1. 1 Thermochemistry Chapter 6

2. 2 Energy and change The study of the energy involved in a change is THERMODYNAMICS In thermodynamics, the universe is divided into System (what you are studying) and Surroundings (the rest of the universe!)

3. 3 Systems A system may be OPEN  both matter and energy can be exchanged with the surroundings

4. 4 Systems A system may be CLOSED  only energy can be exchanged with the surroundings

5. 5 Systems A system may be ISOLATED  neither matter nor energy can be exchanged with the surroundings

6. 6 Systems may be open, closed, or isolated

7. 7 Internal energy The internal energy E of a system is all the energy (both kinetic & potential) contained in the system Chemists are especially interested in  thermal energy (energy of random molecular motion)  chemical energy (energy stored in chemical bonds and intermolecular forces)

8. 8 First Law of Thermodynamics The energy a system contains is its internal energy, E Energy can move in or out of a system as heat (q) and/or work (w). Energy is conserved, so all energy lost or gained by a system must be accounted for as heat and/or work:

9. 9 E is a state function A state function (aka function of state) is a property whose value depends only on the state of the system, not how it achieved that state The state of the system is specified by the pressure, temperature, and composition of the system. P, V, & T are state functions. E is a state function. q and w are NOT state functions.

10. 10 The elevation gain is the same by either path: a state function How ∆E is distributed between q and w depends on which path you take: q and w are not state functions

11. 11 fully charged battery fully discharged battery Path 1: short circuit the battery with a wrench. Lots of heat, maybe even sparks and a fire, but no work! Path 2: connect the battery to a motor. Some heat, but also some work. ∆E the same by either path

12. 12 Heat Heat is energy transferred because of a temperature difference  A system does not contain heat; it contains ENERGY  Heat is just a form by which energy is transferred  The other form by which energy is transferred is work Whether q or w, energy IN is positive, energy OUT is negative!

13. 13 Heat The amount of energy transferred as heat, q, is related to  the amount of matter gaining/losing energy (m)  the specific heat of matter gaining/losing energy (c)  the amount of the temperature change (∆T) James Prescott Joule

14. 14 Heat capacity and specific heat Specific heat capacity = c  Intrinsic capacity to gain/lose energy as heat  unit = J/g K or J/g °C  Molar specific heat capacity = J/mol K or J/mol °C Heat capacity of a system = C  Quantity of energy to change temperature by 1 °C  unit = J/K or J/°C

15. 15 Significance of specific heat values Temperature change  change in molecular motion  Low specific heat  less energy to increase motion  High specific heat  more energy to increase motion specific heat substance (J/g °C) air1.00 aluminum0.895 copper0.387 lead0.128 ethanol2.45 water4.18

16. 16 Examples How much heat energy (in kJ) is required to raise the temperature of 237 g ice water from 4.0 °C to 37.0 °C? c water = 4.18 J/g °C How much heat energy (in kJ) is required to raise the temperature of 2.50 kg Hg from –20.0 °C to – 6.0 °C? c Hg = 28.0 J/mol °C

17. 17 Determination of specific heat Heat sample of metal to temperature of boiling water Measure temperature of measured amount of water in insulated beaker Put hot metal in cold water and measure final temperature All energy is assumed to stay in the insulated container  q metal + q water = 0 or q metal = – q water

18. 18 Examples When 1.00 kg Pb (specific heat = 0.13 J/g °C) at 100.0 °C is added to some water at 28.5 °C, the final temperature is 35.2 °C. What is the mass of the water? 100.0 g Cu (specific heat 0.385 J/g °C) at 100.0 °C is added to 50.0 g water at 26.5 °C. What is the final temperature of the mixture?

