1. Thermodynamics

2. Terms/ Definitions Thermodynamics –Deals with the interconversion of heat an other forms of energy Thermochemistry –Deals with heat change in chemical reactions State Function –Function that depends only on the conditions (state) not on how the state was obtained

3. Energy(E) –Internal energy = kinetic + potential energy –Kinetic energy comes from molecular motion, electron motion etc. –Potential energy comes from attractive and repulsive forces in nuclei, and interactions between molecules Enthalpy(H) –H = E + PV –Extensive property –State function –Can only measure the difference in enthalpy between two states

4. Heat (Q,q) –Transfer of thermal energy between two bodies at different temperatures Work (W,w) –Form of energy, can be mechanical or non mechanical –Mechanical work is normally pressure - volume work –W = -P ex  V –Work done by the system on the surrounding = negative –Work done by the surroundings on the system = positive

5. System –Specific thing we are looking at Surroundings –Everything outside the system Universe –System + surroundings Open system –Can exchange both matter and energy with the surroundings Closed system –Can exchange energy but not matter with the surroundings Isolated system –No exchange of matter or energy with the surroundings –Q = ? –Q = 0

6. Thermochemical Equation –Chemical equation that includes the enthalpy change –e.g. H 2 O(s) --> H 2 O(l)  H = 6.01 kJ Heat flow –Endothermic, Q = +, heat absorbed by the system –Exothermic, Q = -, heat given off by the system Units of energy –Joules(J), kilojoules(kJ) –Calories(cal), kilocalories(kcal) –1 cal = 4.184 J –(Liter)(atmosphere)1 L atm = 101.3 J

7. Enthalpy –State function –Heat energy of state –Look at change between states  H Extensive Property - depends on the amount 1 mole of the reaction as written

8. Changes sign as reverse reaction: Follows Hess’ Law

9. Hess’ Law If a reaction can be considered to go by a series of steps,  H of the reaction is the sum of  H of the steps. Reaction  kJ/mol 1/2 N 2 (g) + 1/2 O 2 (g)NO(g) +90.4 NO 2 (g)NO(g) + 1/2 O 2 (g) +56.5 What is the value of  H for the reaction: 1/2 N 2 (g) + O 2 (g)NO 2 (g) NO(g) + 1/2 O 2 (g)NO 2 (g) -56.5 1/2 N 2 (g) + 1/2 O 2 (g) NO(g) +90.4 1/2 N 2 (g) + O 2 (g)NO 2 (g) +33.9

10. Standard States Most stable form at 1 atmosphere and usually at 25 o C Indicate standard state by  H o Allotropes More than one stable form at 1 atmosphere O 2 and O 3 C(s) graphite and C(s) diamond Standard heat of formation (Standard enthalpy of formation) Heat change that occurs when 1 mole of a single product is formed from elements in their standard states. C(s) graph + O 2 (g)CO 2 (g)-393.5 kJ CO 2 = -393.5 kJ/mol e.g. C(s)graphite H 2 (g) N 2 (g) Br 2 (l) Fe(s) S(s) Hg(l) I 2 (s)

11. Write the formation reaction for the following: NH 4 NO 3 (s) benzene aniline N 2 (g) + H 2 (g) + O 2 (g) NH 4 NO 3 (s)2 3/2 C 6 H 6 (l) C(s)graphite + H 2 (g) 6 3 C 6 H 5 NH 2 (l) C(s)graphite + H 2 (g) + N 2 (g) 6 7/2 1/2

12. Value of standard state of elements: By definition,  H o formation of an element in its standard state = 0. Which of the following will NOT have  H formation = 0 H 2 (g)Ne(g)Cl 2 (g)I 2 (l) Hg(s)Br 2 (l)Ca(s)Fe(l)

13. Calculate Heat of a reaction: 2. Heat of reaction can be calculated using Hess’ Law 1. Heat of reaction can be calculated from Heat of Formation data Using Heat of formation data If data for all the reactants and products is given as, then is equal to the sum of of the products minus the sum of of the reactants.

