Chapter 8. Thermochemistry 강의개요. 8.1 Principles of Heat Flow. 2) Magnitude of Heat Flow. 1) State Properties. Basic concepts : study of heat flow
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1 강의개요 Basic concepts : study of heat flow Chapter 8 Thermochemistry Calorimetry : experimental measurement of the magnitude and direction of heat flow Thermochemical Equations Copyright 2005 연세대학교이학계열일반화학및실험 (1) 강의노트 8.1 Principles of Heat Flow 1) State Properties Definition: System : part of the universe on which attention is focused Surroundings : exchange energy with the system ; rest of universe state property : these value is fixed when its temperature, pressure, and composition are specified (depend only on the state of the system not on the way the 2) Magnitude of Heat Flow Basic Equation for heat flow: [joules, kilojoules] (cf. 1 cal = J) q = C t (C = heat capacity) Heat capacity (C): amount of heat required to raise temperature of the system 1 o C [J o C -1 ] q = c m t (c = specific heat) specific heat (c); amount of heat required to raise 1 o C of a substance [J g -1 o C -1 ] e.g.) c = 4.18 J g -1 o C -1 system reaches that state) X = X final X initial
2 2) Magnitude of Heat Flow (cont.) ex) Suppose 652 J of heat is added to 15.0 g of (c = 4.18 J/g C), originally at 20 C. What is the final temperature? t = 652 J 4.18 J/g C 15.0 g = 10.4 C; final t = 30.4 C 2) Magnitude of Heat Flow (cont.) Note that if heat is absorbed by the system (q is positive), temperature increases; if q is negative, temperature drops. For a reaction at constant T and P: endothermic: q > 0 system absorbs heat exothermic: q < 0 system evolves heat 8.2 Measurement of Heat Flow: Calorimetry Measurement of heat flow in a reaction calorimeter q = reaction q calorimeter 1) Coffee cup calorimeter H of the reaction = -q Heat given off by reaction is absorbed by the in the coffee cup. 1 o 1 C = m c = m 4.18J g C q cal reaction = m 4.18J g C 1 o 1 t 1) Coffee Cup Calorimeter (cont.) ex) Suppose heat is absorbed by 412 g of, increasing the temp from to C. What is the heat flow for the reaction system? q = 4.18 J g C 412 g 9.74 C = J q reaction = 1.68 kj (exothermic process)
3 2) Bomb Calorimeter 2) Bomb Calorimeter (cont.) Bomb calorimeter: Some heat is absorbed by the metal as well as the surrounding. ex) Suppose combustion of 1.60 g CH 4 in bomb calorimeter raises the temperature by 5.14 C (C cal = 17.2 kj/ C) What is the heat flow for the reaction system? Equation for bomb calorimeter: q reaction = kj/ C 5.14 C = kj q reaction = -q calorimeter = -C cal t where C cal is the total heat capacity of the bomb and. C cal is determined using the reaction with known heat flow. 8.3 Enthalpy 8.4 Thermochemical Equations At constant pressure; q reaction at constant pressure = H = H products H reactants H 2 (g) + Cl 2 (g) 2HCl(g); H= -185 kj 185 kj of heat evolved when moles of HCl are formed. endothermic: q= H > 0 H products > H reactants exothermic: q = H < 0 H products < H reactants Enthalpy is a state property. 2HgO(s) 2Hg(l) + O 2 (g); H= +182 kj 182 kj of heat must be absorbed to decompose moles HgO.
4 Rules of Thermochemistry Rules of Thermochemistry (cont.) 1 H is directly proportional to amount of reactants or products. H 2 (g) + Cl 2 (g) 2HCl(g) H = -185 kj When one mole of ice melts, 6.00 kj of heat is absorbed, When 1.00 g of Cl 2 reacts: If one gram of ice melts, H = In general, H can be related to amount by the conversion factor approach. H = solid liquid: melting: liquid gas: vaporization: H fus H vap Rules of Thermochemistry (cont.) 2 H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction. H 2 O(s) H 2 O(l); H = kj; H 2 O(l) H 2 O(s); H = kj; Rules of Thermochemistry (cont.) 3 H for a reaction has the same value regardless of the number of steps. Hess law: If equation 1 + equation 2 = equation 3, then H 3 = H 1 + H 2 Often used to calculate H for one step, knowing H for all other steps and for the overall reaction.
5 Rules of Thermochemistry (cont.) 8.5 Enthalpies of Formation C(s) + 1/2O 2 (g) CO(g) H 1 =? CO(g) + 1/2O 2 (g) CO 2 (g) H 2 = kj C(s) + O 2 (g) CO 2 (g) H 3 = kj 1) Meaning of H f ; enthalpy of formation H f of a compound = H when one mole of compound is formed from the elements in their stable states 2Ag(s) + Cl 2 (g) 2AgCl(s) H = kj H f AgCl(s) = 1) Meaning of H f ; enthalpy of formation (cont.) 2) Calculation of H HgO(s) Hg (l) + 1/2O 2 (g) H f HgO(s) = H = kj For any thermochemical equation: H = The heat of formation for an element in a stable state is zero. H 0 f 0 H f 0 Br2 ( l) = H f O2( l) = 0 H + ( aq) = 0 for ions in aqueous solution
6 Heats of Formation (cont.) Heats of Formation (cont.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O H = H f CO 2 (g) + H f H 2 O(g ) - H f CH 4 (g) Can apply to ions: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) = ( kj) + ( kj) - (-74.8 kj) = kj 8.6 Bond Enthalpy 8.7 The First Law of Thermodynamics B.E. = H when one mole of bonds is broken in a gaseous state. Cl 2 (g) 2Cl(g) H = B.E. Cl Cl = 243 kj N 2 (g) N(g) H = B.E. N N = 941 kj E = q + w E = change in energy of system q = heat flow into system w = work done on system In general, multiple bonds are stronger than single bonds: C C = 347 kj C=C = 612 kj C C = 820 kj Constant P Constant V H = E = Mostly constant P (open system) H = E + PV
7 H versus E H = E + (PV) (PV ) i) For liquid & solid; ignore ii) For gas; PV = nrt n g ; the change in the number of moles of gas when the reaction takes place ex) CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(l) H =
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