19. 19 Work Work is done when a force acts for a distance: w = F x d Energy is the capacity to do work  kinetic energy is the energy of motion S the random motion of molecules (thermal energy) is kinetic  potential energy is stored energy, that can do work when it is released S the energy in chemical bonds is potential energy

20. 20 Pressure-volume work Imagine a gas trapped in a cylinder with a moveable lid (a piston) If the piston is not moving, how does the gas pressure inside compare to the external pressure? If the piston is not moving, P gas = P external P external   P gas

21. 21 Pressure-volume work What if P gas increases, so P gas > P external ? P external   P gas

22. 22 Pressure-volume work The piston rises and the gas expands, until once again P gas = P external The gas has moved the piston some distance against the opposing P ext The gas has done work! P external   P gas

23. 23 Pressure-volume work How much work did the gas do?  P gas P external  ∆h

24. 24 Pressure-volume work How much work did the gas do? When the gas did work by expanding, it lost energy  P gas P external  ∆h

25. 25 Pressure-volume work Now what if we increase the external pressure, so P external > P gas ? P external   P gas

26. 26 Pressure-volume work The gas is compressed until once again P gas = P external The piston has been moved some distance against the opposing P gas This time work was done on the gas How much work?  P gas P external  ∆h

27. 27 Pressure-volume (P∆V) work When the gas expands, it expends energy to do work = P ext ∆V  ∆V is positive  energy leaves the system (the gas) When the gas is compressed, it gains the energy used to compress it = P ext ∆V  ∆V is negative  energy enters the system (the gas)

28. 28 Sign conventions for work When energy goes into the system as work, w is positive When energy leaves the system as work, w is negative w = –P∆V when a gas expands ∆V is + and w is – (energy leaves the system) when a gas is compressed ∆V is – and w is + (energy enters the system)

29. 29 Units w = –P∆V  work is in joules (J)  P is in atm and V is in L, so P∆V is in atm L The relationship between J and atm L is 1 atm L = 101.325 J

30. 30 Example What is the work done on a gas (in J) when the gas is compressed from an initial volume of 35.0 L to a final volume of 23.5 L under a constant pressure of 0.987 atm?

31. 31 Calorimetry The process for measuring the amount of heat energy exchanged by system and surroundings is CALORIMETRY Assume total energy is constant: heat lost by system is gained by surroundings q system + q surroundings = 0 q system = – q surroundings

32. 32 Heat of reaction, q rxn If the reaction releases energy, q rxn is negative (the reaction loses energy) The calorimeter gains that energy and gets hotter The reaction is EXOTHERMIC q rxn = – q calorimeter If the reaction absorbs energy, q rxn is positive (the reaction gains energy) The calorimeter loses that energy and gets colder The reaction is ENDOTHERMIC q rxn = – q calorimeter

33. 33 Bomb calorimeter Combustion reaction occurs in sample compartment Reaction releases energy to water in calorimeter q rxn = – q calorimeter = – C∆T  C = heat capacity of calorimeter

34. 34 Examples The combustion of 1.013 g vanillin (C 8 H 8 O 3 ) in a bomb calorimeter with heat capacity = 4.90 kJ/°C causes the temperature to rise from 24.89 °C to 30.09 °C. What is the heat of combustion of vanillin, in kJ/mol? Combustion of 1.176 g benzoic acid (HC 7 H 5 O 2, heat of combustion –26.42 kJ/g) causes the temperature in a bomb calorimeter to increase by 4.96 °C. What is C for that calorimeter?

35. 35 Coffee cup calorimeter Reactants (usually in aqueous solution) mix and react in the insulated cup Reaction releases energy to (or absorbs energy from) liquid in which it is dissolved q rxn = – q cal = – (mc∆T) cal  Calorimeter = liquid in the cup (cup assembly is usually ignored)

36. 36 Example 100.0 mL 1.00 M AgNO 3 (aq) and 100.0 mL 1.00 M NaCl (aq), both initially at 22.4 °C, are mixed in a coffee cup calorimeter. The temperature rises to 30.2 °C. What is the heat of reaction, in kJ per mol AgCl, for the reaction: Ag 1+ (aq) + Cl 1– (aq)  AgCl (s)