14. Calculate  H o rxn for the reaction: C 2 H 6 (g) + O 2 (g)CO 2 (g) + H 2 O(l) 23 7/2 = {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (?)} = {2(-90.1) + 3(-68.3)} - {1(-20.2) + 7/2 (0)} = -364.9 kcal/ mol

15. Using Hess’ Law If all information is not given as  H f, then need to go a different way Use Hess’ Law Given:Reaction  H o (kJ/mol) Find  H for the formation of C 2 H 2 1. Write a target reaction:--> C 2 H 2 (g) 2C(s) graph + H 2 (g) 2. Rearrange reactions so that when added together you get the target --> 2C 2 H 2 (g) 4C(s) graph + 2H 2 (g) +453.2 --> C 2 H 2 (g) 2C(s) graph + H 2 (g) +453.2/2 = 226.6 kJ

16. Which of the following would have = 0? A)C(s) diamond B)Hg(g) C)Xe(g) D)Br 2 (g) E)Cl - (aq)

17. Energy(E) - Internal Energy (KE + PE) Kinetic energy: comes from the molecular motion and the electronic motion Potential Energy: comes from attractive and repulsive forces in nuclei and from interactions between molecules. First Law: Energy can be converted from one form to another, it can not be created or destroyed. i.e. The energy of the universe is constant.  E universe =  E system +  E surroundingss = 0  E system = -  E surroundingss All energies do not have to be the same form Since we are primarily interested in what happens to a chemical system, we use the form:  E = q + w Q = heat W = work

18. Sign convention -+ Heat(q)from system tofrom surroundings surroundingsto system (exothermic)(endothermic) Work(w)by the system on by surroundings the surroundingson the system Work includes all kinds of work, both mechanical and non-mechanical. We limit to mechanical work at this time. Mechanical work: W = -P ex  V = - P  V at constant P A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done?

19. A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work(in J) is done? A)10.0 J B)- 10.0 J C)1.01 x 10 3 J D)20.0 J E)- 1.01 x 10 3 J

20. A gas expands against a constant external pressure of 5.0 atm from 2.0 L to 4.0 L. How much work is done? = - (5.0 atm)(4.0 L - 2.0 L) = - 10.0 L atm 1 L atm = 101.3 J = (- 10.0 L atm) (101.3 J/ Latm) = - 1013 J = - 1.01 x 10 3 J W = - P  V

21. Relationship between  E and q  E = q + w w = - P ex  V  E = q - P ex  V At constant volume DV = ? = 0  E = q v Relationship between  E and  H By definition: H = E + PV So:  H =  E +  (PV) At constant P:  (PV) = P  V  H =  E + P  V When you are looking at a reaction dealing with gases: PV = nRT So: P  V =  nRT And:  H =  E +  nRT

22. Relationship between  H and q  H =  E + P  V  E = q + w  H = q + w + P  VW = - P ex  V At constant P: P ex = P W = - P  V  H = q - P  V + P  V  H = q P

23. A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is  E for the process? A)+ 1000 J B)+ 400 J C)- 400 J D)- 1000 J

24. A system at 1 atm and 11.0 L absorbs 300 J of heat and has 700 J of work performed on it. What is  E for the process?  E = q + w q = + 300 Jw = + 700 J  E = 300J + 700 J = 1000 J

25. Calorimetry Deals with the transfer of heat energy Heat Capacity - Capacity of a system to store heat - Heat needed to raise the temperature of a system by 1 o C ( 1K) = J/K q = (capacity of system)(  T) Specific Heat (c) - Heat needed to raise the temperature of 1 gram of material by 1 o C (1K) = J/g K Molar Heat Capacity (C) - Heat needed to raise the temperature of 1 mole of material by 1 o C (1K) = J/ mol K q = ms  T q = nC  T

