Thermodynamics: Entropy, Free energy, and …dfard.weebly.com/uploads/1/0/5/3/10533150/ch16.pdfThird...
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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 ProcessesSpontaneous 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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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 (ΔH): The heat change in a reaction or process at constant pressure; ΔH = ΔE + PΔV.
Entropy (S): The amount of molecular randomness in a system.
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EnthalpyEnthalpy
= +40.7 kJΔHvap
CO2(g) + 2H2O(l)CH4(g) + 2O2(g)
H2O(l)H2O(s)
H2O(g)H2O(l)
2NO2(g)N2O4(g)
= +3.88 kJΔH°
= +6.01 kJΔHfusion
= -890.3 kJΔH°
= +57.1 kJΔH°
Endothermic:
Exothermic:
Na1+(aq) + Cl1-(aq)NaCl(s)H2O
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EntropyEntropyΔS = Sfinal - Sinitial
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EntropyEntropy
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EntropyEntropy
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The sign of entropy change, ΔS, 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; ΔSmust 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 ThermodynamicsThe 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 (ΔSsys)
When gases are produced (or consumed)
• If a reaction produces more gas molecules than it consumes, ΔS0 > 0.
• If the total number of gas molecules diminishes, ΔS0 < 0.
• If there is no net change in the total number of gas molecules, then ΔS0 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, ΔS is negative.
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Entropy and ProbabilityEntropy and Probability
S = k ln W
k = Boltzmann’s constant
= 1.38 x 10-23 J/K
W = The number of ways that the state can be achieved.
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Standard Molar Entropies and Standard Entropies of ReactionStandard Molar Entropies and Standard Entropies of ReactionStandard 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 ΔH° for a reaction:
ΔH° = ΔH°f (Products) – ΔH°f (Reactants)
• For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.
aA + bB cC + dDΔH° = [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°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 (ΔSsys)
aA + bB cC + dD
ΔS0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
ΔS0rxn nS0
(products)= Σ mS0 (reactants)Σ-
The standard entropy change of reaction (ΔS0 ) 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
ΔS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
ΔS0rxn = 427.2 – [395.8 + 205.0] = -173.3J/K•mol
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Consider 2 HConsider 2 H22(g) + O(g) + O22(g) (g) ------> 2 H> 2 H22O(liq)O(liq)∆∆SSoo = 2 S= 2 Soo (H(H22O) O) -- [2 S[2 Soo (H(H22) + S) + Soo (O(O22)])]∆∆SSoo = 2 mol (69.9 J/= 2 mol (69.9 J/KK••molmol) ) --
[2 mol (130.7 J/[2 mol (130.7 J/KK••molmol) + ) + 1 mol (205.3 J/1 mol (205.3 J/KK••molmol)])]
∆∆SSoo = = --326.9 J/K326.9 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 Calculating ∆∆S for a ReactionS for a Reaction
∆So = Σ So (products) - Σ So (reactants)∆∆SSoo = = ΣΣ SSoo (products) (products) -- ΣΣ SSoo (reactants)(reactants)
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Calculate ΔS° for the equation below using the standard entropy data given:
2 NO(g) + O2(g) → 2 NO2(g)ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205.
1. +176 J/mol2. +147 J/mol
3. −147 J/mol4. −76 J/mol
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Correct Answer:
ΔS° = 2(240) − [2(211) + 205]ΔS° = 480 − [627]
ΔS° = − 147
( ) ( )ΣΣ °° Δ−Δ=°Δ reactantsproducts SSS
1. +176 J/K2. +147 J/K
3. −147 J/K4. −76 J/K
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Standard Molar Entropies 01Standard Molar Entropies 01
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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 (ΔSsurr)
Exothermic ProcessΔSsurr > 0
Endothermic ProcessΔSsurr < 0
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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 > 0Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0
• The system is what you observe; surroundings are everything else.
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Entropy and the Second Law of ThermodynamicsEntropy and the Second Law of Thermodynamics
Δssurr α - ΔH
Δssurr α T1
Δssurr = T
- ΔH
See example of tossing a rock into a calm waters vs. rough watersPage 661 of your book
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2 H2 H22(g) + O(g) + O22(g) (g) ------> 2 H> 2 H22O(liq) O(liq) ∆∆H H °° = = --571.7 kJ571.7 kJ
∆∆SSoosystemsystem = = --326.9 J/K326.9 J/K
ΔSosurroundings = - (-571.7 kJ)(1000 J/kJ)
298.15 K
2nd Law of Thermodynamics2nd Law of Thermodynamics
∆∆SSoosurroundingssurroundings = +1917 J/K= +1917 J/K
ΔSsurr =−Δ 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
ΔS sur =−Δ Hsys
T•∆Stotal = ∆Ssys + ( ) > 0−ΔH sys
T
-T (∆Stotal = ∆Ssys + ( ) ) < 0−Δ H sys
T
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Gibbs Free Energy 02Gibbs Free Energy 02
• 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
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Calculate ∆Go rxn for the following:Calculate Calculate ∆∆GGo o rxnrxn for the following:for the following:CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) (g) ----> 2 CO> 2 CO22(g) + H(g) + H22O(g)O(g)
Use enthalpies of formation to calculateUse enthalpies of formation to calculate∆∆HHoo
rxnrxn = = --1238 kJ1238 kJUse standard molar entropies to calculate Use standard molar entropies to calculate ∆∆SSoo
rxnrxn ( see page ( see page 658)658)∆∆SSoo
rxnrxn = = --97.4 J/K or 97.4 J/K or --0.0974 kJ/K0.0974 kJ/K
∆∆GGoorxnrxn = = --1238 kJ 1238 kJ -- (298 K)((298 K)(--0.0974 kJ/K)0.0974 kJ/K)
= = --1209 kJ1209 kJ
Reaction is Reaction is productproduct--favoredfavored in spite of negative in spite of negative ∆∆SSoorxnrxn. .
