1. CHM 112 Summer 2007 M. Prushan Chapter 17 Thermodynamics: Entropy, Free Energy, and Equilibrium

2. CHM 112 Summer 2007 M. Prushan Questions What factors determine the direction and extent of a chemical reaction? –Combustion of hydrocarbon fuels goes almost to completion –Combination of gold and oxygen occurs hardly at all –Industrial synthesis of ammonia from N 2 and H 2 at 400- 500ºC results in an equilibrium mixture Extent of particular reaction described by K –But what determines value of K? → thermodynamics

3. CHM 112 Summer 2007 M. Prushan Spontaneous Processes & Entropy The 2 nd Law explains why chemical reactions tend to favor a particular direction. It is important to predict whether a reaction will occur. A reaction that does occur under specific conditions is called a spontaneous reaction. A reaction that does not occur under specific conditions is called a nonspontaneous reaction.

4. CHM 112 Summer 2007 M. Prushan Spontaneous Processes & Entropy A spontaneous reaction will always move a reaction mixture toward equilibrium. Reaction spontaneity is independent of reaction rate.

5. CHM 112 Summer 2007 M. Prushan Spontaneous Processes & Entropy To predict spontaneity we need to know the energy change and the entropy. Entropy (S) is a “measure” of the randomness or disorder of a system. The greater the disorder, the greater the entropy. Nature tends to the greatest entropy. S solid < S liquid < S gas

6. CHM 112 Summer 2007 M. Prushan Standard Entropy is the absolute entropy of a substance at 1 atm and 25°C. Change in entropy is given by ∆S = S f – S i Spontaneous Processes & Entropy

7. CHM 112 Summer 2007 M. Prushan The decomposition of N 2 O 4 (O 2 N–NO 2 ) is also accompanied by an increase in randomness. Whenever molecules break apart, randomness increases. Spontaneous Processes & Entropy

8. CHM 112 Summer 2007 M. Prushan Spontaneous Processes & Entropy When NaCl dissolves in water, the crystal breaks up, and the ions are surrounded by water molecules.

9. CHM 112 Summer 2007 M. Prushan Example (a)The rusting of cars is spontaneous, but it occurs slowly (b)Spontaneous reaction occurs slowly if it has high E a

10. CHM 112 Summer 2007 M. Prushan Example The combustion of natural gas (mainly CH 4 ) in air is a spontaneous, exothermic reaction

11. CHM 112 Summer 2007 M. Prushan Question: Consider the gas phase reaction of A 2 molecules (red) with B atoms (blue). –(a) Write a balanced equation for the reaction. –(b) Predict the sign of ∆S for the reaction.

12. CHM 112 Summer 2007 M. Prushan Entropy and Temperature Entropy is associated with molecular motion. As temperature increases, entropy increases. Third Law of Thermodynamics: Entropy of a perfectly ordered crystalline substance at 0 K is zero. At the melting and boiling point there is a discontinuous jump in entropy.

13. CHM 112 Summer 2007 M. Prushan Entropy and Temperature

14. CHM 112 Summer 2007 M. Prushan Entropy and Temperature

15. CHM 112 Summer 2007 M. Prushan Standard Molar Entropy, S°: The entropy of 1 mol of the pure substance at 1 atm pressure and a specified temperature, usually 25°C. Standard molar entropies are absolute entropies measured against an absolute reference point. Standard entropy of reaction: –∆S° = ∑nS° (products) – ∑nS° (reactants) Standard Molar Entropies

16. CHM 112 Summer 2007 M. Prushan Standard Molar Entropies

17. CHM 112 Summer 2007 M. Prushan Standard Molar Entropies Calculate the standard entropy of reaction at 25°C for the synthesis of ammonia: –N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Calculate the standard entropy of reaction at 25°C for the decomposition of calcium carbonate: –CaCO 3 (s)  CaO(s) + CO 2 (g)

18. CHM 112 Summer 2007 M. Prushan Standard Molar Entropies Predict the entropy change, and then calculate the standard entropy change for the following reactions at 25°C. –a. 2 CO(g) + O 2 (g)  2 CO 2 (g) –b. 3 O 2 (g)  2 O 3 (g) –c. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g)

19. CHM 112 Summer 2007 M. Prushan 2 nd Law of Thermodynamics The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. –Spontaneous: ∆S total = ∆S sys + ∆S sur > 0 –Equilibrium: ∆S total = ∆S sys + ∆S sur = 0 The system is what you observe; surroundings are everything else.

20. CHM 112 Summer 2007 M. Prushan 2 nd Law of Thermodynamics For ∆S total, we need to know ∆S sys and ∆S surr. ∆S sys is often determined from the standard entropy of reaction, ∆S° rxn : ∆S° rxn = ∑nS° (Products) – ∑nS° (Reactants)

21. CHM 112 Summer 2007 M. Prushan 2nd Law of Thermodynamics Entropy Changes to the Surroundings: Exothermic  increase ∆S surr Endothermic  decrease ∆S surr  S surr   H sys T

22. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy The 2nd law tells us a process will be spontaneous if ∆S total > 0 which requires a knowledge of ∆S surr. Since we only concern ourselves with the system we derive an expression using only ∆S sys. –T∆S total = ∆H sys – T∆S sys < 0

23. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy The expression –T∆S total is equated as Gibbs free energy change (∆G), or simply free energy change: ∆G = ∆H – T∆S sys ∆G 0 Reaction is spontaneous in reverse direction.

24. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S. If both ∆H and ∆S are positive, ∆G will be negative only when the temperature value is large. If ∆H is positive and ∆S is negative, ∆G will always be positive. If ∆H is negative and ∆S is positive, ∆G will always be negative. If both ∆H and ∆S are negative, ∆G will be negative only when the temperature value is small.

25. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy

26. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe 2 O 3 (s) + 3 H 2 (g)  2 Fe(s) + 3 H 2 O(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?

27. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?

28. CHM 112 Summer 2007 M. Prushan Standard Free Energy (∆G˚ rxn ) is the free energy for a reaction occurring under standard state conditions. Reactants in their standard states are converted to products in their standard states. ∆G˚ rxn = ∑ n ∆G˚ ƒ (products) – ∑ n ∆G˚ ƒ (reactants) ∆G˚ ƒ is the standard free energy of formation of a compound from its elements in their standard states. Gibbs Free Energy

29. CHM 112 Summer 2007 M. Prushan Gibbs Free Energy

30. CHM 112 Summer 2007 M. Prushan Free Energy and Chemical Equilibrium The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. Under nonstandard conditions, ∆G˚ becomes ∆G. ∆G = ∆G˚ + RT lnQ The reaction quotient is obtained in the same way as an equilibrium expression.

31. CHM 112 Summer 2007 M. Prushan Free Energy and Chemical Equilibrium If ∆G˚ is a large negative value, RT lnQ will not be positive enough to match ∆G˚ until a significant amount of product is formed. If ∆G˚ is a large positive value, RT lnQ will only be more negative when very little product has been formed.

32. CHM 112 Summer 2007 M. Prushan Free Energy and Chemical Equilibrium

33. CHM 112 Summer 2007 M. Prushan Using the solubility product of silver chloride at 25°C (1.6 x 10 –10 ), calculate ∆G˚ for the process: AgCl(s) Ag + (aq) + Cl – (aq) The ∆G˚ for the reaction H 2 (g) + I 2 (g) 2 HI(g) is 2.60 kJ at 25°C. In one experiment, the initial pressures are P H 2 = 4.26 atm, P I 2 = 0.024 atm, and P HI = 0.23 atm. Calculate ∆G for the reaction and predict the reaction direction.