1. Chemistry 101 : Chap. 5 Thermochemistry (1) The Nature of Energy (2) The First Law of Thermodynamics (3) Enthalpy (4) Enthalpy of Reaction (5) Calorimetry (6) Hess’s Law (7) Enthalpy of Formation NOTE: We will discuss chapter 19, “Chemical Thermodynamics”, after this chapter

2. Nature of Energy  Thermodynamics : The study of energy and its transformation Energies can be in many different forms. These energy changes can be as simple as those associated with a falling object to complexes processes like metabolism  Thermochemistry : Thermodynamics associated with chemical processes. The study of energy changes accompanying chemical reactions

3. Nature of Energy  Energy : The capacity to do work or transfer heat  Work : Energy used to move an object with mass work = force  distance  Heat : Energy used to increase the temperature of an object An object can do work against the environment or transfer heat to the environment. In such case, the energy of an object needs to be converted to work or heat. An object can possess two fundamentally different kinds of energy : Kinetic energy and Potential energy

4. Nature of Energy  Kinetic Energy : The energy an object has due to its motion mass velocity Ex. (1) A moving car (macroscopic object) (2) Atoms and molecules are constantly moving.  This is the source of thermal energy (temperature increase or decrease)

5. Nature of Energy  Potential Energy : The energy an object has due to its position relative to other objects All objects in the universe are interacting with each other. In other words, they push or pull surrounding objects (exerting force). Potential energy arises when there is any kind of force acting on an object. Ex. (1) Gravitation Energy : Gravitational Force : For the gravitational energy between an object (m 1 = m) on earth and earth (m 2 = m earth ) itself, r ~ radius of earth = r earth. F g, earth = m  (Gm earth /r earth 2 ) = mg This is the formal definition of “weight” Gravitation potential energy (on earth) = mgh (or gravitational potential energy)

6. Nature of Energy (2) Electrostatic Energy : Electrostatic interaction is either attractive (charges with opposite signs) or repulsive (charges with same signs). (3) Chemical Energy : Potential energy stored in the arrangement of the atoms in the substance of interest (mostly electrostatic) One of the important goal in chemistry is to relate the energy change that we see in macroscopic world (i.e. in laboratory experiment) to the kinetic energy and potential energy of substances at the atomic and molecular level. electrical charge (in C)

7. The Units of Energy  joule (J) : 1 J = 1 kg  m 2 /s 2 The kinetic energy of a 2kg object with the speed of 1 m/s is equal to 1J: E k = (1/2)mv 2 = (1/2)  2 kg  (1 m/s) 2 = 1 kg  m 2 /s 2 = 1 J James Joule (1818 ~ 1889) NOTE: kg, m and s are the SI units of mass, distance and time. Therefore, joule is the SI unit of energy.  calorie (cal) : This is an older unit of energy. But, it is still frequently used in chemistry. 1 cal = 4.184 J

8. The Units of Energy  Example : If a hot dog is burned, it produces 180 dietary Calories (1 Cal = 1000 cal). How many joules is this?

9. System and Surrounding We divide the universe into two parts: system and surrounding System = The portion of the universe that we are interested in Surrounding = Everything else system surrounding * No exchange of particles between system and surrounding (closed system) * Energy (heat and work) exchange is allowed between system and surrounding

10. Transferring Energy Energy can be transferred between a system and its surrounding in the form of heat (q) and/or work (w)  Work = Force  Distance ( w = f  d ) Energy (work) is transferred from surrounding to an object (system) when the object is moved by external force.  Heat : Energy (heat) is transferred from an object at higher temperature to one at a lower temperature

11. Transferring Energy Heat Transfer: Whenever two objects at different temperature are brought into contact, heat (thermal energy) will flow from the hot object to the cold object. hotcold heat flow Heat is transferred from system to surrounding Heat is transferred from surrounding to system

12. Transferring Energy Work Transfer: Work can be done on a system by surrounding or work can be done by the system to surrounding. Work done on the systemWork done by the system

13. The First Law of Thermodynamics Energy is conserved Energy can neither be created nor destroyed, only converted from one form to another (Law of conservation of energy) Energy that is lost by the system must be gained by the surrounding, and visa versa.  Internal Energy of a system (E): The sum of all the kinetic and potential energy of all its components. e.g, translation, vibrational and rotational motions of atoms, molecules

14. The First Law of Thermodynamics  Internal Energy change (  E) : The change in E that accompanies a change in the system  E = E final - E initial E final < E initial Heat (q) transferred from system to surroundings:  E < 0 E final > E initial Heat (q) transferred from surroundings to system:  E > 0 q q

15. The First Law of Thermodynamics  First Law of Thermodynamics and internal energy change  E = q + w heat (thermal energy) transferred in +q endothermic heat (thermal energy) transferred out -q exothermic work performed by surrounding on system +w work performed by system on surrounding -w

16. The First Law of Thermodynamics A(g) + B(g) C (s) E E q = -1200J heat work w = +500 J  E = q + w = -1200 J + 500 J = -700 J The system lost 700 J of energy to surrounding

17. The First Law of Thermodynamics Most of the work involved in chemical or physical changes are associated with volume change.  V > 0  w < 0 w done by the system  V 0 w done on the system

18. State Functions  State Function : A property of a system that is determined by the system’s present condition, or state. The value of a state function depends only on the present state of the system, not on the path the system took to reach that state. Example of state function : Internal Energy (E).

