1. 6-1 Thermochemistry: Energy Flow and Chemical Change Chapter 6

2. 6-2 화학반응의 에너지 관계 에너지의 본성과 에너지의 형태 화학반응에서의 에너지 변화, ΔE 열역학. 계와 상태함수 열역학 제 1 법칙, 일 (w) 과 열 (q) 화학반응에서의 엔탈피, H 열계량법 표준생성 (ΔH 0 f ) 및 반응엔탈피 (ΔH 0 rxn )

3. 6-3 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 )

4. 6-4 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.

5. 6-5 isolated Various Systems systemmatterenergy open( 열린 계 ) allowed closed( 닫힌 계 ) forbiddenallowed isolated( 고립 계 ) forbidden Some other thermodynamics terms state, state functions, path, process, extensive properties, intensive properties. closed open

6. 6-6 계물질에너지 열린계○○ 닫힌계 × ○ 고립계 ×× 우주 = 계 + 주위 Universe = System + Surroundings 열역학의 용어들 상태 1 (state) State functions P, V, T, S, H, E, m, n, … 상태 2 경로 (path) q, w 과정 (process) 단열, 정압, 정적, 가역, 비가역, 발열, 흡열, …

7. 6-7  E univ =  E sys +  E surr Units of Energy Joule (J) Calorie (cal) British Thermal Unit 1 cal ≡ 4.184J 1 J = 1N∙m = 1 kg∙m 2 /s 2 1 Btu = 1055 J

8. 6-8 -- Heat and work are forms of energy transfer and energy is conserved. The First Law of Thermodynamics  E = Q + W work done on the system change in total internal energy heat added to system State Function Process Functions

9. 6-9 Calculating the change in internal energy We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating. Suppose someone does 622 kJ of work on an exercise bicycle and loses 82 kJ of energy as heat. What is the change in internal energy of the person? Disregard any matter loss by perspiration. Solution w = −622 kJ (622 kJ is lost by doing work on the bicycle), q = −82 kJ (82 kJ is lost by heating the surroundings). Then the first law of thermodynamics gives us  E = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ

10. 6-10 Work, and the expansion(p-V) work FF dW p ex =F/A Increase in volume, dV (p-V work) Work = force distance force acting on the piston = p ex × Area distance when expand = dy +y

11. 6-11 Total Work Done To evaluate the integral, we must know how the pressure depends (functionally) on the volume. We will consider the following cases 0. Constant volume work 1.Free expansion 2.Expansion against constant pressure

12. 6-12 0. Work for the constant volume process For the constant volume process, there is no p-V work

13. 6-13 Heat and Internal Energy  E = q + w = q V For the constant volume process, there is no p-V work, so If we add heat q to the system, the temperature of the system increases by  T, and the internal energy increases by  E. Constant volume heat capacity

14. 6-14 A constant-volume bomb calorimeter. The constant-volume heat capacity is the slope of a curve showing how the internal energy varies with temperature. The slope, and therefore the heat capacity, may be different at different temperatures.

15. 6-15 Determining the Change in Internal Energy of a System 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: 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

16. 6-16 molar constant volume heat capacity of monatomic gases (at 1 atm, 25 °C) Monatomic gasC V, m (J/(mol·K))C V, m /R He 12.51.50 Ne 12.51.50 Ar 12.51.50 Kr 12.51.50 Xe 12.51.50

17. 6-17 1. Work for free expansion case For the free expansion process, there is no p-V work  E = q + w = q

18. 6-18 2. Work for expansion against constant pressure  E = q + w = q - p  V q p =  E + p  V =  (E + p V ) If we define Enthalpy as, H ≡ E + pV q p =  H Enthalpy change is the heat given to the system at constant pressure.

19. 6-19 Heat and Enthalpy  H = q P For the constant pressure process, If we add heat q to the system, the temperature of the system increases by  T, and the internal energy increases by  H. Constant pressure heat capacity

20. 6-20 Enthalpy and Internal Energy q p = ΔE + P Δ V = Δ H Δ H ≈ ΔE 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 q V = ΔE ΔE = q + w

21. 6-21 An exothermic process 발열반응 Fe 2 O 3 (s) + 2Al(s) → Al 2 O 3 (s) + 2Fe (s) if the reaction carried out at constant pressure, and the heat out -q p = -  H > 0, it is called an exothermic reaction. Fe 2 O 3 (s) + 2Al(s) heat out Al 2 O 3 (s) + 2Fe (s) Enthalpy, H

22. 6-22 An endothermic process 흡열반응 NH 4 NO 3 (s) + H 2 O (l) → NH 4 + (aq) + NO 3 − (aq) if the reaction carried out at constant pressure, and the heat in q p =  H > 0, it is called an endothermic reaction. Enthalpy, H heat in

23. 6-23 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

24. 6-24 Finding the Quantity of Heat from Specific Heat Capacity 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. q = 0.389 J/gK × 125 g × (300-25) 0 C= 1.33 × 10 4 J

25. 6-25 Coffee-cup calorimeter. 온도계 막대젓개 이중 보온 컵 뚜껑

26. 6-26 Determining the Heat of a Reaction You place 50.0 mL of 0.500 M NaOH in a 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) 50.0 mL of 0.500 M NaOH 25.0 mL of 0.500 M HCl 75.0 mL 27.21 0 C25.00 0 C

27. 6-27 total volume after mixing = 0.0750 L 0.0750 L × 10 3 mL/L × 1.00 g/mL= 75.0 g of water q = mass × specific heat ×  T = 75.0 g × 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 For NaOH 0.500 M × 0.0500 L = 0.0250 mol OH - For HCl0.500 M × 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.

28. 6-28 Calculating the Heat of Combustion 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? q calorimeter = heat capacity ×  T = 8.151 kJ/K × 4.937 K = 40.24 kJ 40.24 kJ ×( kcal/4.184 kJ) = 9.63 kcal or Calories The manufacturer’s claim is true.

29. 6-29 Using the Heat of Reaction (  H rxn ) to Find Amounts 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 1.000 × 10 3 kJ of heat is transferred? Al 2 O 3 (s) → 2Al(s) + 3/2 O 2 (g)  H rxn = 1676 kJ 1.000x10 3 kJ ×( 2 mol Al / 1676 kJ )×( 26.98 g Al /mol) = 32.20 g Al

30. 6-30 Using Hess’s Law to Calculate an Unknown  H 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: (1) CO( g ) + 1/2 O 2 ( g ) → CO 2 ( g )  H A = -283.0 kJ (2) N 2 ( g ) + O 2 ( g ) → 2NO( g )  H B = 180.6 kJ Multiply Equation (2) 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 )

31. 6-31 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

32. 6-32 Writing Formation Equations 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. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (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 )

33. 6-33 The general process for determining  H 0 rxn from  H 0 f values. aA + bB → cC + dD Δ H 0 rxn = {c Δ H 0 f (C) + d Δ H 0 f (D) } – {a Δ H 0 f (A) + b Δ H 0 f (B) } products reactants

34. 6-34 Calculating the Heat of Reaction from Heats of Formation 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: 4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( g ) Calculate  H 0 rxn from  H 0 f.  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

35. 6-35 Figure 6.11 The trapping of heat by the atmosphere.