1. Topics 5 and 15

2. Hess’s Law Calorimetry Enthalpy Enthalpy of Formation Bond Energy

3. Assessment Statements 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. 5.2.2 Design suitable experimental procedures for measuring the heat/energy changes of reactions. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. 5.2.4 Evaluate results of experiments to determine enthalpy changes.

4. 4 ways to find H Using calorimeter Using Hess’s Law Enthalpies of formation Bond enthalpies

5. Calorimetry The measurement of heat flow Calorimeter Device used to measure heat flow We can not measure heat flow directly….

6. Enthalpy of combustion experiment

7. Enthalpy of neutralization experiment

8. Using experimental results

9. Calorimetry calculations Enthalpy change can be calculated using the following equation: q = mc ∆T q = enthalpy change in joules To work out the enthalpy of neutralization, the density and specific heat capacities of the acid and base used are taken to be the same as for pure water. ∆T = change of temperature in Kelvin. c = specific heat capacity in joules per Kelvin per gram (4.18 JK -1 g -1 for water) m = mass of substance being heated (often water) in grams

10. Calorimetry calculation examples

11. Calorimetry calculation problems

12. The bomb calorimeter

13. Specific heat (c or s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius (or K). Heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms  t q = C  t  t = t final - t initial 6.5

14. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms  t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J 6.5

15. Constant-Pressure Calorimetry No heat enters or leaves! 6.5

16. Heat Changes The heat equation may be stated as  Q = m C  T where:  Q = Change in heat m = mass in grams C = specific heat in J g -1 o C -1  T = Temperature change 16

17. Calorimetry When measuring heat lost or gained by the reaction, we measure it using water as the basis for the calculations. Q w = m x s x  t m of water Q of water s of water  T of water

18. Temperature Changes Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat. By plotting a graph of time v temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings. 18 A time v temperature graph

19. Calorimetry The heat gained or lost by the reaction is equal to, but opposite in sign, to the heat gained or lost by the water -q water = q rxn water gained = rxn lost water lost = rxn gained

20. Heat Transfer Problem 1 Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is 0.902 J g -1 o C -1. Solution  Q = mC  T = (400 g) (0.902 J g -1 o C -1 )(200 o C – 20 o C) = (400 g) (0.902 J g -1 o C -1 )(200 o C – 20 o C) = 64,944 J = 64,944 J 20

21. Heat Transfer Problem 2 What is the final temperature when 50 grams of water at 20 o C is added to 80 grams water at 60 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 o C -1 Solution:  Q (Cold) =  Q ( hot ) mC  T= mC  T Let T = final temperature (50 g) (4.184 J g -1 o C -1 )(T- 20 o C) = (80 g) (4.184 J g -1 o C -1 )(60 o C- T) (50 g)(T- 20 o C) = (80 g)(60 o C- T) 50T -1000 = 4800 – 80T 130T =5800 T = 44.6 o C 21

22. Sample problem 0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.0 0 C to 29.9 0 C. Calculate the enthalpy change for the formation of 1.0 mol of water in this rxn. Assume no heat is lost the surroundings and the density of the solution is the same as the water. H + (aq) + OH - (aq) H 2 O (l) Step 1: find q water

23. Sample problem 0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.0 0 C to 29.9 0 C. Calculate the enthalpy change for the formation of 1.0 mol of water in this rxn. Assume no heat is lost the surroundings and the density of the solution is the same as the water. H + (aq) + OH - (aq) H 2 O (l) Step 1: find q water q = m s  t q = (150g)(4.184 J/g 0 C)(29.9 0 C – 23.0 0 C)

24. Sample problem 0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.0 0 C to 29.9 0 C. Calculate the enthalpy change for the formation of 1.0 mol of water in this rxn. Assume no heat is lost the surroundings and the density of the solution is the same as the water. H + (aq) + OH - (aq) H 2 O (l) Step 1: find  q water q = m s  t  q = (150g)(4.184 J/g 0 C)(23.0 0 C – 29.9 0 C)  q = +4330 J

25. 2 nd Step: Find  q rxn Water absorbed heat : heat gained by water is lost in the reaction -q water = q rxn

26. 2 nd Step: Find  q rxn Water absorbed heat : heat gained by water is lost in the reaction -q water = q rxn -(+4330 J) = q rxn

27. 2 nd Step: Find  q rxn Water absorbed heat : heat gained by water is lost in the reaction -q water = q rxn -(+4330 J) = q rxn -4330 J = q rxn This was an exothermic reaction

28. Step 3: find the  H of the reaction H + (aq) + OH - (aq) H 2 O (l)  H rxn =  q rxn = -4330 J = -5.8x10 4 J n H2O 0.075 mol H 2 O  H rxn = -58 kJmol -1

29.  H rxn vs.  q rxn  H rxn = heat lost or gained in the balanced chemical equation  q rxn = heat lost or gained in the experiment that took place in the calorimeter

30. Phase Changes & Heat Energy is required to change the phase of a substance The amount of heat necessary to melt a substance is called the Heat of fusion (  H fus ).The heat of fusion is expressed in terms of 1 mole or 1 gram It takes 6.00 kJ of energy to melt 1 mole (18 grams) of ice into liquid water. This is equivalent to about 335 J per gram The amount of heat necessary to boil a substance is called the Heat of vaporization (  H vap ) It may be expressed in terms of 1 mole or 1 gram It takes 40.6 kJ of energy to boil away 1 mole (18 grams) of water. This is equivalent to about 2240 J per gram. 30

31. Substance  H fus  H vap Mercury, Hg 2.29kJ/mol59.1kJ/mol Ethanol, C 2 H 5 OH 5.02kJ/mol38.6kJ/mol Water, H 2 O 6.00kJ/mol40.6kJ/mol Ammonia, NH 3 5.65kJ/mol23.4kJ/mol Helium, He 0.02kJ/mol0.08kJ/mol Acetone5.72kJ/mol29.1kJ/mol Methanol, CH 3 OH 3.16kJ/mol35.3kJ/mol Molar Heat Data for Some Common Substances 31

32. mC liquid wax  T  Q tota = (50g)(2.31J g -1 ˚C -1 )(62˚C-85˚C) (50g)(2.31J g -1 ˚C -1 )(62˚C-85˚C) + (50g/352.7gmol -1 )(-70,500J mol -1 ) + (50g)(2.18J g -1 ˚C -1 )(25˚C-62˚C) How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C? (C solid wax = 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C liquid wax =2.31 J/g˚C; MM = 352.7 g/mol,  H fusion =70,500 J/mol) Heat Transfer Problem 3  Q =  Q total =  Q liquid wax +  Q solidification +  Q solid wax + n(  Q fusion ) + mC solid wax  T  Q total= = mC liquid wax  T +n(  Q fusion ) + mC solid wax  T  Q total = (-2656.5 J) + (-9994.3 J)+ (-4033 J)  Q total = -16,683.8 J 32

33. Steam at 175°C that occupies a volume of 32.75 dm 3 and a pressure of 2.60 atm. How much energy would it need to lose to end as liquid water at 20 o C? Heat Transfer Problem 4 Solution: n = PV/RT = (2.60 atm)(32.75 dm 3 ) (0.0821 dm 3 atm mol -1 K -1 )(448 K -1 ) = 2.315 mol  Q = (2.315 mol) (37.47 J mol -1 K -1 )(175 o C-100 o C) +(2.315 mol)(40600 J mol -1 ) +(2.315 mol)(75.327 J mol -1 K -1 )(100 o C-20 o C)  Q = 6505.7J + 93989 J + 13950.6 J = 114445.3 J = 114.445 kJ