1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-1 Chapter 6 Thermochemistry: Energy Flow and Chemical Change

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-2 Thermochemistry: Energy Flow and Chemical Change 6.1 Forms of Energy and Their Interconversion 6.2 Enthalpy: Heats of Reaction and Chemical Change 6.3 Calorimetry: Laboratory Measurement of Heats of Reaction 6.4 Stoichiometry of Thermochemical Equations 6.5 Hess’s Law of Heat Summation 6.6 Standard Heats of Reaction (  H 0 rxn )

3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-3 Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. Thermodynamics is the study of heat and its transformations. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings. The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes.

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-4  E = E final - E initial = E products - E reactants Figure 6.1 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.

5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-5 Figure 6.2 A system transferring energy as heat only.

6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-6 Figure 6.3 A system losing energy as work only. Energy, E Zn(s) + 2H + (aq) + 2Cl - (aq) H 2 (g) + Zn 2+ (aq) + 2Cl - (aq)  E<0 work done on surroundings

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-7 Table 6.1 The Sign Conventions* for q, w, and  E qw += EE + + - - - - + + + - depends on sizes of q and w * For q : + means system gains heat; - means system loses heat. * For w : + means word done on system; - means work done by system.

8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-8  E universe =  E system +  E surroundings Units of Energy Joule (J) Calorie (cal) British Thermal Unit 1 cal = 4.18J 1 J = 1 kg*m 2 /s 2 1 Btu = 1055 J

9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-9 Sample Problem 6.1Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (  E) in J, kJ, and kcal. SOLUTION: PLAN: Define system and surroundings, assign signs to q and w and calculate  E. The answer should be converted from J to kJ and then to kcal. q = - 325 J w = - 451 J  E = q + w = -325 J + (-451 J) =-776 J 10 3 J kJ = -0.776kJ -0.776kJ 4.18kJ kcal = -0.185 kcal

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-10 Figure 6.4 Two different paths for the energy change of a system.

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-11 Figure 6.5 Pressure-volume work.

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-12 The Meaning of Enthalpy w = - P  V  H =  E + P  V q p =  E + P  V =  H  H ≈  E in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> P  V. H = E + PV where H is enthalpy

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-13 Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. Enthalpy, H CH 4 + 2O 2 CO 2 + 2H 2 O H initial H final H 2 O( l ) H 2 O( g ) heat outheat in  H < 0  H > 0 A Exothermic processB Endothermic process CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O( g )H 2 O( l ) H 2 O( g )

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-14 Sample Problem 6.2Drawing Enthalpy Diagrams and Determining the Sign of  H PROBLEM: In each of the following cases, determine the sign of  H, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. SOLUTION: PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction (a) H 2 ( g ) + 1/2 O 2 ( g ) H 2 O( l ) + 285.8kJ (b) 40.7kJ + H 2 O( l ) H 2 O( g ) (a) The reaction is exothermic. H 2 ( g ) + 1/2 O 2 ( g )(reactants) H 2 O( l )(products) EXOTHERMIC (products)H 2 O( g ) (reactants)H 2 O( l )  H = -285.8kJ  H = +40.7kJ ENDOTHERMIC (b) The reaction is endothermic.

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-15 Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials Specific Heat Capacity (J/g*K) SubstanceSpecific Heat Capacity (J/g*K) Substance Compounds water, H 2 O( l ) ethyl alcohol, C 2 H 5 OH( l ) ethylene glycol, (CH 2 OH) 2 ( l ) carbon tetrachloride, CCl 4 ( l ) 4.184 2.46 2.42 0.864 Elements aluminum, Al graphite,C iron, Fe copper, Cu gold, Au 0.900 0.711 0.450 0.387 0.129 wood cement glass granite steel Materials 1.76 0.88 0.84 0.79 0.45

16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-16 Sample Problem 6.3Finding the Quantity of Heat from Specific Heat Capacity PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25 0 C to 300. 0 C? The specific heat capacity ( c ) of Cu is 0.387 J/g*K. SOLUTION: PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x  T to find the answer.  T in 0 C is the same as for K. q = 0.387 J g*K 125 g(300-25) 0 Cxx = 1.33x10 4 J

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-17 Figure 6.7 Coffee-cup calorimeter.

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-18 Sample Problem 6.4 Determining the Heat of a Reaction PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00 0 C and carefully add 25.0 mL of 0.500 M HCl, also at 25.00 0 C. After stirring, the final temperature is 27.21 0 C. Calculate q soln (in J) and  H rxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.18 J/g*K) PLAN: We need to determine the limiting reactant from the net ionic equation. The moles of NaOH and HCl as well as the total volume can be calculated. From the volume we use density to find the mass of the water formed. At this point q soln can be calculated using the mass, c, and  T. The heat divided by the M of water will give us the heat per mole of water formed.

