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Chapter 6 Thermochemistry: Energy Flow and Chemical Change 1 Slide 6-1 Chapter 6: First Law 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 (ΔH0rxn) 2 Slide 6-2 Chapter 6: First Law Thermodynamics: First Law Thermodynamics: Study and measure of the nature of heat and energy and its transformations Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. ✔Energy flow is key in science; understanding it will help us to understand WHY things occur. When energy is transferred from one object to another, it appears as work and/or as heat Thermodynamics was developed independent of theory through observation of bulk matter To deal with large groups of molecules and atoms we need to define THREE parameters: System, Surroundings and Thermodynamic Universe 3 Slide 6-3 Chapter 6: First Law Thermodynamics: First Law System: The part of the universe of immediate interest; composed of particles with their own internal energies (E or U). The system has an internal energy. When a change occurs, the internal energy changes. Can be closed, open or isolated; Sign is from the point of view of the System Surroundings: The portion of the universe that can exchange energy and matter with the system; Thermodynamic Universe: System + Surroundings Open System: Exchanges both E and matter with Surroundings; Closed System: Exchanges only E with Surroundings; Isolated System: Exchanges NEITHER with Surroundings Slide 6-4 Chapter 6: First Law 6-4 Figure 6.1 A chemical system and its surroundings. The system in this case is the contents of the reaction flask. The surroundings comprise everything else, including the flask itself. 5 Slide 6-5 Chapter 6: First Law An Auto Slide 6-6 A Battery Chapter 6: First Law A perfect Thermos 6 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. A. E of system decreasses B. E of system increases ΔE = Efinal - Einitial = Eproducts - Ereactants 7 Slide 6-7 Chapter 6: First Law A system transferring energy as heat only. A E lost as heat B E gained as heat 8 Slide 6-8 Chapter 6: First Law A system losing energy as work only. 9 Slide 6-9 Chapter 6: First Law Work (w) Motion against an opposing force; a directed motion thru a distance; w = F•Δx Capacity of a system to do work is called Internal Energy (U) Most important: ΔU = Ufinal – Uinitial If E transfer is only due to work then ΔU = w Work is either expansion or nonexpansion (ΔV≠ 0 vs. ΔV= 0) w = –PextΔV (since w = – when system does work) 10 Slide 6-10 Chapter 6: First Law Work w = – PextΔV (Pext must be constant) Work is in L-atm: 1 L-atm = 101.325 J If is Pext = 0, then free expansion occurs If Pext does change in infinitesimally small amts, then process is REVERSIBLE: a process in which an infinitesimally small change in the opposite direction will cause the process to reverse direction Chapter 6: First Law 11 Slide 6-11 Figure 6.5 Pressure-volume work. 12 Slide 6-12 Chapter 6: First Law Figure 6.7 Pressure-volume work. An expanding gas pushing back the atmosphere does PV work (w = -PΔV). 13 Slide 6-13 Chapter 6: First Law Heat (q) Energy transfer due to a temperature difference More familiar way to transfer E Sys and Surr are at different T s If no work is done, then ΔU = q Temperature is not heat; it is directly proportional to it (q ∝ ΔT)! Temperature: a measure of the relative hotness or coldness of an object q = –qsurr Exothermic vs Endothermic Chapter 6: First Law Slide 6-14 6-14 Calorimetry q = c x m x ΔT q = heat lost or gained c = specific heat capacity m = mass in g ΔT = Tfinal - Tinitial The specific heat capacity (c) of a substance is the quantity of heat required to change the temperature of 1 gram of the substance by 1 K. 15 Slide 6-15 Chapter 6: First Law Measuring Heat Using a Calorimeter Constant Pressure Slide 6-16 Constant Volume Chapter 6: First Law 6-16 Figure 6.7 Coffee-cup calorimeter. 17 Slide 6-17 Chapter 6: First Law Figure 6.8 A bomb calorimeter. 