1. 1 © 2009 Brooks/Cole - Cengage A piece of chromium metal weighing 24.26 g is heated in boiling water to a temperature of 98.3°C and then dropped into a coffee cup calorimeter containing 82.3g of water at 23.3°C. When thermal equilibrium is reached, the final temperature is 25.6°C. Calculate the C p of chromium. Constant Pressure Calorimetry: Another Example

2. 2 © 2009 Brooks/Cole - Cengage Technique can be used to obtain the heat content of combustion of compoundsTechnique can be used to obtain the heat content of combustion of compounds Used in the food, fuel and pharmaceutical industries to know how much energy would be released by completely consuming the compoundUsed in the food, fuel and pharmaceutical industries to know how much energy would be released by completely consuming the compound Uses a BOMB CalorimeterUses a BOMB Calorimeter Constant Volume Calorimetry: It’s The Bomb!

3. 3 © 2009 Brooks/Cole - Cengage Measuring Heats of Reaction CALORIMETRYCALORIMETRY Place sample of known mass inside the bomb Place oxygen in the sample chamber and immerse bomb into water Ignite the bomb and measure temperature of water Since the volume doesn’t change, no P-V work is done, so the q r is a measurement of the ΔU

4. 4 © 2009 Brooks/Cole - Cengage Calorimetry Some heat from reaction warms water q water = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” q bomb = (heat capacity, J/K)(∆T) Total heat evolved = q total = q water + q bomb

5. 5 © 2009 Brooks/Cole - Cengage Calculate energy of combustion (∆U) of octane. 2C 8 H 18 + 25O 2 --> 16CO 2 + 18H 2 O Burn 1.00 g of octaneBurn 1.00 g of octane Temp rises from 25.00 to 33.20 o CTemp rises from 25.00 to 33.20 o C Calorimeter contains 1200. g waterCalorimeter contains 1200. g water Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Measuring Heats of Reaction CALORIMETRY

6. 6 © 2009 Brooks/Cole - Cengage Step 1 Calc. energy transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. energy transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3 Total energy evolved 41,200 J + 6860 J = 48,060 J 41,200 J + 6860 J = 48,060 J Energy of combustion (∆U) of 1.00 g of octane = - 48.1 kJ Measuring Heats of Reaction CALORIMETRY

7. 7 © 2009 Brooks/Cole - Cengage The First Law of Thermodynamics Up until now, we have only considered the changes in the internal energy of a system as functions of a single change: either work or heat However, these changes rarely occur singly, so we can describe the change in internal energy as:  U = q + w (The 1st Law) The change in internal energy is dependent upon the work done by the system and the heat gained or lost by the system

8. 8 © 2009 Brooks/Cole - Cengage The First Law: Put Another Way 1 st Law of Thermodynamics A system can store energy. A change in the energy of a system means that there must be a change in the heat or the work done BY or TO the system. OR The Total Energy of the Universe is Constant

9. 9 © 2009 Brooks/Cole - Cengage The First Law Internal energy is an example of a state function A state function is a property that only depends on the current state of the system and is independent of how that state was reached Pressure, Volume, Temperature and density are all examples of state functions

10. 10 © 2009 Brooks/Cole - Cengage State Functions and Things that Aren’t Work and heat ARE NOT state functions The amount of work done depends on how the change was brought about

11. 11 © 2009 Brooks/Cole - Cengage Change in Internal Energy (Implications of a State Function) It doesn’t matter what path we take to get to the final point, the change in internal energy is only dependent on where we started and where we finished Let’s think about this on a molecular level… –If we expand an ideal gas isothermally, the molecules will have the same kinetic energy and will move at the same speed –Despite the fact that the volume has increased, the potential energy of the system remains the same because there are no forces between molecules (KMT) –Since neither the kinetic nor potential energy has changed, the change in internal energy is…

12. 12 © 2009 Brooks/Cole - Cengage Zero!  U = 0 for the isothermal expansion of an ideal gas

13. 13 © 2009 Brooks/Cole - Cengage Enthalpy In a constant volume system in which no work is done (neither expansion nor non-expansion), we can rearrange the first law to:  U = q + w(but w=0)  U = q Most systems are constant pressure systems which can expand and contract When a chemical reaction takes place in such a system, if gas is evolved, it has to push against the atmosphere in order to leave the liquid or solid phase –Just because there’s no piston, it doesn’t mean that no work is done!

