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. 6277 Sea Harbor Drive, Orlando, Copyright Florida 32887-6777 2001 by Harcourt, Inc. All rights reserved. 8.1
Thermochemistry Basic concepts Calorimetry Thermochemical Equations Copyright 2001 by Harcourt, Inc. All rights reserved. 8.2
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
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
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
Enthalpy reaction for exothermic reactions Copyright 2001 by Harcourt, Inc. All rights reserved. 8.6
Enthalpy reaction for endothermic reactions Copyright 2001 by Harcourt, Inc. All rights reserved. 8.7
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
Calorimetry (cont.) Suppose heat is absorbed by 412 g of water, increasing the temp from 20.12 to 29.86 C. What is the H? q water = 4.18 J g C 4.12 g 9.74 C = 1.68 10 3 J H = 1.68 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.9
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
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
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 = -17.2 kj/ C 5.14 C = -88.4 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.12
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
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 = +6.00 kj. If one gram of ice melts, H = 6.00 kj/18.02 = +0.333 kj. In general, H can be related to amount by the conversion factor approach. Copyright 2001 by Harcourt, Inc. All rights reserved. 8.14
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 2 2-185 kj = -2.61 kj 70.90 g Cl 2 1 mol Cl 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 8.15
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 = +6.00 kj; 6.00 kj absorbed H 2 O(l) H 2 O(s); H = -6.00 kj; 6.00 kj evolved Copyright 2001 by Harcourt, Inc. All rights reserved. 8.16
H independent of path Copyright 2001 by Harcourt, Inc. All rights reserved. 8.17
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
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 = -283.0 kj C(s) + O 2 (g) CO 2 (g) H 3 = -393.5 kj H 1 = -110.5 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.19
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 = -254.0 kj H f AgCl(s) = -127.0 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.20
Heats of Formation (cont.) HgO(s) Hg (l) + 1/2O 2 (g) H = +90.8 kj H f HgO(s) = -90.8 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.21
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
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) = -393.5 kj + 2(-285.8 kj) - (-74.8 kj) = -890.3 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.23
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) = 152.4 kj Copyright 2001 by Harcourt, Inc. All rights reserved. 8.24
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
Comparison of bond energies in H 2 and Cl 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 8.26
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
First law of thermodynamics Copyright 2001 by Harcourt, Inc. All rights reserved. 8.28