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Chapter 2 Elements of Chemical Thermodynamics Atkins: Ch6 & 7

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1 Chapter 2 Elements of Chemical Thermodynamics Atkins: Ch6 & 7
Silberberg: Ch 6 & 20

2 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 (U or E). Therefore the system has an internal energy. When a change occurs, the internal energy changes.

3 A chemical system and its surroundings.
the surroundings the system

4 System & Surroundings System - the specific part of the universe that is of interest in the study. often the reacting vessel in the chemistry lab A beaker of NaOH and HCl Surroundings - the rest of the universe When heat is given off from the beaker, the energy is transferred out of the system


6 open closed isolated Exchange: mass & energy energy nothing

7 Exothermic & Endothermic Reactions
Exothermic - transfer of heat from the system to the surroundings 2H2(g) + O2(g)  2H2O(l) + energy Endothermic - the transfer of heat from the surroundings to the system energy + 2HgO(s)  2Hg(l) + O2(g)

8 Endothermic vs. Exothermic

9 Copyright McGraw-Hill 2009
Internal Energy Internal energy: 2 components Kinetic energy - molecular motion Potential energy - attractive/repulsive interactions Copyright McGraw-Hill 2009

10 Copyright McGraw-Hill 2009
Internal Energy We cannot calculate the changes in internal energy with any certainty We can calculate the changes in energy of the system experimentally U = U(products) - U(reactants)‏ Copyright McGraw-Hill 2009

11 DU = Ufinal - Uinitial = Uproducts - Ureactants
Energy diagrams for the transfer of internal energy (U) between a system and its surroundings. DU = Ufinal - Uinitial = Uproducts - Ureactants

12 DUuniverse = DUsystem + DUsurroundings
Units of Energy Joule (J) 1 J = 1 kg*m2/s2 Calorie (cal) 1 cal = 4.18J British Thermal Unit 1 Btu = 1055 J

13 First Law of Thermodynamics
Energy can be converted from one form to another but cannot be created nor destroyed (the Law of Energy Conservation)‏ Copyright McGraw-Hill 2009

14 Energy and the Universe
Usystem + Usurroundings = 0 When a system releases heat, some of the chemical energy is released as thermal energy to the surroundings but this does not change the total energy of the universe. Copyright McGraw-Hill 2009

15 Energy and the Universe
Usystem = - Usurroundings When a system undergoes a change in energy, the surroundings must undergo a change in energy of equal magnitude but opposite in sign. Copyright McGraw-Hill 2009

16 Copyright McGraw-Hill 2009
Work and Heat Usys = q + w “q” = heat “w” = work Copyright McGraw-Hill 2009

17 Pressure-volume work.

18 + = + + + + - - + - - - q w DU The Sign Conventions* for q, w and DU
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 work done on system; - means work done by system.

19 Copyright McGraw-Hill 2009
Work and Heat Calculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the system. “q” = + (heat absorbed) “w” = - (work done) Usys = q + w = (+125 J) + (-141 J)‏ = -16 J Copyright McGraw-Hill 2009

20 State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DU = Ufinal - Uinitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

21 Copyright McGraw-Hill 2009
Constant Volume Constant volume: (V = 0) U = q - PV qv = U (not easy to achieve in the chem lab) Copyright McGraw-Hill 2009

22 Copyright McGraw-Hill 2009
Constant Pressure Constant pressure: (common in the chem lab)‏ U = q + w U = qp - PV qp = U + PV Copyright McGraw-Hill 2009

23 Enthalpy and Enthalpy Changes
Enthalpy (H) is a state function Enthalpy measures heat content Constant pressure process: (most are)‏ qp = H Enthalpy of reaction: H = H(products) - H(reactants) H = - (exothermic) H = + (endothermic) Copyright McGraw-Hill 2009

24 Thermochemical Equations
Equations that represent mass and enthalpy relationships H2O(s)  H2O(l) rH = kJ/mol This is an endothermic process. It requires 6.01 kJ to melt one mole of ice. (the “per mole” means per mole of reaction as written - respect stoichiometry) Copyright McGraw-Hill 2009

25 Thermochemical Equation Guidelines
Always specify state of reactants and products. When multiplying an equation by a factor (n), multiply the H value by same factor. Reversing an equation changes the sign but not the magnitude of the H. Copyright McGraw-Hill 2009

26 Enthalpy and the First Law of Thermodynamics
DU = q + w q = DH and w = -PDV At constant pressure: DU = DH - PDV DH = DU + PDV

27 The Meaning of Enthalpy
w = - PDV DH ≈ DU in H = U + 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 = DU + PDV 3. Reactions in which the number of moles of gas does change but q is >>> PDV. qp = DU + PDV = DH

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


30 The standard enthalpy of reaction (DrH0 ) is the enthalpy of a reaction carried out at 1 atm.
aA + bB cC + dD DrH0 dDfH0 (D) cDfH0 (C) = [ + ] - bDfH0 (B) aDfH0 (A) DrH0 nDfH0 (products) = S mDfH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

31 Sample Problem 2.1 Using Hess’s Law to Calculate an Unknown DH 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 = kJ/mol Equation B: N2(g) + O2(g) NO(g) DHB = kJ/mol 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. CO(g) + 1/2O2(g) CO2(g) DHA = kJ/mol NO(g) /2N2(g) + 1/2O2(g) DHB = kJ/mol CO(g) + NO(g) CO2(g) + 1/2N2(g) DHrxn = kJ/mol

32 Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DfH0) as a reference point for all enthalpy expressions. Standard enthalpy of formation (DfHm) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. Df H m (O2) = 0 DfH0 (C, graphite) = 0 DfH0 (O3) = 142 kJ/mol DfH0 (C, diamond) = 1.90 kJ/mol

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

34 The general process for determining DrH0 from DfH0 values.
Elements decomposition formation -DfH0 DfH0 Enthalpy, H Reactants Hinitial DrH0 Products Hfinal DrH0 = S mDfH0 (products) - S nDfH0 (reactants)

35 Sample Problem 2.2 Calculating the Heat of Reaction from Heats of Formation 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: Calculate DH0rxn from DH0f values. 4NH3(g) + 5O2(g) NO(g) + 6H2O(g) PLAN: Look up the DH0f values and use Hess’s Law to find DHrxn. SOLUTION: DHrxn = S mDfH0 (products) - S nDfH0 (reactants) DrH = [4(DfH0 NO(g) + 6(DfH0 H2O(g)] - [4(DfH0 NH3(g) + 5(DfH0O2(g)] = (4)(90.3 kJ/mol) + (6)( kJ/mol) - [(4)(-45.9 kJ/mol) + (5)(0 kJ/mol)] DrH = -906 kJ/mol

36 Some Important Types of Enthalpy Change
heat of combustion (DcombH) C4H10(l) /2O2(g) CO2(g) + 5H2O(g) heat of formation (DfH) K(s) + 1/2Br2(l) KBr(s) heat of fusion (DfusH) NaCl(s) NaCl(l) heat of vaporization (DvapH) C6H6(l) C6H6(g)

37 Limitations of the First Law of Thermodynamics
DU = q + w Uuniverse = Usystem + Usurroundings DUsystem = -DUsurroundings DUsystem + DUsurroundings = 0 = DUuniverse The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.

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