37. 37 Example 100.0 mL 1.020 M HCl and 50.0 mL 1.988 M NaOH, both initially at 24.52 °C, are mixed in a coffee cup calorimeter. If q neutralization = –56 kJ/mol H 2 O, what will be the final temperature in the mixture?  Hint: this is a limiting reactant problem

38. 38 Energy and Enthalpy ∆E = q + w  w = –P∆V  At constant volume, ∆V = 0 so w = 0 and ∆E = q V We define H = E + PV  H is called enthalpy  At constant pressure, ∆H = ∆E + P∆V  But ∆E = q P – P∆V, so ∆H = q P – P∆V + P∆V  ∆H = q P

39. 39 ∆E and ∆H At constant volume, measured heat = ∆E At constant pressure, measured heat = ∆H  Constant pressure is the more common lab condition  ∆H is the heat of reaction for a chemical change carried out at constant pressure

40. 40 How different are ∆E and ∆H? C (s) + 1/2 O 2 (g)  CO (g) ∆H = – 110.5 kJ ∆H = ∆E + ∆(PV) or ∆E = ∆H – ∆(PV)  ∆(PV) = ∆(nRT) = ∆n gas RT (solids are not significant)  ∆n gas = 1/2 mol  At 298 K,  ∆E is – 111.7 kJ, only 1% difference from ∆H

41. 41 Enthalpy Like E, absolute values of H cannot be measured, but changes in H can be measured: ∆H = q P ∆H is part of the chemical reaction stoichiometry  ∆H is extensive (depends on amount of material)  ∆H is directional (sign reverses if direction of reaction reverses) ∆H is a state function  Value of ∆H same whether reaction occurs in a single step or a series of steps, if final result is the same

42. 42 Reactions add together Look at these reactions: C (s) + O 2 (g)  CO (g) + 1/2 O 2 (g) CO (g) + 1/2 O 2 (g)  CO 2 (g) The reactions add to give C (s) + O 2 (g)  CO 2 (g) The CO and 1/2 O 2 terms on both sides cancel

43. 43 Heats of reaction add, too  H values add just like reactions: ∆H (kJ/mol) C (s) + O 2 (g)  CO (g) + 1/2 O 2 (g) –111 kJ/mol CO (g) + 1/2 O 2 (g)  CO 2 (g) –283 kJ/mol C (s) + O 2 (g)  CO 2 (g) –394 kJ/mol

44. 44 Hess’ Law Germain Hess discovered that heats of reaction add together in 1840 The additivity of ∆H values is called Hess’ Law Hess’ Law establishes enthalpy as a state function  State functions depend only on the state of the system  It doesn’t matter how the system achieved that state

45. 45 Hess’ Law Hess’ Law allows us to calculate ∆H for a reaction from measured ∆H values for other reactions The rules of the game  Multiply reaction by a factor to get desired coefficients  Multiply ∆H by the same factor  Reverse a reaction to get a substance on the desired side  If you reverse the reaction, reverse the sign of ∆H

46. 46 Example Given these heats of reaction,  H 2 (g) + 1/2 O 2 (g)  H 2 O (l) ∆H = –286 kJ  2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (l) ∆H = –3123 kJ  2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O (l) ∆H = –2602 kJ Find ∆H for this reaction: C 2 H 2 (g) + 2 H 2 (g) ––> C 2 H 6 (g)

47. 47 Example First look at the target reaction: C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g) You want 1 C 2 H 2 on the left. We need to multiply the reaction with C 2 H 2 by 1/2:  1/2 [2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O (l)]  C 2 H 2 (g) + 5/2 O 2 (g)  2 CO 2 (g) + H 2 O (l) Multiply ∆H by 1/2: ∆H = 1/2[–2602 kJ] = –1301 kJ

