1. Thermochemistry

2. Energy Energy is the ability to do work or transfer heat. – Energy used to cause an object that has mass to move is called work. – Energy used to cause the temperature of an object to rise is called heat.

3. Units of Energy The SI unit of energy is the joule (J). An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = 4.184 J 1 J = 1  kg m 2 s2s2

4. Definitions: System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

5. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

6. Internal Energy By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial

7. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

8.  E, q, w, and Their Signs

9. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic.

10. Examples of endothermic and exothermic reactions 2 NH 4 SCN(s) + Ba(OH) 2 * 8 H 2 O(s) → Ba(SCN) 2 (aq) + 2 NH 3 (aq) + 10 H 2 O(l). As a result, the temperature of the system drops from about 20 °C to -9 °C. Endothermic reaction

11. The reaction of powdered aluminum with Fe 2 O 3 is highly exothermic. The reaction proceeds vigorously to form Al 2 O 3 and molten iron: 2 Al(s) + Fe 2 O 3 (s) → Al 2 O 3 (s) + 2 Fe(l ). Examples of endothermic and exothermic reactions Exothermic reaction

12. Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

13. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = -P  V

14. Enthalpy If a process takes place at constant pressure and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

15. Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

16. Enthalpy Since  E = q + w and w = -P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q+w) − w  H = q So, at constant pressure, the change in enthalpy is the heat gained or lost.

17.  H =  E + P  V  E  H - P  V For Gas Materials P  V = n f RT – n i RT = ( n f – n i )RT =  nRT Where n is the numbers of moles of gas molecules

18. Olive oils is completely burned in oxygen at 100.3 ˚C according to: ∆H= -31150 kJ Calculate the change in the internal energy ∆E (in kJ) for this combustion process.  n = n f – n i = (52+57) – 80 = 29 P  V =  nRT = 29 x 8.314 x 373 = 89932.5 J = 89.93 kJ  E  H - P  V = -31150 – 89.93 = -31240 kJ

19. Endothermicity and Exothermicity A process is endothermic when  H is positive.

20. Endothermicity and Exothermicity A process is endothermic when  H is positive. A process is exothermic when  H is negative.

21. Enthalpy of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

22. Enthalpy of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

23. The Truth about Enthalpy 1.Enthalpy is an extensive property. 2.  H for a reaction in the forward direction is equal in size, but opposite in sign, to  H for the reverse reaction. 3.  H for a reaction depends on the state of the products and the state of the reactants.

24. Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity.

25. Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

26. Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change C S = q m  Tm  T

27. (a)How much heat is needed to warm 250 g of water from 22 °C to 98 °C? (b) What is the molar heat capacity of water? (a)  T= 98 o C – 22 o C = 76 o C = 76 K q = C S x m x  T = (4.18 J/g-K)(250 g)(76 K)= 7.9 x 10 4 J (b) 1 mol H 2 O = 18 g H 2 O C m = (4.18 J/g-K)(18 g/mol) = 75.2 J/mol-K C S = q m  Tm  T

28. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow.

29. Bomb Calorimetry (Constant-Volume Calorimetry) Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

30. Total heat capacity of the calorimeter C cal q rnx = - C cal x ∆T CH 6 N 2 is used as a liquid rocked fuel. The combustion of it with O 2 produces N 2, CO 2 and H 2 O. When 4 g of CH 6 N 2 is combusted in a bomb calorimeter, the temperature of the calorimeter increase from 25 ˚C to 39.5 ˚C. C cal of calorimeter is 7.794 KJ/ ˚C. calculate the heat of reaction for the combustion of a mole of CH 6 N 2. ∆T = 14.5 ˚C q rnx = - C cal x ∆T = -(7.794 KJ/ ˚C) x (14.5 ˚C ) = -113 KJ = -113 KJ / (4 g / 46 g mol -1 ) = -1.3 x 10 3 KJ /mol

31. Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

32. Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure  H for the reaction with this equation: q = m  C S   T

33. When a student mixes 50 mL of 1 M HCl and 50 mL of NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increase from 21 o C to 27.5 o C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses a negligible quantity of heat, that its density is 1 g/mL and that its specific heat is 4.18 J/g-K. Total mass = 100 mL x 1 g/mL = 100 g ΔT = 27.5 o C – 21 o C = 6.5 o C = 6.5 K q rnx = -C s x m x ΔT = -4.18 J/g-K x 100 g x 6.5 K = 2.7 x 10 3 J = -2.7 kJ ΔH = q p = -2.7 kJ. No of HCl moles = 0.05 L x 1 mol / L = 0.05 mol ΔH = -2.7 kJ/0.05 mol = -54 kJ/mol

34. Hess’s Law  H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using published  H values and the properties of enthalpy.

35. Hess’s Law Hess’s law states that “If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

36. Hess’s Law Because  H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

37. The enthalpy of reaction for the combustion of C to CO 2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO 2 is –283.0 kJ/mol CO: (1) C (s) + O 2(g) CO 2(g)  H 1 = -393.5 kJ (2) CO (g) + 0.5 O 2(g) CO 2(g)  H 2 = -283.0 kJ Using these data, calculate the enthalpy for combustion of C to CO: (3) C (s) + 0.5 O 2(g) CO (g)  H 3 = ? kJ

38. Enthalpies of Formation An enthalpy of formation,  H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

39. Standard Enthalpies of Formation Standard enthalpies of formation,  H f °, are measured under standard conditions (25 °C and 1.00 atm pressure).

40. Calculation of  H C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g)

41. Calculation of  H C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g)

42. Calculation of  H C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l)

43. Calculation of  H C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l)

44. Calculation of  H We can use Hess’s law in this way:  H =   n  H f ° products –   m  H f ° reactants where n and m are the stoichiometric coefficients.

45. Calculation of  H C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l)  H= [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