Chapter 6: First Law. 6.1 Forms of Energy and Their Interconversion. 6.2 Enthalpy: Heats of Reaction and Chemical Change

Transcription

1 Chapter 6 Thermochemistry: Energy Flow and Chemical Change Slide 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 (ΔH 0 rxn ) Slide

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 To deal with large groups of molecules and atoms we need to define THREE parameters: System, Surroundings and Thermodynamic Universe Slide Thermodynamics: First Law Sign is from the point of view of the System 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 Open System: Exchanges both E and matter with Surroundings; Closed System: Exchanges only E with Surroundings; Isolated System: Exchanges NEITHER with Surroundings Slide

3 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 Slide 6-6 An Auto A Battery A perfect Thermos 6 3

4 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 = E final - E initial = E products - E reactants Slide A system transferring energy as heat only. A E lost as heat B E gained as heat Slide

5 A system losing energy as work only. Slide 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 = U final U initial If E transfer is only due to work then ΔU = w Work is either expansion or nonexpansion (ΔV 0 vs. ΔV= 0) w = P ext ΔV (since w = when system does work) Slide

6 Work w = P ext ΔV (P ext must be constant) Work is in L-atm: 1 L-atm = J If is P ext = 0, then free expansion occurs If P ext 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 Slide Figure 6.5 Pressure-volume work. Slide

7 Figure 6.7 Pressure-volume work. Slide 6-13 An expanding gas pushing back the atmosphere does PV work (w = -PΔV). 13 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 = q surr Exothermic vs Endothermic Slide

8 Calorimetry 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 Measuring Heat Using a Calorimeter Constant Pressure Constant Volume Slide

9 Figure 6.7 Coffee-cup calorimeter. Slide Figure 6.8 A bomb calorimeter. Slide

10 Slide Selected Heat Capacities (C s and C m ) Amt of heat to raise 10.0 g of water 5 C: J Temp increase if same amount of heat is added to 10.0 g of gold (C s = J/g-C): C! Slide

11 21 Slide 6-21 C m,v vs. C m,p Since q = nc m ΔT then C m = q n!t = q!t (n = 1) However, q = ΔU at constant volume or q = ΔH at constant pressure, So: C m,v =!U n!t OR and C m,p = C m,v + R C m,p =!H n!t For Ar: C m,v =12.8 J/mol-K and C m,p = 21.1 J/mol-K Slide

12 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 25 o C to 300. o C? The specific heat capacity (c) of Cu is 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 o C is the same as for K. SOLUTION: q = J/g K x 125 g x (300 o C - 25 o C) = 1.33 x 10 4 J Slide Figure 6.4 Two different paths for the energy change of a system. Slide

13 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) Slide ΔE universe = ΔE system + ΔE surroundings Units of Energy Joule (J) 1 J = 1 kg m 2 /s 2 calorie (cal) 1 cal = J British thermal unit (Btu) 1 Btu = 1055 J Slide

14 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. PLAN: 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 = J w = J ΔE = q + w = -325 J + (-451 J) = -776 J -776 J x kj 10 3 J = kj kj x kcal kj = kcal Slide 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 Slide

15 The Meaning of Enthalpy w = - PΔV H = E + PV where H is enthalpy ΔH = ΔE + PΔV ΔH ΔE in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. q p = ΔE + PΔV = ΔH 3. Reactions in which the number of moles of gas does change but q is >>> PΔV. Slide 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 3 + Slide

16 Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. Slide ΔH of Physical Processes Slide

17 Heating Curves o ΔH fus ΔH vap q q q o Slide 6-33 T 1 ( 5 C)) T 2 (+120 C) 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) Slide

18 ΔH of Chemical Change Enthalpy change during chemical reactions important to understanding chemistry. Reaction Enthalpies: ΔH or ΔH rxn ΔH associated with Thermochemical Equations: 2Al (s) + 6HCl (aq) 2AlCl 3 (aq) + H 2 (g) ΔH = 1049 kj ΔH rxn associated with certain types of reactions; corrected to a per mole basis; used in tabular collections of data: ΔH rxn = kj/mol for the HCl oxidation of Al ΔH combustion = 1560 kj/mol for ethane (C 2 H 6 ) Relation between ΔH and ΔU: For reactions in which NO gas is generated: ΔH = ΔU For reactions in which gas is generated: ΔH = ΔU + Δn gas RT Slide 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! Slide

19 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 0 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) Slide 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) 40.7 kj + H 2 O(l) H 2 O(g) 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: Slide

