Chapter 7: Thermochemistry Chemistry 140 Fall 2002 General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8th Edition Chapter 7: Thermochemistry Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 (modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi) Thermochemistry branch of chemistry concerned with heat effects accompanying chemical reactions. Direct and indirect measurement of heat. Answer practical questions: why is natural gas a better fuel than coal, and why do fats have higher energy value than carbohydrates and protiens.

A couple of Announcements MIDTERM TEST IS WRITTEN ON WEDNESDAY OCTOBER 26TH, 2005 DURING CLASSTIME IN THIS ROOM MIDTERM TEST WILL COVER MATERIAL FROM CHAPTERS 1-7. YOU MUST BE REGISTERED IN 59-140 SECTION 1 IN ORDER TO BE ALLOWED TO WRITE IN THIS ROOM. EVERYONE IN SECTION 2 MUST WRITE IN ER 1120

Contents 7-1 Getting Started: Some Terminology 7-2 Heat 7-3 Heats of Reaction and Calorimetry 7-4 Work 7-5 The First Law of Thermodynamics 7-6 Heats of Reaction: U and H 7-7 The Indirect Determination of H: Hess’s Law

Contents 7-7 The Indirect Determination of H, Hess’s Law 7-8 Standard Enthalpies of Formation 7-9 Fuels as Sources of Energy Focus on Fats, Carbohydrates, and Energy Storage.

Standard States and Standard Enthalpy Changes Chemistry 140 Fall 2002 Standard States and Standard Enthalpy Changes Define a particular state as a standard state. Standard enthalpy of reaction, H° The enthalpy change of a reaction in which all reactants and products are in their standard states. Standard State The pure element or compound at a pressure of 1 bar and at the temperature of interest. Temperature must be specified because H varies with temperature.

Enthalpy Diagrams

7-7 Indirect Determination of H: Hess’s Law H is an extensive property. Enthalpy change is directly proportional to the amount of substance in a system. N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ H changes sign when a process is reversed NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ

Hess’s Law ½N2(g) + O2(g) → NO2(g) H = +33.18 kJ Hess’s law of constant heat summation If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ ½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

Hess’s Law Schematically

Example 7-4 Applying Hess’s Law. Use the heat of combustion data listed below to determine ΔH° for the reaction: 3 C(graphite) + 4 H2(g)  C3H8(g) ΔH° = ?? (1) C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ΔH° = -2219.9 kJ (2) C(graphite) + O2(g)  CO2(g) ΔH° = -393.5 kJ (3) H2(g) + ½ O2(g)  H2O(l) ΔH° = -285.8 kJ

Example 7-4 Find a reaction to produce C3H8(g) Use the reverse of (1) [(ie. –(1)) Remember to change sign of ΔH°] -(1) 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) ΔH° = -(-2219.9 kJ) = +2219.9kJ Need to get the proper number of moles of reactants. We take equation (2) and multiply it by 3, then (3) and multiply by 4 3 (2) 3 C(graphite) + 3 O2(g)  3 CO2(g) ΔH° = 3(-393.5 kJ) = -1181 kJ 4 (3) 4 H2(g) + 2 O2(g)  4 H2O(l) ΔH° = 4(-285.8 kJ) = -1143 kJ

Example 7-4 Add the resulting equations to get the final equation of interest and calculate the overall enthalpy change for this reaction (ΔH°) -(1) 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) ΔH°=2219.9kJ 3 (2) 3 C(graphite) + 3 O2(g)  3 CO2(g) ΔH°=-1181 kJ 4 (3) 4 H2(g) + 2 O2(g)  4 H2O(l) ΔH°=-1143 kJ 3 C(graphite) + 4 H2(g)  C3H8(g) ΔH°=-104 kJ

7-8 Standard Enthalpies of Formation Chemistry 140 Fall 2002 7-8 Standard Enthalpies of Formation Hf° The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. The standard enthalpy of formation of a pure element in its reference state is 0. [ Na(s), H2(g), C(graphite), Br2(l)] Absolute enthalpy cannot be determined. H is a state function so changes in enthalpy, H, have unique values. Reference forms of the elements in their standard states are the most stable form of the element at one bar and the given temperature. The superscript degree symbol denotes that the enthalpy change is a standard enthalpy change and The subscript “f” signifies that the reaction is one in which a substance is formed from its elements.

Standard Enthalpies of Formation Formation of formaldehyde from its elements in their standard forms

Standard Enthalpies of Formation

Standard Enthalpies of Reaction Hoverall = -2Hf°NaHCO3+ Hf°Na2CO3 + Hf°CO2 + Hf°H2O

Enthalpy of Reaction Hrxn = Hf°products- Hf°reactants

Example 7-11 Hrxn = Hf°products- Hf°reactants Calculating ΔH° from tabulated ΔHf° values. Calculate the standard enthalpy of combustion of ethane (C2H6(g)) C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l) ΔHf° kJmol-1 -84.7 -393.5 -285.8 Hrxn = Hf°products- Hf°reactants Hrxn = [2(Hf°CO2) + 3(Hf°H2O)] - [1(Hf°C2H6) + 7/2(Hf°O2)] Hrxn=[2(-393.5 kJmol-1) + 3(-285.8 kJmol-1)] - [1(-84.7 kJmol-1) + 7/2(0 kJmol-1)] Hrxn=[(-787.0 kJ) + (-857.4 kJ)] – [(-84.7 kJ)] Hrxn= -1559.7 kJ

Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions Enthalpies of Formation of Ions in Solution

7-9 Fuels as Sources of Energy Fossil fuels. Combustion is exothermic. Non-renewable resource. Environmental impact. Biomass fuels Fossil fuels

Chapter 7 Questions 1, 2, 3, 11, 14, 16, 22, 23, 24, 29, 33, 37, 49, 52, 63, 69, 73, 75, 81