1. Chapter 15

2. System - substances involved in the chemical and physical changes under investigation (for us this is what is happening inside the beaker) Surroundings - rest of the universe (outside the beaker) Universe - system plus surroundings Thermodynamic State of a System - set of conditions that describe and define the system Ex. number of moles of each substance; temperature; pressure; physical states of each substance

3. State Functions - properties of a system that depend only on the state of the system Normally capital letters Heat is a state function: it is a variable of a system that is independent of the path it took to get there. Most thermodynamic quantities are state functions. Properties that depend only on values of state functions are also state functions Ex: T, P, V

4. Enthalpy (heat changes) H Entropy (order and disorder) S Gibbs Free energy (thermodynamically favored or not) G We use thermo to control rxns Ex. raw food + heat  cooked

5. The total amount of energy in the universe is constant. ΔE rxn = 0 Energy involved in a chemical rxn is neither created nor destroyed. Energy can be converted from one form to another but cannot be created. Also known as Law of Conservation of Energy

6. I. systems tend toward a state of minimum potential energy

7. II. systems tend toward a state of maximum disorder

8. What is energy? The capacity to do work or transfer heat; hot stuff which changes other stuff. Energy causes changes in stuff. Examples: atomic rearrangement (rxns) C + O 2  CO 2 splitting/forming nuclei changes to structure nearly every kind of change

9. Potential – stored energy P.E. = mgh Kinetic – energy of moving objects K.E. = ½ mv 2 Energy depends on mass Bond energy (B.E.) – the amount of energy needed to break one mole of bonds in a covalent gas substance to form new gaseous products

10. In gas phase reactions  H o values may be related to bond energies of all species in the reaction. Use the bond energies listed in Table 15-2 & 15-3 to estimate the heat of reaction for 1) CCl 2 F 2 + F 2  CF 4 + Cl 2 2) CH 4 + O 2  CO 2 + H 2 O

11. Use the bond energies listed in Table 15-2 & 15-3 to estimate the heat of reaction for 1) CCl 2 F 2 + F 2  CF 4 + Cl 2

12. Use the bond energies listed to estimate the heat of reaction for 2) CH 4 + O 2  CO 2 + H 2 O

13. Reaction Diagrams – show amount of activation energy needed for the rxn to occur. Exothermic reactions release specific amounts of heat as products Potential energies of products are lower than potential energies of reactants.

14. Chemistry is done at constant pressure open beakers on a desk top are at atmospheric pressure  H - enthalpy change change in heat content at constant pressure  H = q p  H rxn - heat of reaction  H rxn = H products - H reactants  H rxn = H substances produced - H substances consumed

15. Change in enthalpy,  H, or heat of reaction is amount of heat absorbed or released when a reaction occurs at constant pressure. When:  H is > 0, the reaction is endothermic (heat is a reactant)  H is < 0, the reaction is exothermic (heat is a product)

16. Thermochemical standard state conditions T = 298.15 K P = 1.0000 atm Thermochemical standard states pure substances in their liquid or solid phase - standard state is the pure liquid or solid gases - standard state is the gas at 1.00 atm of pressure gaseous mixtures - partial pressure must be 1.00 atm aqueous solutions - 1.00 M concentration

17. Standard molar enthalpy of formation symbol is  H f o defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements for example:

18. Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text. Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which  H o rxn = -1281 kJ. P in standard state is P 4

19. Hess’s Law of heat summation – the overall enthalpy change in a rxn is equal to the sum of the enthalpy changes for the individual steps in the process. enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps ~ true because  H is a state function

20. 3. Using Hess’s law find ΔH rxn for 2HCl (g) + F 2(g)  2HF (l) + Cl 2(g) from the following rxns ΔH 4HCl (g) + O 2(g)  2H 2 O (l) + 2Cl 2(g) -202.4 kJ/mol ½ H 2(g) + ½ F 2(g)  HF (l) -600.0 kJ/mol H 2(g) + ½ O 2(g)  H 2 O (l)­ -285.8 kJ/mol

21. 4. ΔH for 4FeO (s) + O 2(g)  2Fe 2 O 3(s) is -560 kJ. Use the ΔH for the following two rxns to verify. ΔH 2Fe (s) + O 2(g)  2FeO-544 kJ 4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) -1648 kJ

22. Hess’s Law in a more useful form any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants

23. 2 nd Law of Thermodynamics - The universe tends toward a state of greater disorder and low energy in a thermodynamically favored (spontaneous) reaction…exothermic rxns tend to lower energy states and order requires more energy than disorder. Spontaneous ~ continues on its own after given activation energy to start the rxn. Happens without any continuing outside influences. Spontaneous processes require: ~ free energy change of system must be negative ~ entropy of universe must increase

24. rusting of iron - thermodynamically favored so it occurs spontaneously Have you ever seen rust turn into iron metal without help? melting of ice at room temperature - thermodynamically favored, occurs spontaneously Will water spontaneously freeze at room temperature? * Exothermicity does not ensure spontaneity Ex. Freezing of water exothermic and thermodynamically favored (spontaneous) only below 0 o C * Increase in disorder of the system also does not ensure spontaneity

