6.5 Hess s Law of Heat Summation. 2 Chapter 6: First Law. Slide 6-2

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1 1/3/11 Thermochemistry: Energy Flow and Chemical Change Chapter Forms of Energy and Their Interconversion 6.2 Enthalpy: Heats of Reaction and Chemical Change Thermochemistry: Energy Flow 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 (ΔH rxn ) 1 2 Slide 6-1 Slide 6-2 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 Sign is from the point of view of the System 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; Surroundings: The portion of the universe that can exchange energy and matter with the system; Thermodynamic Universe: System + Surroundings To deal with large groups of molecules and atoms we need to define THREE parameters: System, Surroundings and Thermodynamic Universe Open System: Exchanges both E and matter with Surroundings; Closed System: Exchanges only E with Surroundings; Isolated System: Exchanges NEITHER with Surroundings 3 Slide 6-3 Slide 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. Slide An Auto A Battery A perfect Thermos Slide

2 1/3/11 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. A. E of system decreasses A system transferring energy as heat only. B. E of system increases A E lost as heat B E gained as heat ΔE = Efinal - Einitial = Eproducts - Ereactants 7 Slide Slide 6-8 Work (w) A system losing energy as work only. 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 vs. ΔV= ) w = PextΔV (since w = when system does work) 9 Slide Slide 6-1 Figure 6.5 Work Pressure-volume work. w = PextΔV (Pext must be constant) Work is in L-atm: 1 L-atm = J If is Pext =, 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 11 Slide Slide

3 1/3/11 Figure 6.7 Pressure-volume work. Heat (q) Energy transfer due to a temperature difference Slide 6-13 An expanding gas pushing back the atmosphere does PV work (w = -PΔV). Slide 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 = q surr Exothermic vs Endothermic 6-14 Calorimetry Measuring Heat Using a Calorimeter q = c x m x ΔT q = heat lost or gained c = specific heat capacity m = mass in g ΔT = T final - T initial 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. Slide Constant Pressure Constant Volume Slide Figure 6.7 Coffee-cup calorimeter. Figure 6.8 A bomb calorimeter Slide 6-17 Slide

4 1/3/11 Selected Heat Capacities (Cs and Cm) Amt of heat to raise 1. g of water 5 C: 29.2 J Temp increase if same amount of heat is added to 1. g of gold (Cs =.16 J/g-C): C! 19 Slide 6-19 Slide Cm,V vs. Cm,P Cm = Since q = ncmδt then q q = n!t!t (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 21 Slide 6-21 Sample Problem 6.3 Slide Finding the Quantity of Heat from Specific Heat Capacity Figure 6.4 Two different paths for the energy change of a system. 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 3.oC? The specific heat capacity (c) of Cu is.387 J/g K. 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. q =.387 J/g K x 125 g x (3oC - 25oC) = 1.33 x 14 J 23 Slide Slide

5 1/3/11 First Law of Thermodynamics The internal energy of an isolated system is constant ΔU = q + w Isolated system, ΔU = Container walls can be either adiabatic (q = ) or diathermic (w = ) Both q and w are path dependent ΔE universe = ΔE system + ΔE surroundings Units of Energy Joule (J) 1 J = 1 kg m 2 /s 2 calorie (cal) 1 cal = J Internal Energy (U) is a State Function British thermal unit (Btu) 1 Btu = 155 J To determine the change in a state function, choose any convenient path (even if the real path is too complicated to calculate) 26 Slide Slide 6-26 Sample Problem 6.1 Determining 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 (ΔE) in J, kj, and kcal J x 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. q = J w = J ΔE = q + w = -325 J + (-451 J) = kj 1 3 J = kj kj x -776 J kcal kj = kcal 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 P ext ΔV + PΔV = q P ext ΔV + P ext ΔV So ΔH = q Implication 1: At constant volume you measure ΔU; at constant pressure you measure ΔH Implication 2: The molar heat capacity C m = q/(nδt) Therefore it will be different for a constant volume process than for a constant pressure process 27 Slide 6-27 Slide The Meaning of Enthalpy w = - PΔV H = E + PV where H is enthalpy ΔH = ΔE + PΔV q p = ΔE + PΔV = ΔH ΔH 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 >>> PΔV. 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 = Σn i X pdts Σn j X rxtnts The value for the reverse process is equal in magnitude BUT opposite in sign: ΔX rev = ΔX fwd Multiplication of a process by number multiplies the X value by that number: If A 2B ΔX 1 = 7 then 3A 6B ΔX 2 = 3ΔX 1 = 21 The sum of all of the X values making up a process is equal to the X overall for that process: ΔX total = ΔX step 1 + ΔX step 2 + ΔX step Slide 6-29 Slide