26. Constant volume Calorimetry Bomb Calorimeter Q =  E Isolated system q system = ? = 0 q system = q sample + q calorimeter q sample = - q calorimeter q sample not directly measurable q calorimeter can be measured q calorimeter = heat capacity of bomb(  T) = - q sample Since reaction carried out at constant volume q sample =  E

27. Constant Pressure Calorimetry Coffee Cup Calorimeter Q =  H Isolated system q system = ?= 0 Generally look at two different kinds of systems 1. Reaction in aqueous solution 2. Metal solid in water q system = q reaction + q solution q reaction = - q solution q reaction not directly measurable q solution can be measured 1. q solution = mc  T = -q reaction 2. q system = q metal + q water - q metal = q water -(m)(c)  T = (m)(c)  T metalwater Since at constant pressure q reaction =  H reaction

28. 0.500 L of 1.0 M Ba(NO 3 ) 2 at 25 o C was mixed with 0.500 L of 1.0 M Na 2 SO 4 at 25 o C. A white precipitate forms (BaSO 4 ) and the Temperature rises to 28.1 o C. What is  H/ mol of BaSO 4 formed? Specific heat mix = 4.184 J/g Kd of mix = 1.0 g/ mL Assume a coffee cup calorimeter, P = constant q transfer from system to surroundings = ? = 0 Why? Coffee cup calorimeter assumes an isolated system q sys = q rxn + q mix = 0 q rxn = - q mix Can’t measure heat of the ppt’n directly but can measure heat of mixing q mix = ms  T = (1000mL)(1.0g/mL)(4.184J/g K)(28.1-25.0) = 12970J = 12.970kJ q ppt = - q mix = - 12.970kJ But this is only for the amount of BaSO 4 made in this reaction and we want per mole

29. 0.500 L of 1.0 M Ba(NO 3 ) 2 at 25 o C was mixed with 0.500 L of 1.0 M Na 2 SO 4 at 25 o C. A white precipitate forms (BaSO 4 ) and the Temperature rises to 28.1 o C. What is  H/ mol of BaSO 4 formed? Specific heat mix = 4.184 J/g Kd of mix = 1.0 g/ mL Need to determine the amount of BaSO 4 that was formed in the reaction Rxn: Ba(NO 3 ) 2 + Na 2 SO 4 --> BaSO 4 + 2NaNO 3 Mole Ba(NO 3 ) 2 = VM = (0.500L)(1.0M) = 0.50 mol Mole Na 2 SO 4 = VM = (0.500L)(1.0M) = 0.50 mol Ba(NO 3 ) 2 + Na 2 SO 4 --> BaSO 4 + 2NaNO 3 0.50 mol 0.50 mol --> 0.50 mol So, the reaction produced 0.50 mol BaSO 4 and that released - 12.97kJ Therefore the production of 1 mol of BaSO 4 will have  H = ?  H = 2(- 12.97) = - 25.94 kJ = - 25.9 kJ

30. 28.2 g Ni metal at 99.8 o C was placed in 150. g water at 23.50 o C. The final temperature at equilibrium was 25.00 o C. Find the specific heat of Ni in cal/ g K. A)0.107 cal/ g K B)0.1066 cal/ g K C)- 0.1066 cal/ g K D)- 0.107 cal/g K E)0.1267 cal/g K

31. 28.2 g Ni metal at 99.8 o C was placed in 150. g water at 23.50 o C. The final temperature at equilibrium was 25.00 o C. Find the specific heat of Ni in cal/ g K. q system = q metal + q solution q metal = mc  T for the Ni q solution = mc  T for the water q metal = (28.2 g)c  o C q solution = (150 g)(1 cal/g o C)(25.0-23.5 o C) - q metal = q solution - (28.2 g)c  o C = (150 g)(1 cal/g o C)(25.0-23.5 o C) c Ni = 0.1066 cal/g o C = 0.107 cal/ g K