Chapter 16 Chapter 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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In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.
Spontaneous ProcessesSpontaneous ProcessesSpontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.
Universe: System + SurroundingsUniverse: System + Surroundings
The system is what you observe; surroundings are everything else.
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.
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
EntropyEntropyΔS = Sfinal - Sinitial
EntropyEntropy
EntropyEntropy
The sign of entropy change, ΔS, associated with the boiling of water is _______.
1. Positive
2. Negative
3. Zero
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
Entropy and Temperature 02Entropy and Temperature 02
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)
Ssolid < Sliquid < Sgas
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.
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.
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
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
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)
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
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
Gibbs Free Energy 04Gibbs Free Energy 04
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)?
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
Gibbs Free Energy 05Gibbs Free Energy 05
• Iron metal can be produced by reducing iron(III) oxide with hydrogen: