1. 1 Energy and Thermochemistry

2. 2 Energy The ability to do work The ability to do work 2 types 2 types Potential: stored energy Potential: stored energy Kinetic: energy in motion Kinetic: energy in motion

3. 3 Thermochemistry Changes of heat content and heat transfer Changes of heat content and heat transfer Follow Law of Conservation of Energy Follow Law of Conservation of Energy Or, 1 st Law of Thermodynamics Or, 1 st Law of Thermodynamics Energy can neither be created nor destroyed Energy can neither be created nor destroyed Thus, total energy of universe constant Thus, total energy of universe constant

4. 4 Temperature & Heat Heat not same as temperature Heat not same as temperature Heat = energy transferred to one system by another due to temperature difference Heat = energy transferred to one system by another due to temperature difference Temperature = measure of heat energy content & ability to transfer heat Temperature = measure of heat energy content & ability to transfer heat Thermometer Thermometer Higher thermal energy, greater motion of constituents Higher thermal energy, greater motion of constituents Sum of individual energies of constituents = total thermal energy Sum of individual energies of constituents = total thermal energy

5. 5 Systems and Surroundings System = the object in question System = the object in question Surrounding(s) = everything outside the system Surrounding(s) = everything outside the system When both system and surrounding at same temperature  thermal equilibrium When both system and surrounding at same temperature  thermal equilibrium When not When not Heat transfer to surrounding = exothermic Heat transfer to surrounding = exothermic (you feel the heat)  hot metal! (you feel the heat)  hot metal! Heat transfer to system = endothermic Heat transfer to system = endothermic (you feel cold)  cold metal! (you feel cold)  cold metal!

6. 6 Math! Joules (J) used for energy quantities Joules (J) used for energy quantities But usually kJ (1000 J) used But usually kJ (1000 J) used Ye Royal Olde School used calorie (cal) Ye Royal Olde School used calorie (cal) cal = amt of heat required to raise the temperature of 1.00 g of water by 1  C cal = amt of heat required to raise the temperature of 1.00 g of water by 1  C 1 cal = 4.184 J (SI-unit) 1 cal = 4.184 J (SI-unit) But…Calorie (Cal) = 1000 cal But…Calorie (Cal) = 1000 cal Used in nutrition science and on food labels Used in nutrition science and on food labels

7. 7 Heat Capacity Specific heat capacity Specific heat capacity Quantity of heat required to raise the temp of 1 gram of any substance by 1 K Quantity of heat required to raise the temp of 1 gram of any substance by 1 K Molar heat capacity Molar heat capacity Quantity of heat required to raise the temp of 1 mole of any substance by 1 K

8. 8 Calculating heat transfer FYI FYI Specific heat capacity of metals is very low Specific heat capacity of metals is very low  < 1.000 J/(g  K)  < 1.000 J/(g  K) What does this tell us about heat transfer in metals? What does this tell us about heat transfer in metals?

9. 9 Let’s do an example In your backyard, you have a swimming pool that contains 5.19 x 10 3 kg of water. How many kJ are required to raise the temperature of this water from 7.2 °C to 25.0 °C? In your backyard, you have a swimming pool that contains 5.19 x 10 3 kg of water. How many kJ are required to raise the temperature of this water from 7.2 °C to 25.0 °C?

10. 10 Example solved Trick:  T in K =  T in °C Trick:  T in K =  T in °C

11. 11 Practice How many kJ are required to raise the temperature of 25.8 g of quicksilver from 22.5 °C to 28.0 °C? C Hg = 0.1395 J/(g  K) How many kJ are required to raise the temperature of 25.8 g of quicksilver from 22.5 °C to 28.0 °C? C Hg = 0.1395 J/(g  K)

12. 12 But what if there’s a change of state? Temperature constant throughout change of state Temperature constant throughout change of state Added energy overcomes inter-molecular forces Added energy overcomes inter-molecular forces

13. 13 Change of state What do the flat areas represent? What do the flat areas represent?