19. State Functions Because E is a state function,  E is also a state function NOTE: Although  E is a state function, q and w are NOT state functions

20. Enthalpy Most of chemical reactions we study take place in open container under constant pressure condition (atmospheric pressure)  Enthalpy Change (  H) : A measure of the amount of heat exchanged when a reaction takes place under constant pressure  H = q p NOTE: Enthalpy is also a state function (  H = H final – H initial ) and it is defined mathematically as H = E + PV where P is the pressure.

21. Enthalpy  Properties of enthalpy change : H 2 O (l) H 2 O (g)  H = 44kJ (endothermic) H 2 O (g) H 2 O (l)  H = -44kJ (exothermic) 2H 2 O (l) 2H 2 O (g)  H = 88kJ (endothermic) Enthalpy changes, like all energy changes, are extensive properties. The amount of heat (or enthalpy change) liberated or absorbed depends on the amount of material undergoing reaction

22. Enthalpy  Example: What is the enthalpy change (in kJ) for the sublimation of 15 g of iodine I 2 (s) I 2 (g)  H sub = 62.4kJ MW=254

23. Enthalpy of Reactions  Example : How much heat is given off by burning 3.4 g of H 2 gas? 2 H 2 (g) + O 2 (g) 2 H 2 O (l)  H = - 483.6 kJ

24. Enthalpy of Reactions  Example : What is the enthalpy change for the formation 12 g of CO 2 (g) from the combustion of CH 4 (g)? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g)  H = - 806 kJ MW=44

25. H 2 H 2 (g) + O 2 (g) 2 H 2 O (l) q given off  H < 0 2 H 2 (g) + O 2 (g) → 2 H 2 O (l)  H = - 483.6 kJ 2 H 2 O (l) → 2 H 2 (g) + O 2 (g)  H = +483.6 kJ q absorbed  H > 0 Enthalpy of Reactions

26. Calorimetry  Calorimetry : The measure of heat flow during physical and chemical changes  Calorimeter : A device for measuring heat flow during physical and chemical processes + NaOH dissolves heat (q) How can we measure q ?

27. Calorimetry We need to carry out the reaction in a container that is thermally insulated heat (q) NaOH dissolves simple calorimeter This is under constant pressure condition

28. Calorimetry When NaOH is dissolved in water, the temperature of the solution will increase. But, how much?  Heat Capacity : The amount of energy required to raise the temperature of an object by 1 o C Heat capacity is an extensive property  Specific Heat Capacity : The amount of energy required to raise the temperature of 1 g of a substance by 1 o C Specific heat capacity is an unique property of each substance

29. Calorimetry  Unit of specific heat capacity 1 g of H 2 O 11 o C 1 g of H 2 O 12 o C + 4.184 J of heat heat masstemperature change unit of specific heat capacity

30. Calorimetry  Example :Which of the following substances requires the smallest amount of energy to increase the temperature of 50.0 g of that substance by 10K? CH 4 (g) Hg(l) H 2 O(l) specific heat 2.20 0.14 4.18 capacity (J/g  K)

31. Calorimetry  Example :Which of the following substances requires the smallest amount of energy to increase the temperature by 1K? CH 4 (g) Hg(l) H 2 O(l) C (J/g  K) 2.20 0.14 4.18 mass (g) 10 100 2

32. Calorimetry  Example : Determine the specific heat of Al if it takes 16 J of thermal energy to raise the temperature 6.0g of Al by 3.0 o C

33. Calorimetry  Example : What is the change in thermal energy (in kJ) for 300 g iron rod when it is cooled from 250 o C to 50 o C ? (C for iron is 0.451 J/g  K)

34. Calorimetry  Molar Heat Capacity (C m ) : The amount of energy required to raise the temperature of 1 mol of substance by 1 o C heat moltemperature change unit of molar heat capacity

35. Calorimetry  Example : The specific heat capacity of iron is 0.451 J/g  o C. Determine the molar heat capacity of iron.

36. Calorimetry  Example : A 5.00 g pellet of copper at 75.0 o C is placed in a beaker containing 35.0g of water at 25.0 o C. What is the final temperature of the water and copper? (Cu = 0.385 J/g o C, H 2 O = 4.18 J/g o C)

37. Calorimetry  Example : A 2.00 g of sodium is placed in 150 g of water at 22.0 o C in a calorimeter. When the reaction is complete, the temperature of the solution is found to be 41.0 o C. Calculate  H for the reaction between Na and water: 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq) Strategy to solve the problem: (1) compute the total amount of heat gained by water (2) That is the -  H value for 2.00 g of sodium (3) We need  H value for 2 mol of sodium (conversion from 2.00 g to 2 mol of sodium for  H )

38. Hess’s Law If a reaction is carried out in a series of steps,  H for the overall reaction will equal the sum of the enthalpy changes for the individual steps H H 2 O (g) H 2 O (l) H 2 O (s) -44kJ -6 kJ -50kJ

39. Hess’s Law Consider the following imaginary reactions. (A) A + B → C  H = -125 kJ (B) D 3 → C + B  H = -41 kJ How can we calculate the  H for the following reaction? (C) A + 2B → D 3 To use Hess’s law, we have to rearrange reaction (A) and (B) so that the overall reaction becomes (C) A+2B  D 3. Be sure that you make appropriate changes for  H’s when you rearrange chemical equations.

40. Hess’s Law This is how we can rearrange the reactions. (1) Compare (A) and (C): (A) A + B → C (C) A + 2B → D 3 This is what we have. This is what we want. We need one more B on the left and one D 3 on the right and remove C on the right (2) Consider (B) : D 3 → C + B  H = -41 kJ If we reverse (B), it pretty much satisfies our need: one B on the left and one D 3 on the right ( B’) C + B  D 3  H = +41 kJ

41. Hess’s Law (3) Add (A) and (B’) A + B  C  H = -125kJ C + B  D 3  H = + 41kJ A + C + 2B  C + D 3  H = -84 kJ A + 2B  D 3  H = -84 kJ

42. Hess’s Law  Example : Given the enthalpy changes of following two reactions, calculate the enthalpy change for the combustion of methane to form liquid water and CO 2 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ H 2 O(g)  H 2 O(l)  H = -44 kJ

43. Hess’s Law  Example : Given the enthalpy changes of following two reactions, calculate the enthalpy change for making diamond from graphite : C (graphite)  C (diamond) C (graphite) + O 2 (g)  CO 2 (g)  H = -393.5 kJ C (diamond) + O 2 (g)  CO 2 (g)  H = -395.4 kJ graphitediamond

44. Hess’s Law  Example : Given the enthalpy changes of following two reactions, calculate the enthalpy change for the reaction between hydrogen and ozone : 3 H 2 (g) + O 3 (g) → 3 H 2 O (g) 2 H 2 (g) + O 2 (g) → 2 H 2 O (g)  H = -483.6 kJ 3 O 2 (g) → 2 O 3 (g)  H = +284.6 kJ

45. Standard Enthalpy of Formation  H for the formation of one mole of compound from elements, with all substances in their standard states [25 o C, 1atm] H 2 (g) + ½ O 2 (g) → H 2 O (l)  H o f = -285.8 kJ ½ N 2 (g) + ½ O 2 (g) → NO (g)  H o f = 90.4 kJ 2 C (s) + 3 H 2 (g) + ½ O 2 (g) → C 2 H 5 OH (g)  H o f = -277.7 kJ graphite NOTE :  H o f of the most stable form of any element is zero i.e.  H o f for O 2 (g), N 2 (g), H 2 (g), Br 2 (l) etc. = 0

46. Standard Enthalpy of Formation  Example : For which of the following reactions would  H represent a standard enthalpy of formation? 2 Na (s) + ½ O 2 (g) → Na 2 O (s) K (l) + ½ Cl 2 (g) → KCl (s) CO (g) + ½ O 2 (g) → CO 2 (g) 2 Na (s) + Cl 2 (g) → 2 NaCl (s)

47. Standard Enthalpy of Formation Enthalpies of formations can be used to calculate enthalpies of reactions (under standard conditions)  H o rxn = sum of all  H o f (products) - sum of all  H o f (reactants)  H o rxn = Σ n  H o f (products) – Σ m  H o f (reactants) sum moles of products moles of reactants

48. Standard Enthalpy of Formation

49.  Example : Determine the standard enthalpy change for the following reaction C 2 H 4 (g) + H 2 O(g)  C 2 H 5 OH(l)  H o f [C 2 H 4 (g)] = 52.30kJ,  H o f [H 2 O(g)] = -241.82 kJ,  H o f [C 2 H 6 O(l)] =-277.7kJ

50. Standard Enthalpy of Formation  Example : What is the enthalpy change of the following reaction at standard condition? C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O (l)  H o f [CO 2 (g)] = -394kJ,  H o f [C 3 H 8 (g)]= -104kJ,  H o f [H 2 O(l)] = -286kJ