Chapter 5:

1. Chapter 5: Thermochemistry

2. Thermochemistry: • Energy • Kinetic & Potential • First Law of Thermo • internal energy, heat & work • endothermic & exothermic processes • state functions • Enthalpy • Enthalpies of Reaction

3. Calorimetry • heat capacity and specific heat • constant-pressure calorimetry • bomb calorimetry (constant-volume calorimetry) • Hess’s Law • Enthalpies of Formation • for calculation of enthalpies of reaction • Foods and Fuels

4. Energy • work is a form of energy w = F x d • energy is the capacity to do work or transfer heat • Kinetic Energy • energy of motion E = ½ mv2 • potential energy • energy of position • applies to electrostatic energy • applies to chemical energy (energy of bonds)

5. energy units • one joule = energy of a 2 kg mass moving at 1 m/s • E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J • 1 cal = 4.184 J 1 kcal = 1 food calorie (Cal) Systems & Surroundings • system -- chemicals in the reaction • surroundings -- container & all outside environment • closed system can exchange energy (but not matter) with its surroundings

6. Closed System energy(as heat or work) 2H2(g) + O2(g) ¯ 2H2O(l)+ energy (system) no exchg of matterwith surroundings

7. First Law of Thermo. • Energy is always conserved • any energy lost by system, must be gained by surroundings • Internal Energy -- total energy of system • combination of all potential and kinetic energy of system • incl. motions & interactions of of all components • we measure the changes in energy E = Efinal - Einitial

8. + D E = Efinal > Einitial system has gained E from surroundings • - D E = Efinal < Einitial system has lost E to surroundings • Relating D E to heat and work • D E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings

9. +q +w - q - w surroundings surroundings system system heat heat work work

10. Endothermic • system absorbs heat or heat flows into the system Exothermic • system gives off heat or heat flows out of the system State Function • a property of a system that is determined by specifying its condition or state (T, P, etc.) • internal energy is a state function, \DE depends only on Efinal & Einitial

11. Enthalpy • for most reactions, most of the energy exchanged is in the form of heat, that heat transfer is called enthalpy, H • enthalpy is a state function • like internal energy, we can only measure the change in enthalpy, DH • DH = qp when the process occurs under constant pressureDH = Hfinal - Hinitial = qp • - DH Þ exothermic process • +DH Þ endothermic process

12. system DH > 0 surroundings system surroundings DH < 0

13. Enthalpies of Reaction • D Hrxn = Hprod - Hreact • enthalpy is an extensive property • magnitude of D H depends directly on the amount of reactant • C(s) + 2H2(g)® CH4(g)D H = -74.8 kJ/mol • 2C(s) + 4H2(g)® 2CH4(g)D H = -149.6 kJ/2mol

14. enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn • CH4(g) ® C(s) + 2H2(g)D H = +74.8 kJ/mol • C(s) + 2H2(g)® CH4(g) D H = - 74.8 kJ/mol • enthalpy change for a reaction depends on the state of the reactants and products • C(g) + 2H2(g)® CH4(g)D H = -793.2 kJ/mol • 2H2(g) + O2(g)® 2H2O(g)D H = -486.6 kJ/mol • 2H2(g)+ O2(g)® 2H2O(l)D H = -571.7 kJ/mol

15. DH = Hfinal - Hinitial H2O(g) -241.8 kJ 44 kJ - + Enthalpy H2O(l) -285.8 kJ

16. Practice Ex. 5.2: • Hydrogen peroxide can decompose to water and oxygen . Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure. • 2H2O2(l)® 2H2O(l) + O2(g)D H = -196 kJ • 5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l) 34.0 g H2O2(l) • 0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ 2 mol

17. Calorimetry • experimental determination of D H using heat flow heat capacity • measures the energy absorbed using temperature change • the amount of heat required to raise its temp. by 1 K • molar heat capacity -- heat capacity of 1 mol of substance

18. specific heat • heat energy required to raise some mass of a substance to some different temp. • specific heat = quantity of heat trans. (g substance) (temp. change) • = q . m DT • S.H. = joule g K • q = (S.H.) (g substance) (D T) remember: this is change in temp.

19. Practice Ex. 5.3: • Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temp. increases by 12.0 °C if the specific heat of the rocks is 0.82 J/gK. • S.H. x g x DT = joules • What unit should be in the solution? • joules -- quantity of heat • 0.82 J x 50.0 x 103 g x 12.0 K = 4.9 x 105 Jg K

20. Constant-Pressure Calorimetry • D H = qp at constant pressure as in coffee cup calorimeter • heat gained by solution = qsoln • \qsoln = (S.H.soln)(gsoln)(DT) • heat gained by solution must that which is given off by reaction • \qrxn = - qsoln = - (S.H.soln)(gsoln)(DT) must be opposite in sign if DT is positive then qrxn is exothermic

21. Practice Ex. 5.4: • When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30°C to 23.11°C. Calculate D H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g °C. • AgNO3(aq) + HCl(aq) ® AgCl(s) + HNO3(aq) • qsoln = 4.18 J x 100.0 g soln x 0.81°C = 3.39 x 102 J g °C • qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 mol - 68 kJ/mol

22. insulating cup rxn q soln

23. Bomb Calorimetry (Constant-Volume) • bomb calorimeter has a pre-determined heat capacity • sample is combusted in the calorimeter and D T is used to determine the heat change of the reaction • qrxn = - Ccalorimeter x D T heat capacity of calorimeter because rxn is exothermic

24. thermometer insulation water rxn

25. Practice Ex. 5.5: • A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a calorimeter with C = 4.812 kJ/°C. Temp. increases from 23.10°C to 24.95°C. Calculate heat of combustion per gram and per mole. D T = +1.85°C • qrxn = - (4.812 kJ/°C) (1.85°C) = - 8.90 kJ per 0.5865 g lactic acid-8.90 kJ = - 15.2 kJ/g 0.5865 g • - 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol

26. Hess’s Law • rxns in one step or multiple steps are additive because they are state functions • eg. • CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)D H = - 802 kJ • CH4(g) + 2O2(g)® CO2(g) + 2H2O(l)D H = - 890 kJ 2H2O(g)® 2H2O(l)D H = - 88 kJ

27. Practice Ex. 5.6: • Calculate D H for the conversion of graphite to diamond: • Cgraphite® Cdiamond • Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ • Cdiamond + O2(g)® CO2(g)D H = -395.4 kJ • Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ • CO2(g)® Cdiamond + O2(g)D H = 395.4 kJ Cgraphite® CdiamondD H = + 1.9 kJ

28. Enthalpies of Formation • enthalpies are tabulated for many processes • vaporization, fusion, formation, etc. • enthalpy of formation describes the change in heat when a compound is formed from its constituent elements, DHf • standard enthalpy of formation, DHfo, are values for a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm

29. For elemental forms: • eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc. • DHfo, for any element is = 0 • used for calculation of enthalpies of reaction, DHrxn • DHrxn = S DHfo prod - S DHfo react

30. Practice Ex. 5.9: • Given this standard enthalpy of reaction, use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO(s)( • CuO(a) + H2(g)® Cu(s) + H2O(l)DHo = -130.6 kJ • DHrxn = S DH f o prod - S DH f o react • -130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)] • DHfo CuO = -155.2 kJ/mol