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-19 total volume after mixing = 0.0750 L 0.0750 L x10 3 mL/L x1.00 g/mL= 75.0 g of water q = mass x specific heat x  T = 75.0 g x 4.18 J/g* 0 C x (27.21-25.00) 0 C = 693 J (693 J/0.0125 mol H 2 O)(kJ/10 3 J) = 55.4 kJ/ mol H 2 O formed SOLUTION: For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH - For HCl0.500 M x 0.0250 L = 0.0125 mol H + HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) HCl is the limiting reactant. 0.0125 mol of H 2 O will form during the rxn. Sample Problem 6.4 Determining the Heat of a Reaction continued

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-20 Figure 6.8 A bomb calorimeter

21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-21 Sample Problem 6.5 Calculating the Heat of Combustion PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O 2 (the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.937 0 C. Is the manufacturer’s claim correct? SOLUTION: PLAN: - q sample = q calorimeter q calorimeter = heat capacity x  T = 8.151 kJ/K x 4.937 K = 40.24 kJ 40.24 kJkcal 4.18 kJ = 9.63 kcal or Calories The manufacturer’s claim is true.

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-22 AMOUNT (mol) of compound A AMOUNT (mol) of compound B HEAT (kJ) gained or lost molar ratio from balanced equation  H rxn (kJ/mol) Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction.

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-23 Sample Problem 6.6 Using the Heat of Reaction (  H rxn ) to Find Amounts SOLUTION:PLAN: PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by If aluminum is produced this way, how many grams of aluminum can form when 1.000x10 3 kJ of heat is transferred? Al 2 O 3 (s) 2Al(s) + 3/2 O 2 (g)  H rxn = 1676 kJ heat(kJ) mol of Al g of Al 1676kJ=2mol Al x M 1.000x10 3 kJ x 2 mol Al 1676 kJ 26.98 g Al 1 mol Al = 32.20 g Al

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-24 Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown  H SOLUTION: PLAN: PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO( g ) + NO( g ) CO 2 ( g ) + 1/2 N 2 ( g )  H = ? Given the following information, calculate the unknown  H: Equation A: CO( g ) + 1/2 O 2 ( g ) CO 2 ( g )  H A = -283.0 kJ Equation B: N 2 ( g ) + O 2 ( g ) 2NO( g )  H B = 180.6 kJ Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation. Multiply Equation B by 1/2 and reverse it.  H B = -90.3 kJ CO( g ) + 1/2 O 2 ( g ) CO 2 ( g )  H A = -283.0 kJ NO( g ) 1/2 N 2 ( g ) + 1/2 O 2 ( g )  H rxn = -373.3 kJ CO( g ) + NO( g ) CO 2 ( g ) + 1/2 N 2 ( g )

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-25 Table 6.3 Selected Standard Heats of Formation at 25 0 C(298K) Formula  H 0 f (kJ/mol) calcium Ca( s ) CaO( s ) CaCO 3 ( s ) carbon C(graphite) C(diamond) CO( g ) CO 2 ( g ) CH 4 ( g ) CH 3 OH( l ) HCN( g ) CS s ( l ) chlorine Cl( g ) 0 -635.1 -1206.9 0 1.9 -110.5 -393.5 -74.9 -238.6 135 87.9 121.0 hydrogen nitrogen oxygen Formula  H 0 f (kJ/mol) H( g ) H2(g)H2(g) N2(g)N2(g) NH 3 ( g ) NO( g ) O2(g)O2(g) O3(g)O3(g) H 2 O( g ) H 2 O( l ) Cl 2 ( g ) HCl( g ) 0 0 0 -92.3 0 218 -45.9 90.3 143 -241.8 -285.8 107.8 Formula  H 0 f (kJ/mol) silver Ag( s ) AgCl( s ) sodium Na( s ) Na( g ) NaCl( s ) sulfur S 8 (rhombic) S 8 (monoclinic) SO 2 ( g ) SO 3 ( g ) 0 0 0 -127.0 -411.1 2 -296.8 -396.0

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-26 Sample Problem 6.8 Writing Formation Equations PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include  H 0 f. SOLUTION: PLAN: (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. Use the table of heats of formation for values. (c) Hydrogen cyanide, HCN, a gas at standard conditions.  H 0 f = -127.0 kJ (a) Ag( s ) + 1/2 Cl 2 ( g ) AgCl( s )  H 0 f = -1206.9 kJ (b) Ca( s ) + C( graphite ) + 3/2 O 2 ( g ) CaCO 3 ( s )  H 0 f = 135 kJ (c) 1/2 H 2 ( g ) + C( graphite ) + 1/2 N 2 ( g ) HCN( g )

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-27 Figure 6.10 The general process for determining  H 0 rxn from  H 0 f values. Enthalpy, H Elements Reactants Products  H 0 rxn =  m  H 0 f(products) -  n  H 0 f(reactants) decomposition -H0f-H0f H0fH0f formation  H 0 rxn H initial H final

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-28 Sample Problem 6.9Calculating the Heat of Reaction from Heats of Formation SOLUTION: PLAN: PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: Calculate  H 0 rxn from  H 0 f values. 4NH 3 ( g ) + 5O 2 ( g ) 4NO( g ) + 6H 2 O( g ) Look up the  H 0 f values and use Hess’s Law to find  H rxn.  H rxn =  m  H 0 f (products) -  n  H 0 f (reactants)  H rxn = [4(  H 0 f NO( g ) + 6(  H 0 f H 2 O( g )]- [4(  H 0 f NH 3 ( g ) + 5(  H 0 f O 2 ( g )] = (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) - [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]  H rxn = -906 kJ

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 6-29 Figure 6.11 The trapping of heat by the atmosphere.