18 Slide 6-18 Chapter 6: First Law 19 Slide 6-19 Chapter 6: First Law Selected Heat Capacities (Cs and Cm) Slide 6-20 Amt of heat to raise 10.0 g of water 5°C: 209.2 J Temp increase if same amount of heat is added to 10.0 g of gold (Cs = 0.106 J/g-C): 197.4 °C! Chapter 6: First Law 6-20 21 Slide 6-21 Chapter 6: First Law Cm,V vs. Cm,P q q Cm = = n!T !T Since q = nCmΔT then (n = 1) However, q = ΔU at constant volume or q = ΔH at constant pressure, So: Cm,V !U = n!T OR Cm, P !H = n!T and Cm,P = Cm,V + R For Ar: Cm,V =12.8 J/mol-K and Cm,P = 21.1 J/mol-K Chapter 6: First Law Slide 6-22 6-22 Sample Problem 6.3 Finding 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 25oC to 300.oC? The specific heat capacity (c) of Cu is 0.387 J/g•K. PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x ΔT to find the answer. ΔT in oC is the same as for K. SOLUTION: q = 0.387 J/g•K x 125 g x (300oC - 25oC) = 1.33 x 104 J 23 Slide 6-23 Chapter 6: First Law Figure 6.4 Two different paths for the energy change of a system. 24 Slide 6-24 Chapter 6: First Law First Law of Thermodynamics The internal energy of an isolated system is constant • ΔU = q + w • Isolated system, ΔU = 0 • Container walls can be either adiabatic (q = 0) or diathermic (w = 0) • Both q and w are path dependent Internal Energy (U) is a State Function To determine the change in a state function, choose any convenient path (even if the real path is too complicated to calculate) Chapter 6: First Law Slide 6-25 6-25 ΔEuniverse = ΔEsystem + ΔEsurroundings Units of Energy Joule (J) calorie (cal) 1 J = 1 kg•m2/s2 1 cal = 4.184 J British thermal unit (Btu) 1 Btu = 1055 J 26 Slide 6-26 Chapter 6: First Law Sample Problem 6.1 PROBLEM: PLAN: Determining the Change in Internal Energy of a System 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 (ΔE) in J, kJ, and kcal. Define system and surroundings, assign signs to q and w and calculate ΔE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J ΔE = q + w = -325 J + (-451 J) = -776 J x kJ 103J = -0.776 kJ -0.776 kJ x -776 J kcal 4.184 kJ = -0.185 kcal 27 Slide 6-27 Chapter 6: First Law Enthalpy (H) The heat released or absorbed by a system at constant pressure • For Constant Volume systems ΔU = q • Most Processes WE study are Constant Pressure systems • Enthalpy is a State Function that allows us to keep track of q at constant P; By definition: H = U + PV • ΔH = ΔU + PΔV = q – PextΔV + PΔV = q – PextΔV + PextΔV • So ΔH = q • Implication 1: At constant volume you measure ΔU; at constant pressure you measure ΔH • Implication 2: The molar heat capacity Cm = q/(nΔT) Therefore it will be different for a constant volume process than for a constant pressure process Chapter 6: First Law Slide 6-28 6-28 The Meaning of Enthalpy w = - PΔV H = E + PV ΔH ≈ ΔE in 1. Reactions that do not involve gases. where H is enthalpy ΔH = ΔE + PΔV qp = ΔE + PΔV = ΔH 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 >>> PΔV. 29 Slide 6-29 Chapter 6: First Law Mathematical Aspects of State Functions Given X is a state function, then: • The change in X is equal to the sum of the X values for the products minus the reactants: ΔX = ΣniXpdts – ΣnjXrxtnts • The value for the reverse process is equal in magnitude BUT opposite in sign: ΔXrev = –ΔXfwd • Multiplication of a process by number multiplies the X value by that number: If A → 2B ΔX1 = 7 then 3A → 6B ΔX2 = 3ΔX1 = 21 • The sum of all of the X values making up a process is equal to the X overall for that process: ΔXtotal = ΔXstep 1 + ΔXstep 2 + ΔXstep 3 + … Chapter 6: First Law Slide 6-30 6-30 Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. 31 Slide 6-31 Chapter 6: First Law ΔH of Physical Processes Chapter 6: First Law Slide 6-32 6-32 Heating Curves o ΔHvap ΔHfus q q q o T1 (–5 C)) T2 (+120 C) Chapter 6: First Law Slide 6-33 6-33 Heating Curve Calculations To take 0.5 mole (9.0 g) of ice from –5ºC to vapor at 120ºC: Total heat = (heat to take 0.5 mol of ice from –5ºC to 0ºC) + (heat to melt 0.5 mol of ice) + (heat to take 0.5 mol of water from 0ºC to 100ºC) + (heat to vaporize 0.5 mol of ice) + (heat to take 0.5 mol of vapor from 100ºC to 120ºC) Chapter 6: First Law Slide 6-34 6-34 ΔH of Chemical Change • Enthalpy change during chemical reactions important