C H E M I S T R Y Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium

Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

Spontaneous process Spontaneity reaction always moves a system toward equilibrium ◦ Both forward and reverse reaction depends on  Temperature  Pressure  Composition of reaction mixture Q < K; reaction proceeds in the forward direction Q>K; reaction proceeds in the reverse direction Spontaneity of a reaction does not identify the speed of reaction

Spontaneous Processes

Enthalpy, Entropy, and Spontaneous Processes  S = S final  S initial

Enthalpy, Entropy, and Spontaneous Processes

Entropy 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.

Entropy and Temperature Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero.

Entropy and Temperature ΔS increases when increasing the average kinetic energy of molecules Total energy is distributed among the individual molecules in a number of ways Botzman- the more way (W) that the energy can be distributed the greater the randomness of the state and higher the entropy

Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (S o ): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

Standard Molar Entropies and Standard Entropies of Reaction cC + dDaA + bB  S o = S o (products) - S o (reactants )  S o = [cS o (C) + dS o (D)]  [aS o (A) + bS o (B)] ReactantsProducts

Standard Molar Entropies and Standard Entropies of Reaction Using standard entropies, calculate the standard entropy change for the decomposition of N 2 O 4. 2NO 2 (g)N2O4(g)N2O4(g)

Example Calculate the standard entropy of reaction at 25 o C for the decomposition of calcium carbonate: CaCO 3 (s)  CaO(s) + CO 2 (g)

16.6 Entropy and the Second Law of Thermodynamics First Law of Thermodynamics: In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant. Helps keeping track of energy flow between system and the surrounding Does not indicate the spontaneity of the process Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases. Provide a clear cut criterion of spontaneity Direction of spontaneous change is always determined by the sign of the total entropy change

Entropy and the Second Law of Thermodynamics  S total > 0The reaction is spontaneous.  S total < 0The reaction is nonspontaneous.  S total = 0The reaction mixture is at equilibrium.  S total =  S sys +  S surr or  S total =  S system +  S surroundings

Entropy and the Second Law of Thermodynamics  S surr    H  S surr  T 1  S surr = T  H H

Example Consider the oxidation of iron metal 4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s) Determine whether the reaction is spontaneous at 25 o C S o (J/K mol)ΔH o f (kJ/mol) Fe(s)27.30 O2(g) Fe2O3(s)

Example Consider the combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) a.Calculate the entropy change in the surrounding associated with this reaction occurring at 25.0 o C b.Determine the sign of the entropy change for the system c.Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?

S o (J/K mol)ΔH o f (kJ/mol) C 3 H 8 (g) O 2 (g) CO 2 (g) H 2 O(g)

16.7 Free Energy  G = -T  S total Free Energy: a thermodynamic quantity defined by G = H - TS  G =  H - T  S  S total =  S sys - Using:  s surr = T -  H  S total =  S sys +  S surr HH T Constant pressure and temperature

Free Energy  G < 0The reaction is spontaneous.  G > 0The reaction is nonspontaneous.  G = 0The reaction mixture is at equilibrium. Using the second law and  G =  H - T  S = -T  S total

16.8 Standard Free-Energy Changes for Reactions Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution.  G° =  H° - T  S°

The effect of H, S and T on Spontaneity ∆H∆SLow TemperatureHigh Temperature -+Spontaneous (G<0) +-Nonspontaneous (G > 0) --Spontaneous (G< 0)nonSpontaneous (G>0) ++Nonspontaneous (G>0)Spontaneous (G<0)

Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:  S o =  J/K 2NH 3 (g)N 2 (g) + 3H 2 (g)  H o =  92.2 kJ

Example What is the standard free-energy change, Δ o G, for the following reaction at 25 o C? 2KClO 3 (g)  2KCl(s) + 3O 2 (g) ΔH o = -78.0kJΔS o =493.9J/K Is the reaction spontaneous at standard-state condition? Does the reaction become nonspontaneous at higher temperature?

Standard Free Energies of Formation  G o =  G o f (products)   G o f (reactants)  Go = [c  Go f (C) + d  Go f (D)]  [a  Go f (A) + b  Go f (B)] ReactantsProducts cC + dDaA + bB

Standard Free Energies of Formation Ch aptr 16/ 28

Standard Free Energies of Formation Using table values, calculate the standard free-energy change at 25 °C for the reduction of iron(III) oxide with carbon monoxide: 2Fe(s) + 3CO 2 (g)Fe 2 O 3 (s) + 3CO(g) Fe 2 O 3 (s)3CO(g)2Fe(s)3CO 2 (g) G o f (kJ/mol)

16.10 Free Energy Changes and the Reaction Mixture  G =  G° + RT ln Q  G = Free-energy change under nonstandard conditions. For the Haber synthesis of ammonia: 2NH 3 (g)N 2 (g) + 3H 2 (g) Q p = NH 3 P 2 H2H2 P N2N2 P 3

Free Energy Changes and the Reaction Mixture C2H4(g)C2H4(g)2C(s) + 2H 2 (g) Calculate  G for the formation of ethylene (C 2 H 4 ) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H 2 and 0.10 atm C 2 H 4. Is the reaction spontaneous in the forward or the reverse direction? C(s)H 2 (g)C 2 H 4 (g) ∆ G o f (kJ/mol)0061.8

16.11 Free Energy and Chemical Equilibrium  G =  G° + RT ln Q Q >> 1RT ln Q >> 0  G > 0 The total free energy decreases as the reaction proceeds spontaneously in the reverse direction. When the reaction mixture is mostly products: Q << 1RT ln Q << 0  G < 0 The total free energy decreases as the reaction proceeds spontaneously in the forward direction. When the reaction mixture is mostly reactants:

Free Energy and Chemical Equilibrium At equilibrium,  G = 0 and Q = K.  G o =  RT ln K  G =  G o + RT ln Q

Free Energy and Chemical Equilibrium Calculate K p at 25 oC for the following reaction: CaO(s) + CO 2 (g)CaCO 3 (s) CaO(s)CO 2 (g) G o f (kJ/mol)

Example The value of ΔG o f at 25 o C for gaseous mercury is kJ/mol. What is the vapor pressure of mercury at 25 o C? Hg(l) Hg(g)