14. 14 q tot = q s + q s  l + q l + q l  g + q g q tot = q s + q s  l + q l + q l  g + q g q s  l = heat of fusion q s  l = heat of fusion Heat required to convert solid at melting pt. to liq Heat required to convert solid at melting pt. to liq Ice = 333 J/g Ice = 333 J/g q l  g = heat of vaporization q l  g = heat of vaporization Heat required to convert liq. at boiling pt. to gas Heat required to convert liq. at boiling pt. to gas Water = 2256 J/g Water = 2256 J/g

15. 15 Practice How much heat is required to vaporize 250.0 g of ice at -25.0 °C to 110.0 °C? How much heat is required to vaporize 250.0 g of ice at -25.0 °C to 110.0 °C? Given: Given: Specific heat capacity of ice = 2.06 J/g  K Specific heat capacity of ice = 2.06 J/g  K Specific heat capacity of water = 4.184 J/g  K Specific heat capacity of water = 4.184 J/g  K Specific heat capacity of steam = 1.92 J/g  K Specific heat capacity of steam = 1.92 J/g  K Let’s do this Let’s do this

16. 16 Calorimetry The process of measuring heat transfer in chemical/physical process The process of measuring heat transfer in chemical/physical process q rxn + q soln = 0 q rxn + q soln = 0 q rxn = -q soln q rxn = -q soln Rxn = system Rxn = system Soln = surrounding Soln = surrounding What you’ll do in lab What you’ll do in lab Heat given off by rxn Heat given off by rxn Measured by thermometer Measured by thermometer Figure out q rxn indirectly Figure out q rxn indirectly

17. 17 Enthalpy = heat content at constant pressure = heat content at constant pressure If  H = “+”, process endothermic If  H = “+”, process endothermic If  H = “-”, process exothermic If  H = “-”, process exothermic Enthalpy change dependent on states of matter and molar quantities Enthalpy change dependent on states of matter and molar quantities For example: For example: Is vaporizing ice an exothermic or endothermic process? Is vaporizing ice an exothermic or endothermic process? Thus, will  H be “+” or “-”? Thus, will  H be “+” or “-”?

18. 18 Hess’s Law If a rxn is the sum of 2 or more other reactions,  H = sum of  H’s for those rxns If a rxn is the sum of 2 or more other reactions,  H = sum of  H’s for those rxns So,  H tot =  H 1 +  H 2 +  H 3 + … +  H n So,  H tot =  H 1 +  H 2 +  H 3 + … +  H n

19. 19 Let’s solve a problem C (s) + 2S (s)  CS 2(l) ;  H = ? C (s) + 2S (s)  CS 2(l) ;  H = ? Given: Given: C (s) + O 2(g)  CO 2(g) ;  H = -393.5 kJ/mol C (s) + O 2(g)  CO 2(g) ;  H = -393.5 kJ/mol S (s) + O 2(g)  SO 2(g) ;  H = -296.8 kJ/mol S (s) + O 2(g)  SO 2(g) ;  H = -296.8 kJ/mol CS 2(l) + 3O 2(g)  CO 2(g) + 2SO 2(g) ;  H = -1103.9 kJ/mol CS 2(l) + 3O 2(g)  CO 2(g) + 2SO 2(g) ;  H = -1103.9 kJ/mol How do we manipulate the 3 rxns to achieve the necessary net rxn? How do we manipulate the 3 rxns to achieve the necessary net rxn? Does  H change if the rxns are reversed and/or their mole ratios are changed? Does  H change if the rxns are reversed and/or their mole ratios are changed? Let’s talk about this on the next slide Let’s talk about this on the next slide