Reaction is Reaction is ““enthalpy drivenenthalpy driven””
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Calculate ΔG° for the equation below using the thermodynamic data given:
N2(g) + 3 H2(g) → 2 NH3(g)ΔH° (NH3) = −46 kJ; S° values (J/mol-K) are
NH3 = 192.5, N2 = 191.5, H2 = 130.6.
1. +33 kJ2. +66 kJ
3. −66 kJ4. −33 kJ
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Correct Answer:
ΔH° = −92 kJΔS° = −198.7J/mol . K
ΔG° = ΔH° − TΔS°ΔG° = (−92) − (298)(−0.1987)ΔG° = (−92) + (59.2) = −33 KJ
1. +33 kJ2. +66 kJ3. −66 kJ4. −33 kJ
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Gibbs Free Energy 03Gibbs Free Energy 03
Using ∆G = ∆H – T∆S, we can predict the sign of ∆Gfrom 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
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Gibbs Free Energy 04Gibbs Free Energy 04
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Gibbs Free Energy 06Gibbs Free Energy 06
• What are the signs (+, –, or 0) of ∆H, ∆S, and ∆Gfor the following spontaneous reaction of A atoms (red) and B atoms (blue)?
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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:
ΔS° = -198.7 J/K
2NH3(g)N2(g) + 3H2(g)
= -33.0 kJ1000 J
ΔH° = -92.2 kJ
ΔG° = ΔH° - TΔS°
1 kJK
-198.7 Jx x= -92.2 kJ - 298 K
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Gibbs Free Energy 05Gibbs Free Energy 05
• 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 ΔH° = 178.3 kJ/mol and ΔS° = 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 = ΔH°/ΔS°T = 178.3 kJ/mol/0.159 kJ/mol • K
T = 1121 KT (°C) = 1121 – 273 = 848
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Standard Free Energies of FormationStandard Free Energies of Formation
ΔG° = ΔG°f (products) - ΔG°f (reactants)
ΔG° = [cΔG°f (C) + dΔG°f (D)] - [aΔG°f (A) + bΔG°f (B)]
ReactantsProducts
cC + dDaA + bB
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Gibbs Free Energy 07Gibbs Free Energy 07
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Standard free energy of formation (ΔG0f) is the free-energy
change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atmstates.
f
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aA + bB cC + dDΔG0
rxn dΔG0 (D)fcΔG0 (C)f= [ + ] - bΔG0 (B)faΔG0 (A)f[ + ]
ΔG0rxn nΔG0 (products)f= Σ mΔG0 (reactants)fΣ-
The standard free-energy of reaction (ΔG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.
rxn
ΔG0 of any element in its stable form is zero.
f--
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Calculate ΔG° for the equation below using the Gibbs free energy data given:
2 SO2(g) + O2(g) 2 SO3(g)ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4
1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ
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Correct Answer:
( ) ( )∑ ∑°°
−=° reactantsproducts ff GGG ΔΔΔ
ΔG° = (2 × −370.4) − [(2 × −300.4) + 0]ΔG° = −740.8 − [−600.8] = −140
1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ
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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 ΔG 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.
Calculate ln Qp:
= -11.511002
0.10ln
ΔG = ΔG° + RT ln Q
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Free Energy Changes and the Reaction MixtureFree Energy Changes and the Reaction Mixture
Calculate ΔG:
= 68.1 kJ/mol + 8.314K mol
J
ΔG = 39.6 kJ/mol
(298 K)(-11.51)
ΔG = ΔG° + RT ln Q
1000 J1 kJ
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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 ΔG°:
CaO(s) + CO2(g)CaCO3(s)
- [(1 mol)(-1128.8.0 kJ/mol)]
ΔG° = +130.4 kJ/mol
= [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
ΔG° = [ΔG°f (CaO(s)) + ΔG°f (CO2(g))] - [ΔG°f (CaCO3(s))]
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Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium
Calculate K:
=
ΔG° = -RT ln K
-130.4 kJ/mol
ln K = -52.63
(298 K)1000 J
1 kJ8.314
K molJ
= 1.4 x 10-23
Calculate ln K:
RT-ΔG°
ln K =
-52.63K = e
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Free Energy and ChemicalEquilibrium 04Free Energy and ChemicalEquilibrium 04∆G˚rxn = –RT ln K
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Suppose ΔG° 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:
ΔG° = −RTlnK
Thus, large positive values of ΔG°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 You betchabetcha!!
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Calculate K for the reactionCalculate K for the reactionNN22OO44 ------>2 NO>2 NO22 ∆∆GGoo
rxnrxn = +4.8 kJ/mole = +4.8 kJ/mole ( see page ( see page
667)667)
ΔG° = −RTlnK∆∆GGoo
rxnrxn = +4800 J = = +4800 J = -- (8.31 J/mol.K)(298 K) (8.31 J/mol.K)(298 K) lnln KK
∆∆GGoorxnrxn = = -- RT RT lnKlnK
1.94- = K)J/m.K)(298 (8.31
J/mole 4800- = ln K
Thermodynamics and Thermodynamics and KKeqeq
K = 0.14K = 0.14When When ∆∆GGoo
rxnrxn > 0, then K < 1> 0, then K < 1