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Chapter 6

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

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Chapter 6

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  1. Chapter6 Thermochemistry: Energy Flow and Chemical Change

  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 (DH0rxn)

  3. Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. 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. Figure 6.1 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. DE = Efinal - Einitial = Eproducts - Ereactants

  5. A system transferring energy as heat only. Figure 6.2

  6. Energy, E work done on surroundings DE<0 A system losing energy as work only. Figure 6.3 Zn(s) + 2H+(aq) + 2Cl-(aq) H2(g) + Zn2+(aq) + 2Cl-(aq)

  7. Table 6.1 The Sign Conventions* for q, w, and DE + = DE q w + + + + - depends on sizes of q and w - + 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. DEuniverse = DEsystem + DEsurroundings Units of Energy Joule (J) 1 J = 1 kg*m2/s2 Calorie (cal) 1 cal = 4.18J British Thermal Unit 1 Btu = 1055 J

  9. PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O 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 (DE) in J, kJ, and kcal. kcal kJ -776J -0.776kJ 4.18kJ 103J Sample Problem 6.1 Determining the Change in Internal Energy of a System PLAN: Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J DE = q + w = -325 J + (-451 J) = -776 J = -0.776kJ = -0.185 kcal

  10. Two different paths for the energy change of a system. Figure 6.4

  11. Figure 6.5 Pressure-volume work.

  12. The Meaning of Enthalpy w = - PDV DH ≈ DE in H = E + PV 1. Reactions that do not involve gases. where H is enthalpy 2. Reactions in which the number of moles of gas does not change. DH = DE + PDV 3. Reactions in which the number of moles of gas does change but q is >>> PDV. qp = DE + PDV = DH

  13. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g) Enthalpy, H Enthalpy, H heat out heat in Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. CH4 + 2O2 H2O(g) Hfinal Hinitial DH < 0 DH > 0 CO2 + 2H2O H2O(l) Hfinal Hinitial A Exothermic process B Endothermic process

  14. PROBLEM: In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. (a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ (b) 40.7kJ + H2O(l) H2O(g) H2(g) + 1/2O2(g) (reactants) H2O(g) (products) H2O(l) (products) H2O(l) (reactants) Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of DH 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 SOLUTION: (a) The reaction is exothermic. (b) The reaction is endothermic. EXOTHERMIC DH = -285.8kJ ENDOTHERMIC DH = +40.7kJ

  15. Substance Specific Heat Capacity (J/g*K) Substance Specific Heat Capacity (J/g*K) Elements Materials aluminum, Al 0.900 wood 1.76 graphite,C 0.711 cement 0.88 iron, Fe 0.450 glass 0.84 copper, Cu 0.387 granite 0.79 gold, Au 0.129 steel 0.45 Compounds water, H2O(l) 4.184 ethyl alcohol, C2H5OH(l) 2.46 ethylene glycol, (CH2OH)2(l) 2.42 carbon tetrachloride, CCl4(l) 0.864 Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

  16. 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 250C to 300.0C? The specific heat capacity (c) of Cu is 0.387 J/g*K. 0.387 J g*K Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat Capacity PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x DT to find the answer. DT in 0C is the same as for K. SOLUTION: = 1.33x104 J q = x 125 g x (300-25)0C

  17. Figure 6.7 Coffee-cup calorimeter.

  18. PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and DHrxn (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) Sample Problem 6.4 Determining the Heat of a Reaction 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 qsoln can be calculated using the mass, c, and DT. The heat divided by the M of water will give us the heat per mole of water formed.