to understanding chemistry. • Reaction Enthalpies: ΔH or ΔHrxn • ΔH associated with Thermochemical Equations: 2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + H2 (g) ΔH = –1049 kJ • ΔHrxn associated with certain types of reactions; corrected to a per mole basis; used in tabular collections of data: ΔHrxn = –524.5 kJ/mol for the HCl oxidation of Al ΔHcombustion = –1560 kJ/mol for ethane (C2H6) • Relation between ΔH and ΔU: For reactions in which NO gas is generated: ΔH = ΔU For reactions in which gas is generated: ΔH = ΔU + ΔngasRT Chapter 6: First Law Slide 6-35 6-35 Standard Reaction Enthalpies ΔH° ΔH depends on conditions and phases; to allow standardization for reporting the idea of Standard State was developed. The standard state of a property X (designated as X°) is defined as such: Substances and elements: most stable form of that material at the T of interest and 1 bar of pressure; Solutions: 1.0 M concentration; Allows the tabulation and reporting of data from various sources to be accomplished! Chapter 6: First Law Slide 6-36 6-36 Common ΔHx° s Enthalpy of Formation (ΔHf°) Defined as: the standard reaction enthalpy for the formation of one mole of a substance in its most stable form from its elements in their most stable form at the temperature of interest: 2 Na(s) + C(s,gr) + 3/2 O2(g) Na2CO3 (s) By Definition ΔHf°≡ 0 of an element in its most stable state; Allows ΔHrxn° to be calculated using ΔHf° of pdts - rxtnts: aA + bB dD ΔHf° = [dΔH°f,D] – [aΔH°f,A – bΔH°f,B] Enthalpy of Combustion (ΔHc° or ΔH°comb) Defined as: the standard reaction enthalpy for the combustion of one mole of a substance in its most stable form at T of interest: C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O (l) Chapter 6: First Law Slide 6-37 6-37 Sample Problem 6.2 PROBLEM: Drawing Enthalpy Diagrams and Determining the Sign of ΔH In each of the following cases, determine the sign of ΔH, state whether the reaction is exothermic or endothermic, and draw an enthalpy diagram. 1 (a) H2(g) + O (g) H2O(l) + 285.8 kJ 2 2 (b) 40.7 kJ + H2O(l) 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. H2O(g) SOLUTION: 38 Slide 6-38 Chapter 6: First Law 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 25.00oC and carefully add 25.0 mL of 0.500 M HCl, also at 25.00oC. After stirring, the final temperature is 27.21oC. Calculate qsoln (in J) and ΔHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specific heat capacity as water: d = 1.00 g/mL and c = 4.184 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 qsoln can be calculated using the mass, c, and ΔT. The heat divided by the M of water will give us the heat per mole of water formed. 39 Slide 6-39 Chapter 6: First Law Sample Problem 6.4 Determining the Heat of a Reaction continued SOLUTION: HCl(aq) + NaOH(aq) H+(aq) + OH-(aq) NaCl(aq) + H2O(l) H2O(l) For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH- For HCl 0.500 M x 0.0250 L = 0.0125 mol H+ HCl is the limiting reactant. 0.0125 mol of H2O will form during the reaction. 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 ΔT = 75.0 g x 4.184 J/g•oC x (27.21oC - 25.00oC) = 693 J (693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed 40 Slide 6-40 Chapter 6: First Law Sample Problem 6.5 PROBLEM: PLAN: Calculating the Heat of a Combustion Reaction 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 = 8.151 kJ/K). The temperature increases 4.937oC. Is the manufacturer s claim correct? - q sample = qcalorimeter SOLUTION: qcalorimeter (First Law of Thermodynamics) = heat capacity x ΔT = 8.151 kJ/K x 4.937 K = 40.24 kJ 1 Calorie = 1 kcal = 4.184 kJ 10 Calorie = 41.84 kJ The manufacturer’s claim is true. 41 Slide 6-41 Chapter 6: First Law Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. 