25. Entropy is a measure of the disorder or randomness of a system. The entropy of the universe is increasing. When:  S is positive disorder increases (favors spontaneity)  S is negative disorder decreases (disfavors spontaneity) From 2 nd Law of Thermodynamics, for a spontaneous process In general: S gas > S liquid > S solid

26. 3 rd Law of Thermodynamics The entropy of a hypothetical pure, perfect, crystalline substance at absolute zero temperature is zero. allows us to measure absolute values of entropy for substances cool them down to 0 K, or as close as possible, then measure entropy increase as substance warms up Entropy changes for reactions can be determined similarly to  H for reactions. As with  H, entropies have been measured and tabulated in Appendix K as S o 298. When:

27. J. Willard Gibbs determined the relationship of enthalpy and entropy that best describes the maximum useful energy which can be obtained in the form of work from a process at constant Temperature & Pressure. The relationship also describes the spontaneity of a system. Whether the reaction is thermodynamically favored or not thermodynamically favored This is also a new state function,  G, the Gibbs Free Energy.

28. The change in the Gibbs Free Energy is a reliable indicator of spontaneity of a physical process or chemical reaction. does not tell us the speed of the process (kinetics does - Ch. 16) When:  G is > 0 reaction is not thermodynamically favored (nonspontaneous) reactant favored  G is = 0 system is at equilibrium  G is < 0 reaction is thermodynamically favored (spontaneous) product favored Changes in free energy obey the same type of relationship we have described for enthalpy and entropy changes.

29. Enthalpy and entropy can sometimes reinforce each other – this makes the reaction really go or really not go. Ex. Dynamite has a neg Δ H & a pos Δ S so the rxn really goes once started… If the signs don’t reinforce does a rxn occur? This is where Gibbs Free energy addresses the spontaneity of rxns. Ex. Liquid water to water vapor Δ H is + and Δ S is + Δ G is -, rxn is thermodynamically favored (spontaneous) – reaction will go on its own once started to make the products. Ex. gas burning. Δ G is +, rxn is not thermodynamically favored (nonspontaneous) – rxn won’t go on it’s own, wants to stay as reactants. Ex. batter  cake

30. State function +- Δ HEndothermic (taking in heat) ice melting Exothermic (giving off heat) dynamite Δ STowards disorderTowards order Δ Gnot thermodynamically favored thermodynamically favored Enthalpy (heat changes) H Entropy (order and disorder) S Gibbs Free energy (spontaneity) G We use thermo to control rxns Ex. raw food + heat  cooked

31. The general relationship of  G,  H, and  S is Which gives us 4 possibilities among the signs  H  S  GTherefore - +-forward rxn spontaneous at all T’s - -?forward rxn spontaneous at low T’s + +?forward rxn spontaneous at high T’s + -+forward rxn nonspontaneous at all T’s

32. 5. Find the a) enthalpy, b) entropy, and c) Gibbs free energy for the following rxn using data from Appendix K. C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) Also, using the information for a: is the rxn exothermic or endothermic, for b: more or less ordered, and for c: thermodynamically favored, (spontaneous/product favored), or not thermodynamically favored, (nonspontaneous/reactant favored)?

33. 6. Given that the ΔH f o for O 2 = 0; ΔH f o for SO 2 = -296.8 kJ/mole; ΔH f o for H 2 O = -285.8 kJ/mole and the ΔH rxn for the following balanced equation is = -1124 kJ, what is the Δ H f o for H 2 S? 2H 2 S + 3O 2  2SO 2 + 2H 2 O

34. 7. Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al 2 O 3 at 25 o C and one atmosphere. ΔH rxn is -3352 kJ/mole

35. 8. Calculate ΔS 298 for the combustion of propane. Δ H 298 = -2219.9 kJ, and ΔG 298 = -2108.5 kJ.

36. coffee-cup calorimeter - used to measure the amount of heat produced (or absorbed) in a reaction at constant P measures q P

37. exothermic reaction - heat evolved by reaction is determined from the temperature rise of the solution Amount of heat gained by calorimeter is the heat capacity of the calorimeter or calorimeter constant value determined by adding a specific amount of heat to calorimeter and measuring T rise

38. 9. When 3.425 kJ of heat is added to a calorimeter containing 50.00 g of water the temperature rises from 24.000 o C to 36.540 o C. Calculate the heat capacity of the calorimeter in J/ o C. The specific heat of water is 4.184 J/g o C.

39. 10. A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization of acetic acid and sodium hydroxide. When we add 25.00 mL of 0.500 M NaOH at 23.000 0 C to 25.00 mL of 0.600 M CH 3 COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947 0 C. The heat capacity of the calorimeter had previously been determined to be 27.8 J/ 0 C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g 0 C and that the density of the mixture is 1.02 g/mL. A) Calculate the amount of heat given off in the reaction. B) Determine ΔH for the reaction under the conditions of the experiment. (must determine the number of moles of reactants consumed; use limiting reactant)

40. When it rains an inch of rain, that means that if we built a one inch high wall around a piece of ground that the rain would completely fill this enclosed space to the top of the wall. Rain is water that has been evaporated from a lake, ocean, or river and then precipitated back onto the land. How much heat must the sun provide to evaporate enough water to rain 1.0 inch onto 1.0 acre of land? 1 acre = 43,460 ft 2 vaporization of water = 44.0 kJ/mol HH