Chapter 8. Thermochemistry
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1 Chapter 8 Thermochemistry Copyright 2001 by Harcourt, Inc. All rights reserved. Requests for permission to make copies of any part of the work should be mailed to the following address: Permissions Department, Harcourt, Inc Sea Harbor Drive, Orlando, Copyright Florida by Harcourt, Inc. All rights reserved. 8.1
2 Thermochemistry Basic concepts Calorimetry Thermochemical Equations Copyright 2001 by Harcourt, Inc. All rights reserved. 8.2
3 Basic Concepts Define: system surroundings state property Basic Equation for heat flow: q = c m t (c = specific heat) Copyright 2001 by Harcourt, Inc. All rights reserved. 8.3
4 Basic Concepts Suppose 652 J of heat is added to 15.0 g of water (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 Copyright 2001 by Harcourt, Inc. All rights reserved. 8.4
5 Basic Concepts 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 = H > 0 system absorbs heat exothermic: q = H < 0 system evolves heat Copyright 2001 by Harcourt, Inc. All rights reserved. 8.5
6 Enthalpy reaction for exothermic reactions Copyright 2001 by Harcourt, Inc. All rights reserved. 8.6
7 Enthalpy reaction for endothermic reactions Copyright 2001 by Harcourt, Inc. All rights reserved. 8.7
8 Calorimetry Coffee cup calorimeter: H of the reaction = -q water Heat given off by reaction is absorbed by the water in the coffee cup. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.8
9 Calorimetry (cont.) Suppose heat is absorbed by 412 g of water, increasing the temp from to C. What is the H? q water = 4.18 J g C 4.12 g 9.74 C = J H = 1.68 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.9
10 Calorimetry (cont.) Bomb calorimeter: Some heat is absorbed by the metal as well as the surrounding water. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.10
11 Calorimetry (cont.) Equation for bomb calorimeter: q reaction = -q calorimeter = -(C calorimeter ) t where C calorimeter is the total heat capacity of the bomb and water. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.11
12 Calorimetry (cont.) Suppose combustion of 1.60 g CH 4 in bomb calorimeter raises the temperature by 5.14 C (C calorimeter = 17.2 kj/ C) q reaction = kj/ C 5.14 C = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.12
13 Thermochemical Equations H 2 (g) + Cl 2 (g) 2HCl(g); H= -185 kj 185 kj of heat evolved when 2 moles of HCl are formed. 2HgO(s) 2Hg(l) + O 2 (g); H= +182 kj 182 kj of heat must be absorbed to decompose 2 moles HgO. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.13
14 Rules of Thermochemistry H is directly proportional to amount of reactants or products. When one mole of ice melts, 6.00 kj of heat is absorbed, H = kj. If one gram of ice melts, H = 6.00 kj/18.02 = kj. In general, H can be related to amount by the conversion factor approach. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.14
15 Rules of Thermochemistry (cont.) H 2 (g) + Cl 2 (g) 2HCl(g) H = -185 kj When 1.00 g of Cl 2 reacts: 1 mol Cl H = 1.00 g Cl kj = kj g Cl 2 1 mol Cl 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 8.15
16 Rules of Thermochemistry (cont.) 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; 6.00 kj absorbed H 2 O(l) H 2 O(s); H = kj; 6.00 kj evolved Copyright 2001 by Harcourt, Inc. All rights reserved. 8.16
17 H independent of path Copyright 2001 by Harcourt, Inc. All rights reserved. 8.17
18 Rules of Thermochemistry (cont.) 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. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.18
19 Rules of Thermochemistry (cont.) 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 H 1 = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.19
20 Heats 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) = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.20
21 Heats of Formation (cont.) HgO(s) Hg (l) + 1/2O 2 (g) H = kj H f HgO(s) = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.21
22 Heats of Formation (cont.) For any thermochemical equation: H = Σ H f products - Σ H f reactants The heat of formation for an element in a stable state is zero. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.22
23 Heats of Formation (cont.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O H = H f CO 2 (g) + 2 H f H 2 O(g ) - H f CH 4 (g) = kj + 2( kj) - (-74.8 kj) = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.23
24 Heats of Formation (cont.) Can apply to ions: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) set H f H + (aq) = 0 H = H f Zn 2+ (aq) = kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.24
25 Bond Energies 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 In general, multiple bonds are stronger than single bonds: C C = 347 kj C=C = 612 kj C C = 820 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.25
26 Comparison of bond energies in H 2 and Cl 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 8.26
27 First Law E = q + w E = change in energy of system q = heat flow into system w = work done on system H = E + (PV) To calculate (PV), ignore liquids and solids, use ideal gas law to find PV for gases. CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(l) H = E - RT ; RT = 2.5 kj at 25 C Copyright 2001 by Harcourt, Inc. All rights reserved. 8.27
28 First law of thermodynamics Copyright 2001 by Harcourt, Inc. All rights reserved. 8.28
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