6 1/3/11 ΔH of Physical Processes Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. 31 Slide 6-31 To take.5 mole (9. g) of ice from 5ºC to vapor at 12ºC: Total heat = (heat to take.5 mol of ice from 5ºC to ºC) + (heat to melt.5 mol of ice) + (heat to take.5 mol of water from ºC to 1ºC) + (heat to vaporize.5 mol of ice) + (heat to take.5 mol of vapor from 1ºC to 12ºC) o q 6-32 Heating Curve Calculations Heating Curves ΔHvap ΔHfus Slide 6-32 q q o T1 ( 5 C)) T2 (+12 C) Slide Slide 6-34 ΔH of Chemical Change Standard Reaction Enthalpies ΔH 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 = 149 kj ΔHrxn associated with certain types of reactions; corrected to a per mole basis; used in tabular collections of data: ΔHrxn = kj/mol for the HCl oxidation of Al ΔHcombustion = 156 kj/mol for ethane (C2H6) 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. M concentration; Allows the tabulation and reporting of data from various sources to be accomplished! 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 Slide Slide

7 1/3/11 Common ΔH x s Enthalpy of Formation (ΔH f ) 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 O 2 (g) Na 2 CO 3 (s) By Definition ΔH f of an element in its most stable state; Allows ΔH rxn to be calculated using ΔH f of pdts - rxtnts: aa + bb dd ΔH f = [dδh f,d ] [aδh f,a bδh f,b ] Enthalpy of Combustion (ΔH c 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: C 2 H 2 (g) + 5/2 O 2 (g) 2 CO 2 (g) + H 2 O (l) Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of ΔH PROBLEM: 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) H 2 (g) + O 2 (g) H 2 O(l) kj 2 (b) 4.7 kj + H 2 O(l) 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. H 2 O(g) 38 Slide Slide 6-38 Sample Problem 6.4 Determining the Heat of a Reaction PROBLEM: You place 5. ml of.5 M NaOH in a coffee-cup calorimeter at 25. o C and carefully add 25. ml of.5 M HCl, also at 25. o C. After stirring, the final temperature is o C. Calculate q soln (in J) and ΔH 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 specific heat capacity as water: d = 1. g/ml and c = J/g K) Sample Problem 6.4 Determining the Heat of a Reaction continued HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) For NaOH.5 M x.5 L =.25 mol OH - For HCl.5 M x.25 L =.125 mol H + 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 ΔT. The heat divided by the M of water will give us the heat per mole of water formed. HCl is the limiting reactant..125 mol of H 2 O will form during the reaction. total volume after mixing =.75 L.75 L x 1 3 ml/l x 1. g/ml = 75. g of water q = mass x specific heat x ΔT = 75. g x J/g o C x (27.21 o C o C) = 693 J (693 J/.125 mol H 2 O)(kJ/1 3 J) = kj/ mol H 2 O formed 39 4 Slide 6-39 Slide 6-4 Sample Problem 6.5 Calculating the Heat of a Combustion Reaction PROBLEM: A manufacturer claims that its new dietetic dessert has fewer than 1 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 = kj/k). The temperature increases o C. Is the manufacturer s claim correct? Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kj) transferred during a reaction. - q sample = q calorimeter (First Law of Thermodynamics) q calorimeter = heat capacity x ΔT = kj/k x K = 4.24 kj 1 Calorie = 1 kcal = kj 1 Calorie = kj The manufacturer s claim is true Slide 6-41 Slide