  19. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) 0.500 M x 0.0500 L = 0.0250 mol OH- For NaOH For HCl 0.500 M x 0.0250 L = 0.0125 mol H+ Sample Problem 6.4 Determining the Heat of a Reaction continued SOLUTION: HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn. total volume after mixing = 0.0750 L 0.0750 L x 103 mL/L x 1.00 g/mL = 75.0 g of water q = mass x specific heat x DT = 75.0 g x 4.18 J/g*0C x (27.21-25.00)0C = 693 J (693 J/0.0125 mol H2O)(kJ/103 J) = 55.4 kJ/ mol H2O formed

  20. Figure 6.8 A bomb calorimeter

  21. 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 O2(the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.9370C. Is the manufacturer’s claim correct? 40.24 kJ kcal 4.18 kJ Sample Problem 6.5 Calculating the Heat of Combustion PLAN: - q sample = qcalorimeter SOLUTION: qcalorimeter = heat capacity x DT = 8.151 kJ/K x 4.937 K = 40.24 kJ = 9.63 kcal or Calories The manufacturer’s claim is true.

  22. AMOUNT (mol) of compound A AMOUNT (mol) of compound B HEAT (kJ) gained or lost Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. molar ratio from balanced equation DHrxn (kJ/mol)

  23. PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by Al2O3(s) 2Al(s) + 3/2O2(g) DHrxn = 1676 kJ If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred? 26.98 g Al 2 mol Al 1 mol Al 1676 kJ Sample Problem 6.6 Using the Heat of Reaction (DHrxn) to Find Amounts PLAN: SOLUTION: heat(kJ) 1.000x103 kJ x 1676kJ=2mol Al mol of Al = 32.20 g Al x M g of Al

  24. 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) CO2(g) + 1/2N2(g) DH = ? Given the following information, calculate the unknown DH: Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ Equation B: N2(g) + O2(g) 2NO(g) DHB = 180.6 kJ CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ NO(g) 1/2N2(g) + 1/2O2(g) CO(g) + NO(g) CO2(g) + 1/2N2(g) Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown DH PLAN: 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. SOLUTION: Multiply Equation B by 1/2 and reverse it. DHB = -90.3 kJ DHrxn = -373.3 kJ

  25. Formula DH0f(kJ/mol) Formula DH0f(kJ/mol) Formula DH0f(kJ/mol) calcium silver Cl2(g) 0 Ca(s) 0 Ag(s) 0 -92.3 HCl(g) CaO(s) -635.1 AgCl(s) -127.0 -1206.9 CaCO3(s) hydrogen H(g) 218 sodium carbon H2(g) 0 Na(s) 0 C(graphite) 0 Na(g) 107.8 1.9 C(diamond) nitrogen NaCl(s) -411.1 CO(g) -110.5 N2(g) 0 CO2(g) -393.5 NH3(g) -45.9 sulfur CH4(g) -74.9 NO(g) 90.3 S8(rhombic) 0 CH3OH(l) -238.6 S8(monoclinic) 2 oxygen HCN(g) 135 SO2(g) -296.8 O2(g) 0 87.9 CSs(l) O3(g) 143 SO3(g) -396.0 chlorine H2O(g) -241.8 Cl(g) 121.0 H2O(l) -285.8 Table 6.3 Selected Standard Heats of Formation at 250C(298K)

  26. PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH0f. (a) Ag(s) + 1/2Cl2(g) AgCl(s) (b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s) (c)1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g) Sample Problem 6.8 Writing Formation Equations (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Use the table of heats of formation for values. SOLUTION: DH0f = -127.0 kJ DH0f = -1206.9 kJ DH0f = 135 kJ

  27. decomposition formation Figure 6.10 The general process for determining DH0rxn from DH0f values. Elements -DH0f DH0f Enthalpy, H Reactants Hinitial DH0rxn Products Hfinal DH0rxn = SmDH0f(products) - SnDH0f(reactants)

  28. 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: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Calculate DH0rxn from DH0f values. Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation PLAN: Look up the DH0f values and use Hess’s Law to find DHrxn. SOLUTION: DHrxn = SmDH0f (products) - SnDH0f (reactants) DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f O2(g)] = (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) - [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] DHrxn = -906 kJ

  29. Figure 6.11 The trapping of heat by the atmosphere.

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