14. 14 © 2009 Brooks/Cole - Cengage Enthalpy Let’s look at an example: If we supply 100J of heat to a system at constant pressure and it does 20J of work during expansion, the  U of the system is +80J (w=-20J) –We can’t lose energy like this Enthalpy, H, is a state function that we use to track energy changes at constant pressure H=U + PV The change in enthalpy of a system (  H) is equal to the heat released or absorbed at constant pressure

15. 15 © 2009 Brooks/Cole - Cengage Enthalpy Another way to define enthalpy is at constant pressure:  H = q

16. 16 © 2009 Brooks/Cole - Cengage Enthalpy and Chemical Reactions Enthalpy is a tricky thing to grasp, but we can look at it this way: Enthalpy is the macroscopic energy change (in the form of heat) that accompanies changes at the atomic level (bond formation or breaking) Enthalpy has the same sign convention as work, q and  U –If energy is released as heat during a chemical reaction the enthalpy has a ‘-’ sign –If energy is absorbed as heat from the surrounding during a reaction, the enthalpy has a ‘+’ sign

17. 17 © 2009 Brooks/Cole - Cengage Heat Transfers at Constant Pressure H = U + PV The change in the enthalpy of a system is equal to the heat released or absorbed at constant pressure WTF? 1.In a coffee cup calorimeter (constant pressure calorimeter), the heat, q, that is released or absorbed is equal to the change in enthalpy,  H 2.When we add heat to a constant pressure system, the enthalpy increases by that amount 3.  H<0 for exothermic reactions  H>0 for endothermic reactions

18. 18 © 2009 Brooks/Cole - Cengage Consider the formation of water H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) + 241.8 kJ USING ENTHALPY Exothermic reaction — energy is a “product” and ∆H = – 241.8 kJ

19. 19 © 2009 Brooks/Cole - Cengage Energy Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) f Liquid water q = (heat of fusion)(mass)

20. 20 © 2009 Brooks/Cole - Cengage Energy Transfer and Changes of State Requires energy (heat). This is the reason a)you cool down after swimming b)you use water to put out a fire. + energy Liquid --> Vapor

21. 21 © 2009 Brooks/Cole - Cengage Heating/Cooling Curve for Water Note that T is constant as ice melts

22. 22 © 2009 Brooks/Cole - Cengage Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? Heat & Changes of State +333 J/g +2260 J/g

23. 23 © 2009 Brooks/Cole - Cengage What quantity of energy as heat is required to melt 500. g of ice and heat the water to steam at 100 o C? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 10 5 J q = (500. g)(333 J/g) = 1.67 x 10 5 J 2.To raise water from 0 o C to 100 o C q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J 3.To evaporate water at 100 o C q = (500. g)(2260 J/g) = 1.13 x 10 6 J q = (500. g)(2260 J/g) = 1.13 x 10 6 J 4. Total energy = 1.51 x 10 6 J = 1510 kJ Heat & Changes of State

24. 24 © 2009 Brooks/Cole - Cengage Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. USING ENTHALPY H 2 + O 2 gas Liquid H 2 OH 2 O vapor

25. 25 © 2009 Brooks/Cole - Cengage Hess’s Law The ∆ r H o for a reaction that is the sum of one or more reactions is the sum of the ∆ r H o values for all of the reactions Remember: Since ∆H is a state function, it doesn’t matter how we get to where we end up, all that matters is where we started and where we finished.

26. 26 © 2009 Brooks/Cole - Cengage Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) f H 2 O(g) + 242 kJ H 2 O(g) f H 2 O(liq) + 44 kJ ----------------------------------------------------------------------- H 2 (g) + 1/2 O 2 (g) f H 2 O(liq) + 286 kJ Example of HESS’S LAW — If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. USING ENTHALPY

27. 27 © 2009 Brooks/Cole - Cengage Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed.

28. 28 © 2009 Brooks/Cole - Cengage Making CO from C (graphite) and O 2 is a single step, but the CO reacts with oxygen to form CO 2. 1) C(graphite) + ½O 2 (g) --> CO (g) ? kJ/mole-rxn 2) CO(g) + O 2 (g) --> CO 2 (g) + -283 kJ/mole-rxn ----------------------------------------------------------------------- 3) C (graphite) + ½O 2 (g) --> CO 2 (g) -393.5 kJ/mole-rxn Example of HESS’S LAW — If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. ∆H o 3 = ∆H o 1 + ∆H o 2 USING ENTHALPY

29. 29 © 2009 Brooks/Cole - Cengage Using Enthalpy ∆H o 3 = ∆H o 1 + ∆H o 2 ∆H o 3 = ∆H o 1 + ∆H o 2 -393.5 kJ/mole-rxn = ∆H o 1 + (-283.0 kJ/mole-rxn) ∆H o 1 = -110.5 kJ/mole-rxn

30. 30 © 2009 Brooks/Cole - Cengage Hess’s Law & Energy Level Diagrams Forming CO 2 can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Active Figure 5.16

31. 31 © 2009 Brooks/Cole - Cengage Hess’s Law: Another Example Use Hess’s Law to calculate the enthalpy change for the formation of CS 2 (l) from C(s) and S(s). The overall equation is: C(s) + 2S(s) --> CS 2 (l) C(s) + O 2 (g) --> CO 2 (g) ∆H o = -393.5 kJ/mole-rxn S(s) + O 2 (g) --> SO 2 (g) ∆H o = -295.8 kJ/mole-rxn CS 2 (l) + 3O 2 (g) --> CO 2 (g) +2SO 2 (g) ∆H o = -1103.9 kJ/mole-rxn

