1. Energy & Chemistry 2H 2(g) + O 2(g) → 2H 2 O (g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H 2 → 4 H + + 4 e - Reduction: 4 e - + O 2 + 2 H 2 O → 4 OH - H 2 /O 2 Fuel Cell Energy, page 288

2. Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — lightlight electricalelectrical kinetic and potentialkinetic and potential Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy NaCl — composed of Na + and Cl - ions.

3. Potential & Kinetic Energy Kinetic energy — energy of motion.

4. © 2006 Brooks/Cole - Thomson

5. Internal Energy (E) PE + KE = Internal energy (E or U) Internal Energy of a chemical system depends on number of particles type of particles temperature The higher the T the higher the internal energy So, use changes in T (∆T) to monitor changes in E (∆E).

6. Thermodynamics Thermodynamics is the science of heat (energy) transfer. Heat transfers until thermal equilibrium is established. ∆T measures energy transferred. SYSTEM –The object under study SURROUNDINGS –Everything outside the system

7. Directionality of Heat Transfer Heat always transfer from hotter object to cooler one. EXO thermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down T(surr) goes up

8. Directionality of Heat Transfer Heat always transfers from hotter object to cooler one. ENDO thermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T (surr) goes down

9. Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE. Energy Change in Chemical Processes Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.

10. HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?

11. Specific Heat Capacity How much energy is transferred due to Temperature difference? The heat (q) “lost” or “gained” is related to a)sample mass b) change in T and c) specific heat capacity Specific heat capacity= heat lost or gained by substance (J) (mass, g) (T change, K)

12. © 2006 Brooks/Cole - Thomson Table of specific heat capacities SubstancePhasec p J g -1 K -1 C p J mol -1 K -1 Air (typical room conditions A )gas1.01229.19 Aluminiumsolid0.89724.2 Argongas0.520320.7862 Coppersolid0.38524.47 Diamondsolid0.50916.115 Ethanolliquid2.44112 Goldsolid0.129125.42 Graphitesolid0.7108.53 Heliumgas5.193220.7862 Hydrogengas14.3028.82 Ironsolid0.45025.1 Lithiumsolid3.5824.8 Mercuryliquid0.139527.98 Nitrogengas1.04029.12 Neongas1.030120.7862 Oxygengas0.91829.38 Uraniumsolid0.11627.7 Water gas (100 °C)2.08037.47 liquid (25 °C)4.181375.327 solid (0 °C)2.11438.09 All measurements are at 25 °C unless noted. Notable minimums and maximums are shown in maroon text. A Assuming an altitude of 194 meters above mean sea level (the world–wide median altitude of human habitation), an indoor temperature of 23 °C, a dewpoint of 9 °C (40.85% relative humidity), and 760 mm–Hg sea level–corrected barometric pressure (molar water vapor content = 1.16%). Aluminum

13. Specific Heat Capacity If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al?

14. Heat/Energy Transfer No Change in State q transferred = (sp. ht.)(mass)(∆T)

15. Heat Transfer Use heat transfer as a way to find specific heat capacity, C p 55.0 g Fe at 99.8 ˚C Drop into 225 g water at 21.0 ˚C Water and metal come to 23.1 ˚C What is the specific heat capacity of the metal?

16. Heating/Cooling Curve for Water Note that T is constant as ice melts or water boils

17. Chemical Reactivity But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.

18. FIRST LAW OF THERMODYNAMICS ∆E = q + w heat energy transferred energychange work done by the system Energy is conserved!

19. © 2006 Brooks/Cole - Thomson The First Law of Thermodynamics Exothermic reactions generate specific amounts of heat. This is because the potential energies of the products are lower than the potential energies of the reactants.

20. © 2006 Brooks/Cole - Thomson The First Law of Thermodynamics There are two basic ideas of importance for thermodynamic systems. 1.Chemical systems tend toward a state of minimum potential energy. 2.Chemical systems tend toward a state of maximum disorder. The first law is also known as the Law of Conservation of Energy. –Energy is neither created nor destroyed in chemical reactions and physical changes.

21. heat transfer out (exothermic), -q heat transfer in (endothermic), +q SYSTEM ∆E = q + w w transfer in (+w) w transfer out (-w)

22. © 2006 Brooks/Cole - Thomson Some Thermodynamic Terms Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! Some examples of state functions are: –T (temperature), P (pressure), V (volume),  E (change in energy),  H (change in enthalpy – the transfer of heat), and S (entropy) Examples of non-state functions are: –n (moles), q (heat), w (work)  ∆H along one path =  ∆H along another path This equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not how it got there. V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

23. © 2006 Brooks/Cole - Thomson Some Thermodynamic Terms The properties of a system that depend only on the state of the system are called state functions. –State functions are always written using capital letters. The value of a state function is independent of pathway. An analog to a state function is the energy required to climb a mountain taking two different paths. –E 1 = energy at the bottom of the mountain –E 1 = mgh 1 –E 2 = energy at the top of the mountain –E 2 = mgh 2 –  E = E 2 -E 1 = mgh 2 – mgh 1 = mg(  h)

24. © 2006 Brooks/Cole - Thomson Standard States and Standard Enthalpy Changes Thermochemical standard state conditions –The thermochemical standard T = 298.15 K. –The thermochemical standard P = 1.0000 atm. Be careful not to confuse these values with STP. Thermochemical standard states of matter –For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. –For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm. –For aqueous solutions the standard state is 1.00 M concentration. ∆H f o = standard molar enthalpy of formation the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

