Thermochemistry is the study of heat change in chemical reactions. Energy & Rates of Reaction Unit 3 Thermochemistry is the study of heat change in chemical reactions. Chapter 5 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Chapter 5 Key Concepts: After completing this chapter you will be able to….. compare the energy changes of different substances, perform calorimetry calculations, and describe the relationship between reactions using enthalpy terms. represent thermochemical equations using different methods, and determine whether a reaction is exothermic or endothermic calculate and use bond energies to estimate the enthalpy change of a reaction. define and solve problems using Hess’s law. write formation reactions and calculate enthalpy changes using standard enthalpy of formation values. describe current and future energy sources and explain their advantages and disadvantages

5.1: Energy: is the capacity to do work Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Electrical energy is the energy associated with the flow of electrons Potential energy is the energy available by virtue of an object’s position 5.1

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. thermal energy: the total quantity of kinetic and potential energy in a substance Temperature = Thermal Energy 400C 900C greater thermal energy 5.1

Law of Conservation of Energy: Energy cannot be created or destroyed The system is the specific part of the universe that is of interest in the study. SYSTEM SURROUNDINGS open closed isolated Exchange: mass & energy energy nothing 5.1

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g) 5.1

Nuclear Energy All nuclear reactions are exothermic. Per unit of mass, nuclear reactions release much more energy than exothermic chemical reactions. Two main types of Reactions: Fusion and Fission A fusion reaction occurs when nuclei of small atomic mass combine to form larger molecule. 21H + 31H → 42He + 10n In Fission, large nuclei with high atomic mass are split to smaller, lighter mass by collision with a neutron: 23592U + 10n → 9236Kr + 14156Ba + 310n + 1.9 x 10 10 kJ/mol See P. 290 Table 1 5.1

5.2: Calorimetry and Enthalpy The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. Heat (q) absorbed or released: C = mc q = mcDt q = CDt Dt = tfinal - tinitial 5.2

the heat released = the heat absorbed Calorimetry: the measurement of heat into or out of a system for chemical and physical processes. the heat released = the heat absorbed The device used to measure the absorption or release of heat in chemical or physical processes is called a “Calorimeter” 5.2

a “Calorimeter”

Calorimetry Calculations When analyzing data obtained using a calorimeter, make these assumptions: • Any thermal energy transferred from the calorimeter to the outside environment is negligible. • Any thermal energy absorbed by the calorimeter itself is negligible. • All dilute, aqueous solutions have the same density (1.00 g/mL) and specifi c heat capacity (4.18 J/(g∙°C)) as water thermal energy absorbed or released by a chemical system is given the symbol q . qwater = mcDt 5.2

Calculations Involving Thermal Energy Transfer Thermal Energy transfer by metal How much heat is given off when an 869 g iron bar cools from 940C to 50C? Solution: Given: c of Fe = 0.444 J/g • 0C m = 869 g Dt = tfinal – tinitial = 50C – 940C = -890C q = mcDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J No heat enters or leaves! 5.2

Calculations of metal in water A student places 50.0 mL of liquid water at 21.00 °C into a coffee-cup calorimeter. She places a sample of gold at 100 °C into the calorimeter. The final temperature of the water is 21.33 °C. The specific heat capacity of water is 4.18 J/g∙°C and the density of water, d, is 1.00 g/mL. Calculate the quantity of thermal energy, q , absorbed by the water in the calorimeter SOLUTION: Given: VH2O=50.0 mL; c=4.18 J/(g∙°C); dH2O=1.00 g/mL; Tinitial=21.00 °C;Tfinal=21.33 °C R:quantity of thermal energy absorbed, q d =m/v m= vd, m =50mL x 1.00g/mL = 50 g ∆t = 21.33oC – 21.00oC 5.2

q = mc∆t =50 g x 4.18 J x 0.33oC g. oC = 69 J Determining specific heat capacity of a substance Using the value of q above, calculate the specific heat capacity, c, of gold if its mass is 6.77 g. Final temperature of the gold is same as that of water in the calorimeter. Given:mAu =6.77 g; q = 69 J; t initial =100 oC; t final = 21.33oC ∆t =21.33 oC – 100oC = - 78.67 oC Required: cAu q = mc∆t Rearrage and solve for c

Thermal Energy Transfer during a Neutralization reaction A 50.0 mL sample of a 1.0 mol/L aqueous solution of hydrochloric acid, HCl(aq), was mixed with 50.0 mL of a 1.0 mol/L aqueous solution of sodium hydroxide, NaOH(aq), at 25.0 °C in a calorimeter. After the solutions were mixed by stirring, the temperature was 31.9 °C. (a) Determine the quantity of thermal energy transferred by the reaction to the water, q , Assume that the specific heat capacity and density of both solutions is the same as that of liquid water (b) State whether the reaction was endothermic or exothermic. Given: Total volume = 100 mL; c water= 4.18J/g. oC dwater = m =1g/mL v m = vd; msolution = 100 g ∆t = 31.9 oC – 25.0 oC = 6.9 oC; q = mc∆t =100 g x 4.18 J x 6.9oC g. oC = 2.9 x103 J = 2.9 kJ

Enthalpy (H): Total amount of thermal energy in a substance Enthalpy Change (∆H) : Energy changes in Physical and Chemical Reactions The enthalpy change of the chemical system is equal to the flow of thermal energy in and out of the system, ∆H surrounding = |q system| ∆H surrounding > 0, Endothermic reaction ∆H surrounding < 0, Exothermic reaction 5.2