32. 32 © 2009 Brooks/Cole - Cengage Hess’s Law: Another Example How do we start? -Let’s set our overall reaction up: C(s) + O 2 (g) --> CO 2 (g) ∆H o = -393.5 kJ/mole-rxn S(s) + O 2 (g) --> SO 2 (g) ∆H o = -295.8 kJ/mole-rxn CO 2 (g) +2SO 2 (g) --> CS 2 (l) + 3O 2 (g) ∆H o = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS 2 (l) ∆H o = ? Note: We reversed reaction 3 and changed the sign Now: We need to make sure that all of the molecules balance and take care to multiply their ∆ r H o values if necessary

33. 33 © 2009 Brooks/Cole - Cengage Hess’s Law: Another Example C(s) + O 2 (g) --> CO 2 (g) ∆H o = -393.5 kJ/mole-rxn S(s) + O 2 (g) --> SO 2 (g) ∆H o = -295.8 kJ/mole-rxn CO 2 (g) +2SO 2 (g) --> CS 2 (l) + 3O 2 (g) ∆H o = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS 2 (l) ∆H o = ? We need 2 S(s) atoms, so let’s multiply reaction2 by 2: 2S(s) + 2O 2 (g) --> 2SO 2 (g) ∆H o = -591.6 kJ/mole-rxn

34. 34 © 2009 Brooks/Cole - Cengage Hess’s Law: Another Example Now let’s look at what we’ve got: C(s) + O 2 (g) --> CO 2 (g) ∆H o 1 = -393.5 kJ/mole-rxn 2S(s) + 2O 2 (g) --> 2SO 2 (g) ∆H o 2 = -591.6 kJ/mole-rxn CO 2 (g) +2SO 2 (g) --> CS 2 (l) + 3O 2 (g) ∆H o 3 = 1103.9 kJ/mole-rxn ----------------------------------------------------------------------------------- C(s) + 2S(s) --> CS 2 (l) ∆H o = ∆H o 1 + ∆H o 2 + ∆H o 3 Everything looks good, so let’s add it up: ∆H o = (-393.5 kJ/mole-rxn)+(-591.6 kJ/mole-rxn)+(1103.95 kJ/mole-rxn) ∆H o = 118.9 kJ/mole-rxn

35. 35 © 2009 Brooks/Cole - Cengage Standard Enthalpy Values Most ∆H values are labeled ∆H o Measured under standard conditions P = 1 bar Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas

36. 36 © 2009 Brooks/Cole - Cengage Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆H f o = standard molar enthalpy of formation ∆H f o = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Appendix

37. 37 © 2009 Brooks/Cole - Cengage ∆H f o, standard molar enthalpy of formation Equations for Standard Molar Enthalpy of Formation: KNO 3 K(s) + ½ N 2 (g) + 3/2O 2 (g) --> KNO 3 FeCl 3 Fe(s) + 3/2 Cl 2 (g) --> FeCl 3 (s) C 12 H 22 O 11 12C(s) + 11H 2 (g) + 11/2 O 2 (g) --> C 12 H 22 O 11 (s)

38. 38 © 2009 Brooks/Cole - Cengage Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known, ∆H o =  ∆ f H o (products) -  ∆ f H o (reactants) Remember that ∆ always = final – initial

39. 39 © 2009 Brooks/Cole - Cengage Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆H o for CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o =  ∆H f o (prod) -  ∆H f o (react) ∆H o =  ∆H f o (prod) -  ∆H f o (react)

40. 40 © 2009 Brooks/Cole - Cengage Using Standard Enthalpy Values ∆H r o = ∆H f o (CO 2 ) + 2 ∆H f o (H 2 O) - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} = (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} - {0 + (-201.5 kJ)} ∆H r o = -675.6 kJ per mol of methanol CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆ r H o =  ∆H f o (prod) -  ∆H f o (react) ∆ r H o =  ∆H f o (prod) -  ∆H f o (react)

41. 41 © 2009 Brooks/Cole - Cengage Using Standard Enthalpy Values Calculate the heat of reaction of nitric oxide, i.e., ∆H r o for 4NH 3 (g) + 5 O 2 (g) --> 4NO(g) + 6 H 2 O(g) ∆H o =  ∆H f o (prod) -  ∆H f o (react) ∆H o =  ∆H f o (prod) -  ∆H f o (react)  ∆H o (prod) = 4(∆H f o NO) + 6(∆H f o H 2 O) = 4(90.29) + 6(-241.830) = 4(90.29) + 6(-241.830) = -1089 kJ/mole-rxn = -1089 kJ/mole-rxn  ∆H f o (react) = 4(∆H f o NH 3 ) + 5(∆H f o O 2 ) = 4(-45.90) + 5(0) = -183.6 kJ/mole-rxn ∆H r o =  ∆H f o (prod) -  ∆H f o (react) = -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = -906.2 kJ/mole-rxn = -1089 kJ/mole-rxn + 183.6 kJ/mole-rxn = -906.2 kJ/mole-rxn