25. © 2006 Brooks/Cole - Thomson ENTHALPY Most chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial Heat transferred at constant P = q p q p = ∆H where H = enthalpy

26. © 2006 Brooks/Cole - Thomson If H final < H initial then ∆H is negative Process is EXOTHERMIC If H final > H initial then ∆H is positive Process is ENDOTHERMIC ENTHALPY ∆H = H final - H initial

27. © 2006 Brooks/Cole - Thomson Consider the formation of water H 2 (g) + 1/2 O 2 (g) → H 2 O(g) + 241.8 kJ USING ENTHALPY Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

28. © 2006 Brooks/Cole - Thomson Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. USING ENTHALPY H 2 + O 2 gas Liquid H 2 O H 2 O vapor Making H 2 O from H 2 involves two steps. H 2(g) + 1/2 O 2(g) → H 2 O (g) + 242 kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

29. © 2006 Brooks/Cole - Thomson Enthalpy Values H 2(g) + 1/2 O 2(g) → H 2 O (g) ∆H˚ = -242 kJ 2H 2(g) + O 2(g) → 2H 2 O (g) ∆H˚ = -484 kJ H 2 O (g) → H 2(g) + 1/2 O 2(g) ∆H˚ = +242 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) ∆H˚ = -286 kJ Depend on how the reaction is written and on phases of reactants and products

30. © 2006 Brooks/Cole - Thomson Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Active Figure 6.18

31. © 2006 Brooks/Cole - Thomson Thermochemical Equations Thermochemical equations are a balanced chemical reaction plus the  H value for the reaction. –For example, this is a thermochemical equation. The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles. 1 mol of C 5 H 12 reacts with 8 mol of O 2 to produce 5 mol of CO 2, 6 mol of H 2 O, and releasing 3523 kJ is referred to as one mole of reactions.

32. © 2006 Brooks/Cole - Thomson Hess’s Law Hess’s Law of Heat Summation,  H rxn =  H 1 +  H 2 +  H 3 +..., states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. –Hess’s Law is true because  H is a state function. If we know the following  H o ’s

33. © 2006 Brooks/Cole - Thomson Hess’s Law For example, we can calculate the  H o for reaction [1] by properly adding (or subtracting) the  H o ’s for reactions [2] and [3]. Notice that reaction [1] has FeO and O 2 as reactants and Fe 2 O 3 as a product. –Arrange reactions [2] and [3] so that they also have FeO and O 2 as reactants and Fe 2 O 3 as a product. Each reaction can be doubled, tripled, or multiplied by half, etc. The  H o values are also doubled, tripled, etc. If a reaction is reversed the sign of the  H o is changed.

34. © 2006 Brooks/Cole - Thomson Hess’s Law Given the following equations and  H o  values calculate  H o for the reaction below.

35. © 2006 Brooks/Cole - Thomson Hess’s Law Use a little algebra and Hess’s Law to get the appropriate  H o  values

36. © 2006 Brooks/Cole - Thomson Thermochemical Equations This is an equivalent method of writing thermochemical equations.  H < 0 designates an exothermic reaction.  H > 0 designates an endothermic reaction

37. © 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. –The symbol for standard molar enthalpy of formation is  H f o. The standard molar enthalpy of formation for MgCl 2 is:

38. © 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o 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. Example 15-4: 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 Phosphoric acid in standard state is H 3 PO 4(s)

39. © 2006 Brooks/Cole - Thomson Hess’s Law Hess’s Law in a more useful form. –For 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.

40. © 2006 Brooks/Cole - Thomson Hess’s Law

41. © 2006 Brooks/Cole - Thomson ∆H f o, standard molar enthalpy of formation ½˚ H 2(g) + ½ O 2(g) → H 2 O (g) ∆H f ˚ (H 2 O, g )= -241.8 kJ/mol C (s) + ½ O 2(g) → CO (g) ∆H f ˚ of CO = - 111 kJ/mol By definition, ∆H f o = 0 for elements in their standard states. Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) → H 2 (g) + CO(g)

42. © 2006 Brooks/Cole - Thomson Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆H o rxn for CH 3 OH(g) + 3/2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

43. © 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o Calculate the enthalpy change for the reaction of one mole of H 2(g) with one mole of F 2(g) to form two moles of HF (g) at 25 o C and one atmosphere.

44. © 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o 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.

45. © 2006 Brooks/Cole - Thomson Hess’s Law Calculate the  H o 298 for the following reaction from data in Appendix K.

46. © 2006 Brooks/Cole - Thomson Hess’s Law Application of Hess’s Law and more algebra allows us to calculate the  H f o  for a substance participating in a reaction for which we know  H rxn o, if we also know  H f o  for all other substances in the reaction. Given the following information, calculate  H f o for H 2 S (g).

47. © 2006 Brooks/Cole - Thomson Thermochemical Equations Write the thermochemical equation for the reaction for CuSO 4(aq) + 2NaOH (aq) Cu(OH) 2(s) + Na 2 SO 4(aq) 50.0mL of 0.400 M CuSO4 at 23.35 o C T final 25.23 o C 50.0mL of 0.600 M NaOH at 23.35 o C Density final solution = 1.02 g/mL C H 2 O = 4.184 J/g o C