK = - RT lnK
1.94- = K)J/m.K)(298 (8.31
J/mole 4800- = ln K
Thermodynamics and Thermodynamics and KKeqeq
K = 0.14K = 0.14When ∆GWhen ∆Goo
rxnrxn > 0, then K < 1 > 0, then K < 1