20 Sample Problem 6.4 Determining the Heat of a Reaction PROBLEM: You place 50.0 ml of M NaOH in a coffee-cup calorimeter at o C and carefully add 25.0 ml of M HCl, also at 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 = 1.00 g/ml and c = 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 ΔT. The heat divided by the M of water will give us the heat per mole of water formed. Slide Sample Problem 6.4 continued Determining the Heat of a Reaction SOLUTION: HCl(aq) + NaOH(aq) H + (aq) + OH - (aq) H 2 O(l) NaCl(aq) + H 2 O(l) For NaOH M x L = mol OH - For HCl M x L = mol H + HCl is the limiting reactant mol of H 2 O will form during the reaction. Slide 6-40 total volume after mixing = L L x 10 3 ml/l x 1.00 g/ml = 75.0 g of water q = mass x specific heat x ΔT = 75.0 g x J/g o C x (27.21 o C o C) = 693 J (693 J/ mol H 2 O)(kJ/10 3 J) = kj/ mol H 2 O formed 40 20

21 Sample Problem 6.5 Calculating the Heat of a Combustion Reaction 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 = kj/k). The temperature increases o C. Is the manufacturer s claim correct? PLAN: - q sample = q calorimeter (First Law of Thermodynamics) SOLUTION: q calorimeter = heat capacity x ΔT = kj/k x K = kj 1 Calorie = 1 kcal = kj 10 Calorie = kj Slide 6-41 The manufacturer s claim is true. 41 Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kj) transferred during a reaction. Slide

22 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 x 10 3 kj of heat is transferred? PLAN: SOLUTION: x 10 3 kj x 2 mol Al 1676 kj x g Al 1 mol Al = g Al Slide 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 = 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. Slide

23 Sample Problem 6.8 SOLUTION: 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

24 Table 6.3 Selected Standard Enthalpies of Formation at 25 C (298K) Slide 6-47 Formula Calcium Ca(s) CaO(s) CaCO 3 (s) ΔH f (kj/mol) Carbon C(graphite) 0 C(diamond) 1.9 CO(g) CO 2 (g) CH 4 (g) CH 3 OH(l) HCN(g) 135 CS s (l) 87.9 Chlorine Cl(g) 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) 0 S 8 (monoclinic) 0.3 SO 2 (g) SO 3 (g) Sample Problem 6.9 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 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. 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. Slide

25 Sample Problem 6.9 SOLUTION: (a) Silver chloride, AgCl, a solid at standard conditions. Ag(s) + ½Cl 2 (g) AgCl(s) ΔH f = kj (b) Calcium carbonate, CaCO 3, a solid at standard conditions. Ca(s) + C(graphite) + 3 O2 (g) CaCO 3 (s) ΔH f = kj (c) Hydrogen cyanide, HCN, a gas at standard conditions. ½H 2 (g) + C(graphite) + ½N 2 (g) HCN(g) ΔH f = 135 kj Slide Slide

26 Slide Hess Law of Heat Summation Let's say we want to determine the reaction enthalpy for the following reaction: H 2 (g) + Cl 2 (g)! 2HCl (g) H rxn =?? given the following individual reactions: NH 3 (g) + HCl (g)! NH 4 Cl (s) H 1 = kj N 2 (g) + 3H 2 (g)! 2NH 3 (g) H 2 = kj N 2 (g) + 4H 2 (g) + Cl 2 (g)! 2 NH 4 Cl (s) H 3 = kj Start by adding two reactions that give us the reactants we want on the correct side: Step 1 N 2 (g) + 4H 2 (g) + Cl 2 (g)! 2 NH 4 Cl (s) H 3 2NH 3 (g)! N 2 (g) + 3H 2 (g) -( H 2 ) 2NH 3 (g) + H 2 (g) + Cl 2 (g)! 2 NH 4 Cl (s) H (step 1) = H 3 + -( H 2 ) Step 2 2NH 3 (g) + H 2 (g) + Cl 2 (g)! 2 NH 4 Cl (s) H (step 1) 2[NH 4 Cl (s)! NH 3 (g) + HCl (g)] 2(- H 1 ) H 2 (g) + Cl 2 (g)! 2HCl (g) So H rxn = H (step 1) + 2(- H 1 ) = H 3 + -( H 2 )+ 2(- H 1 ) H rxn = kj Slide

27 Figure 6.10 The general process for determining ΔH o rxn from ΔHo f values. Slide 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: 4NH 3 (g) + 5O 2 (g) Calculate ΔH o rxn from ΔHo f values. 4NO(g) + 6H 2 O(g) PLAN: Look up the ΔH o f values and use Hess s law to find ΔH rxn. SOLUTION: ΔH rxn = Σ mδh o f (products) - Σ nδho f (reactants) ΔH rxn = {4[ΔH o f NO(g)] + 6[ΔHo f H 2 O(g)]} - {4[ΔH o f NH 3 (g)] + 5[ΔHo f O 2 (g)]} = (4 mol)(90.3 kj/mol) + (6 mol)( kj/mol) - [(4 mol)(-45.9 kj/mol) + (5 mol)(0 kj/mol)] Slide 6-54 ΔH rxn = -906 kj 54 27

28 Figure 6.11 Slide 6-55 The trapping of heat by the atmosphere. 55 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 Slide