20. 20 Let’s work it out 1. Switch this rxn: CS 2(l) + 3O 2(g)  CO 2(g) + 2SO 2(g) ;  H = -1103.9 kJ Thus, CO 2(g) + 2SO 2(g)  CS 2(l) + 3O 2(g) ;  H = + 1103.9 kJ Thus, -  H fwd = +  H rev 2. Double this rxn: S (s) + O 2(g)  SO 2(g) ;  H = -296.8 kJ Thus, 2S (s) + 2O 2(g)  2SO 2(g) ;  H = (-296.8 kJ) x 2 = -593.6 kJ Since  H is per mole, changing the stoichiometric ratios entails an equivalent change in  H 3. Keep this rxn: C (s) + O 2(g)  CO 2(g) ;  H = -393.5 kJ 4. Add those on same side of rxns/eliminate those on opposite sides of rxn: CO 2, 2SO 2, 3O 2 5. Net rxn: C (s) + 2S (s)  CS 2(l)  H = 1103.9 kJ - 593.6 kJ – 393.5 kJ = 116.8 kJ Is it an exo- or endothermic rxn?

21. 21 Practice Given: Given: CH 4(g)  C (s) + 2H 2(g) ;  H = 74.6 kJ/mol CH 4(g)  C (s) + 2H 2(g) ;  H = 74.6 kJ/mol C (s) + O 2(g)  CO 2(g) ;  H = -393.5 kJ/mol C (s) + O 2(g)  CO 2(g) ;  H = -393.5 kJ/mol H 2(g) + O 2(g)  H 2 O (g) ;  = -241.8 kJ/mol H 2(g) + O 2(g)  H 2 O (g) ;  = -241.8 kJ/mol CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) ;  H rxn = ? CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) ;  H rxn = ?

22. 22 Standard Energies of Formation Standard molar enthalpies of formation =  H f  = enthalpy change for formation of 1 mol of cmpd directly from component elements in standard states Standard molar enthalpies of formation =  H f  = enthalpy change for formation of 1 mol of cmpd directly from component elements in standard states Standard state = most stable form of substance in physical state that exists @ 1 bar pressure & a specific temp., usually 25  C (298K) Standard state = most stable form of substance in physical state that exists @ 1 bar pressure & a specific temp., usually 25  C (298K) 1 bar = 100kPa 1 bar = 100kPa 101.325 kPa = 1 atm 101.325 kPa = 1 atm So 1 bar  1 atm (an SI unit) So 1 bar  1 atm (an SI unit) Example Example C (s) + O 2(g)  CO 2(g) ;  H =  H f  = -393.5 kJ C (s) + O 2(g)  CO 2(g) ;  H =  H f  = -393.5 kJ  H f  = 0 for elements in standard state  H f  = 0 for elements in standard state

23. 23 Enthalpy Change for a Rxn Must know all std molar enthalpies Must know all std molar enthalpies  H  rxn =  H f  prods -  H f  reactants  H  rxn =  H f  prods -  H f  reactants Given to you in a table in the back of the book Given to you in a table in the back of the book Again, keep in mind the mole ratios for each species involved! Again, keep in mind the mole ratios for each species involved!

24. 24 Example  H  rxn for 10.0 g of nitroglycerin?  H  rxn for 10.0 g of nitroglycerin? 2C 3 H 5 (NO 3 ) 3(l)  3N 2(g) + ½O 2(g) + 6CO 2(g) + 5H 2 O (g) 2C 3 H 5 (NO 3 ) 3(l)  3N 2(g) + ½O 2(g) + 6CO 2(g) + 5H 2 O (g) C 3 H 5 (NO 3 ) 3(l) = -364 kJ/mol C 3 H 5 (NO 3 ) 3(l) = -364 kJ/mol CO 2(g) = -393.5 kJ/mol CO 2(g) = -393.5 kJ/mol H 2 O (g) = -241.8 kJ/mol H 2 O (g) = -241.8 kJ/mol

25. 25 Solution

26. 26 Practice Determine  H° rxn for: Determine  H° rxn for: 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) Given: Given: NH 3(g) = -45.9 kJ/mol NH 3(g) = -45.9 kJ/mol NO (g) = 91.3 NO (g) = 91.3 H 2 O (g) = -241.8 H 2 O (g) = -241.8