42 Slide 6-42 Chapter 6: First Law Sample Problem 6.6 PROBLEM: PLAN: Using the Heat of Reaction (ΔHrxn) to Find Amounts The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by 3 Al2O3(s) 2Al(s) + O2(g) ΔHrxn = 1676 kJ 2 If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is transferred? SOLUTION: 1.000 x 103 kJ x 2 mol Al 1676 kJ x 26.98 g Al 1 mol Al = 32.20 g Al 43 Slide 6-43 Chapter 6: First Law Sample Problem 6.8 PROBLEM: Using Hess’s Law to Calculate an Unknown ΔH Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction: CO(g) + NO(g) → CO2(g) + ½N2(g) ΔH = ? Given the following information, calculate the unknown ΔH: Equation A: CO(g) + ½ O2(g) → CO2(g) ΔHA = -283.0 kJ Equation B: N2(g) + O2(g) → 2NO(g) ΔHB = 180.6 kJ PLAN: Manipulate Equations A and/or B and their ΔH values to get to the target equation and its ΔH. All substances except those in the target equation must cancel. 44 Slide 6-44 Chapter 6: First Law Sample Problem 6.8 SOLUTION: Multiply Equation B by ½ and reverse it: NO(g) → ½ N2(g) + ½ O2(g); ΔH = - 90.3 kJ Add the manipulated equations together: Equation A: CO(g) + ½ O2(g) → CO2(g) ½ Equation B: (reversed) NO(g) ΔH = -283.0 kJ → ½ N2(g) + ½ O2(g) CO(g) + NO(g) → ΔH = - 90.3 kJ CO2(g) + ½ N2(g) ΔHrxn = -373.3 kJ 45 Slide 6-45 Chapter 6: First Law 46 Slide 6-46 Chapter 6: First Law Table 6.3 Selected Standard Enthalpies of Formation at 25°C (298K) Formula ΔH°f (kJ/mol) Formula Calcium Ca(s) CaO(s) CaCO3(s) 0 -635.1 -1206.9 Carbon C(graphite) C(diamond) CO(g) CO2(g) CH4(g) CH3OH(l) HCN(g) CSs(l) Chlorine Cl(g) Slide 6-47 0 1.9 -110.5 -393.5 -74.9 -238.6 135 87.9 121.0 ΔH°f (kJ/mol) Formula ΔH°f (kJ/mol) 0 -92.3 Hydrogen H(g) H 2 (g ) Silver Ag(s) AgCl(s) 218 0 Sodium Nitrogen N 2 (g ) NH3(g) NO(g) 0 -45.9 90.3 Cl2(g) HCl(g) Oxygen O2(g) O3(g) H2O(g) H2O(l) Chapter 6: First Law 0 143 -241.8 -285.8 Na(s) Na(g) NaCl(s) 0 -127.0 0 107.8 -411.1 Sulfur S8(rhombic) 0 S8(monoclinic) 0.3 SO2(g) -296.8 SO3(g) -396.0 47 Sample Problem 6.9 PROBLEM: 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 ΔH°f. (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: Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of ΔH°f in Table 6.3 or Appendix B. 48 Slide 6-48 Chapter 6: First Law Sample Problem 6.9 SOLUTION: (a) Silver chloride, AgCl, a solid at standard conditions. Ag(s) + ½Cl2(g) → AgCl(s) ΔH°f = -127.0 kJ (b) Calcium carbonate, CaCO3, a solid at standard conditions. Ca(s) + C(graphite) + 3 O22(g) → CaCO3(s) ΔH°f = -1206.9 kJ (c) Hydrogen cyanide, HCN, a gas at standard conditions. ½H2(g) + C(graphite) + ½N2(g) → HCN(g) ΔH°f = 135 kJ 49 Slide 6-49 Chapter 6: First Law Slide 6-50 Chapter 6: First Law 6-50 Slide 6-51 Chapter 6: First Law 6-51 Hess Law of Heat Summation Let's say we want to determine the reaction enthalpy for the following reaction: H2 (g) + Cl2 (g) ! 2HCl (g) H°rxn = ?? given the following individual reactions: NH3 (g) + HCl (g) ! NH4Cl (s) H1° = -176.0 kJ N2 (g) + 3H2 (g) ! 2NH3 (g) H2° = -92.22 kJ N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H3° = -628.66 kJ Start by adding two reactions that give us the reactants we want on the correct side: Step 1 N2 (g) + 4H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H3° 2NH3 (g) ! N2 (g) + 3H2 (g) -( H2°) 2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s) H°(step 1) = H3° + -( H2°) Step 2 2NH3 (g) + H2 (g) + Cl2 (g) ! 2 NH4Cl (s) 2[NH4Cl (s) ! NH3 (g) + HCl (g)] H2 (g) + Cl2 (g) ! 2HCl (g) So H° rxn = H°(step 1) + 2(- H1°) = H° rxn = -184.64 kJ Slide 6-52 H°(step 1) 2(- H1°) H3° + -( H2°)+ 2(- H1°) Chapter 6: First Law 6-52 Figure 6.10 The general process for determining ΔHorxn from ΔHof values. 53 Slide 6-53 Chapter 6: First Law Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation PROBLEM: Nitric acid, whose worldwide annual production is about 10 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 ΔHorxn from ΔHof values. PLAN: Look up the ΔHof values and use Hess s law to find ΔHrxn. SOLUTION: ΔHrxn = Σ mΔHof (products) - Σ nΔHof (reactants) ΔHrxn = {4[ΔHof NO(g)] + 6[ΔHof H2O(g)]} - {4[ΔHof NH3(g)] + 5[ΔHof 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)] ΔHrxn = -906 kJ Slide 6-54 Chapter 6: First Law 54 Figure 6.11 The trapping of heat by the atmosphere. 55 Slide 6-55 Chapter 6: First Law End of Chapter 6 Silberberg HW: 10, 14, 15, 19, 35, 37, 39, 41, 48, 52, 54, 57, 59, 64, 65, 67, 71, 77, 79, 81, 82, 92, 97, 98, 100 56 Slide 6-56 Chapter 6: First Law