48. 48 Example Another look at the target reaction: C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g) You need 2 H 2 on the left. We need to multiply the reaction with H 2 by 2:  2 [H 2 (g) + 1/2 O 2 (g)  H 2 O (l)]  2 H 2 (g) + O 2 (g)  2 H 2 O (l) Multiply ∆H by 2: ∆H = 2[–286 kJ] = –572 kJ

49. 49 Example The target reaction: C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g) Finally, you want 1 C 2 H 6 on the right. Multiply the reaction with C 2 H 6 by 1/2, and reverse it:  1/2 [2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O (l)]  reverse C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l)  2 CO 2 (g) + 3 H 2 O (l)  C 2 H 6 (g) + 7/2 O 2 (g) Multiply ∆H by 1/2, and reverse its sign: ∆H = 1/2[+3123 kJ] = +1561 kJ

50. 50 Example Now you have  C 2 H 2 (g) + 5/2 O 2 (g)  2 CO 2 (g) + H 2 O (l) ∆H = –1301 kJ  2 H 2 (g) + O 2 (g)  2 H 2 O (l) ∆H = –572 kJ  2 CO 2 (g) + 3 H 2 O (l)  C 2 H 6 (g) + 7/2 O 2 (g) ∆H = +1561 kJ  Now add the reactions and cancel items identical on both sides The total is C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g) ∆H = – 312 kJ

51. 51 Formation reactions To organize collections of ∆H values for reactions, chemists defined a formation reaction  A formation reaction is the equation to form one mole of compound from its elements in their most stable state The formation reaction for H 2 O is H 2 (g) + 1/2 O 2 (g)  H 2 O (l)  Notice that the reactants are elements in their most common, stable form  Notice that only one mole of product forms

52. 52 Standard enthalpy of formation The enthalpy of a formation reaction is called the standard heat of formation (symbol ∆H f °).  The little f means a formation reaction  The ° means standard conditions, 1 atm and 1 M solutions ∆H f ° values are collected in reference books (see Appendix II in the back of your text)  ∆H f ° for an element in its most stable state is zero  The state (liquid vs. gas or solid) is important  Equations are not given (you are supposed to know how the write the appropriate formation reaction)

53. 53 Heats of formation combine You can combine formation equations and their ∆H f ° values just like any other reactions Same rules of the game:  If you reverse the reaction, reverse the sign of ∆H f °  If you multiply the reaction by some factor, multiply ∆H f ° by the same factor

54. 54 But, of course, there’s a SHORTCUT ∆H° = ∑n∆H f ° products – ∑n∆H f ° reactants  n is the coefficient for the substance, in moles C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g)  ∆H f ° values: C 2 H 2 = +227 kJ/mol, H 2 = 0 kJ/mol (element), C 2 H 6 = –85 kJ/mol  ∆H° = [1(∆H f ° C 2 H 6 )] – [1(∆H f ° C 2 H 2 ) + 2(∆H f ° H 2 )]  ∆H° = [1 mol (–85 kJ/mol)] – [1 mol (+227 kJ/mol) + 2 mol (0 kJ/mol)]  ∆H° = –312 kJ

55. 55 A shortcut for the shortcut An easy way to set up the shortcut is  Write the target equation  Beneath each substance, write its ∆H f ° (watch signs!)  Beneath ∆H f °, write the coefficient for that substance  Multiply each ∆H f ° by its coefficient  Add the answers for the products, add the answers for the reactants  ∆H° = products – reactants

56. 56 The short shortcut C 2 H 2 (g) + 2 H 2 (g)  C 2 H 6 (g) +227 kJ 0 kJ –85 kJ x 1 x 2 x 1 +227 kJ 0 kJ –85 kJ +227 kJ  –85 kJ ∆H° = (–85 kJ) – (+227 kJ) = –312 kJ The same result we got earlier by adding reactions

57. 57 Reactions in aqueous solution Additivity also applies to reactions in solution  ∆H f ° values for ions are given in the Appendix  The baseline for aqueous ions is H 1+ (aq) = 0 kJ/mol instead of the element form  All rules apply as before