8 1/3/11 Sample Problem 6.6 Using the Heat of Reaction (ΔH rxn ) to Find Amounts PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by 3 Al 2 O 3 (s) 2Al(s) + O 2 (g) ΔH rxn = 1676 kj 2 If aluminum is produced this way, how many grams of aluminum can form when 1. x 1 3 kj of heat is transferred? 1. x 1 3 kj x 2 mol Al 1676 kj x g Al 1 mol Al = 32.2 g Al 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) CO 2 (g) + ½N 2 (g) ΔH =? Given the following information, calculate the unknown ΔH: Equation A: CO(g) + ½ O 2 (g) CO 2 (g) ΔH A = kj Equation B: N 2 (g) + O 2 (g) 2NO(g) ΔH B = 18.6 kj 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 Slide 6-43 Slide 6-44 Sample Problem 6.8 Multiply Equation B by ½ and reverse it: NO(g) ½ N 2 (g) + ½ O 2 (g); ΔH = kj Add the manipulated equations together: Equation A: CO(g) + ½ O 2 (g) CO 2 (g) ΔH = kj ½ Equation B: NO(g) ½ N 2 (g) + ½ O 2 (g) ΔH = kj (reversed) CO(g) + NO(g) CO 2 (g) + ½ N 2 (g) ΔH rxn = kj Slide Slide Table 6.3 Selected Standard Enthalpies of Formation at 25 C (298K) Sample Problem 6.9 Writing Formation Equations Slide 6-47 Formula 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) ΔH f (kj/mol) Formula Cl 2 (g) HCl(g) Hydrogen H(g) H 2 (g) Nitrogen N 2 (g) NH 3 (g) NO(g) Oxygen O 2 (g) O 3 (g) H 2 O(g) H 2 O(l) ΔH f (kj/mol) Formula Silver Ag(s) AgCl(s) Sodium Na(s) Na(g) NaCl(s) ΔH f (kj/mol) Sulfur S 8 (rhombic) S 8 (monoclinic).3 SO 2 (g) SO 3 (g) Slide PROBLEM: 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, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. 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 8

9 1/3/11 Sample Problem 6.9 (a) Silver chloride, AgCl, a solid at standard conditions. Ag(s) + ½Cl2(g) AgCl(s) ΔH f = kj (b) Calcium carbonate, CaCO3, a solid at standard conditions. Ca(s) + C(graphite) + 3 O22(g) CaCO3(s) ΔH f = 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 Slide 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 = kj N2 (g) + 3H2 (g)! 2NH3 (g) H2 = kj N2 (g) + 4H2 (g) + Cl2 (g)! 2 NH4Cl (s) H3 = 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) Slide So H rxn = H (step 1) + 2(- H1 ) = H rxn = kj The general process for determining ΔHorxn from ΔHof values. H3 + -( H2 )+ 2(- H1 ) Slide 6-52 Sample Problem 6.9 Figure 6.1 H (step 1) 2(- H1 ) 6-52 Calculating the Heat of Reaction from Heats of Formation PROBLEM: Nitric acid, whose worldwide annual production is about 1 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. Look up the ΔHof values and use Hess s law to find ΔHrxn. 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)(9.3 kj/mol) + (6 mol)( kj/mol) [(4 mol)(-45.9 kj/mol) + (5 mol)( kj/mol)] ΔHrxn = -96 kj 53 Slide 6-53 Slide

10 1/3/11 Figure 6.11 End of Chapter 6 Silberberg HW: 1, 14, 15, 19, 35, 37, 39, 41, 48, 52, 54, 57, 59, 64, 65, 67, 71, 77, 79, 81, 82, 92, 97, 98, 1 The trapping of heat by the atmosphere Slide 6-55 Slide

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