DH = H (products) – H (reactants) Energy Changes during Exothermic & Endothermic Reaction in an Open System DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts > Hreactants Hproducts < Hreactants DH > 0 DH < 0 5.2

Molar Enthalphy Change Energy changes involving 1 mole of a substances… (∆H) Types of molar Enthalpies Solution (∆Hsol) NaBr(s) →Na(aq) + Br(aq) Combustion (∆Hcomb) CH4(g) + 2O2(aq) → CO2(aq) + H2O(l) Vapourization (∆Hvap) CH3OH(l) → CH3OH(g) Freezing (∆Hfr) H2O(l) → H2O(s) Neutralization (∆Hneut) 2NaOH(aq) + H2SO4(aq) →Na2SO4(aq) + H2O(l) Formation (∆Hf) C(s) +2H2(g) + 1/2O2(g) → CH3OH(l) ∆H = n∆Hr Where n is the amount and ∆Hr is the molar enthalpy change of the reaction 5.2

5.2

Molar Enthalpy Calculations Calculate ∆H for Vaporization Reactions Ethanol, CH3CH2OH(l), is used to disinfect the skin prior to an injection. If a 1.00 g sample of ethanol is spread across the skin and evaporated, what is the expected enthalpy change? The molar enthalpy of vaporization of ethanol is 38.6 kJ/mol. Given: Methanol=1.00 g; ∆Hvap=38.6 kJ/mol Methanol= 24 +6 +16 46g/mol R = ∆H? n = 1 g x 1 mol/46 g A= ∆H = n∆Hvap = 0.022 ∆H = 0.022 mol x 38kJ/mol = 0.836 KJ 5.2

Representing Molar Enthalpy Changes 1. The equations we use to represent energy changes are called thermochemical equations. 2. Endothermic enthalpy changes are reported as positive values; H2O(l) → H2 + 1/2O2(g) ∆Hdecomp. = + 285.8 KJ/mol H2O 3. Exothermic enthalpy changes are reported as negative values H2(l) + 1/2O2 (g) → H2O(l) ∆Hcomb = -285.8 KJ/mol H2 Energy changes can be communicated in four different ways. Three of them are thermochemical equations and one uses diagram. These are: 1. By including an energy value as a term in the thermochemical equation e.g., CH3OH(l) + 3/2 O2 → CO2(g) + 2H2O(g) + 726 KJ(Exothermic Reaction)

2. By writing a chemical equation and stating its enthalpy change e.g., CH3OH(l) + 3/2O2(g)→CO2(g)+ 2H2O(g) ∆H = -726 KJ (Exothermic Reaction) 3. By stating the molar enthalpy of a specific reaction e.g., ∆Hcombustion or ∆Hc = - 726 KJ/mol CH3OH 4. By drawing a chemical potential energy diagram (next slides) Note: Fractions are convenient in many thermochemical equations. They apply to fractions of moles of substances e.g., 3/2 mol represents 1.5 mol rather than fractions of actual molecules 5.2

Thermochemical Equations Potential Energy Diagram Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. Thermochemical Equations H2O (s) H2O (l) DH = 6.01 kJ 5.2

Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. Thermochemical Equations CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ 5.2

Calculations in Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ 5.2

P4 (s) + 5O2 (g) P4O10 (s) DHc = -3013 kJ/mol The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DHc = -3013 kJ/mol 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 5.2

E.g. bond energy for methane: 413KJ/mol 5.3:Bond energies Bonds in compounds are broken when energies are supplied. The energy required to break a chemical bond is called bond dissociation energy. Bond dissociation energies have positive values Bond dissociation energies are reported as average bond energies because it depends on types and # of bonds in the molecule. E.g. bond energy for methane: 413KJ/mol Calculated as follows:CH4 →CH3 + H---- (435KJ/mol) CH3 → CH2 + H----(453KJ/mol) CH2 → CH + H ----(425KJ/mol) CH → C + H ----- (339KJ/mol) Total = 1652 4 This value is convenient for calculating Av. = 413 5.3

Multiple Bonds and Bond Energies Multiple bonds are generally stronger than single bonds. c≡c (839 KJ/mol); c=c (614 KJ/mol); c-c (347 KJ/mol) As the # of bonds increases, the bond length decreases, the more the energy required to break it. E.g. Bond length (pm) Bond energy(KJ/mol) c-c 154 347 c=c 134 614 c≡c 120 839 5.3

Enthalpy and Bond Energies During a chemical reaction, the bonds in the reactants must first break, energy must be added —an endothermic process. Bond breaking have positive signs Making new bonds in the products releases energy—an exothermic Process, energy terms associated with bond making have negative signs. ΔH=sum of energies required + to Break old bonds (positive values) Sum of energies released In the formation of new bonds (negative values) This leads to the equation ΔH= Σn x D (bonds broken) – Σn x D(bonds formed) Energy required energy released 5.3

Σ (sigma) means “the sum of,” n is the amount (in moles) of a particular bond type, and D is the bond energy per mole of bonds. D is obtained from reference tables Steps in using bond energies to predict ∆H for a reaction 1. Determine how many of each type of bond must be broken in the reactants. 2. Determine the number of bonds of each type that form in the products. 3. Use bond energy data from Table 1 (page 307) to calculate the total energy required to break the reactant bonds, followed by the total energy released by the formation of product bonds. 4. The energy change, ∆H, of the reaction is the difference between these two sums 5.3

Sample Problem Calculate the enthalpy change for the reaction in which hydrogen gas, H2 (g), is combined with fluorine gas, F2(g), to produce 2 moles of hydrogen fluoride gas, HF(g). This reaction is represented by the balanced chemical equation H2(g)+F2(g) → 2HF(g) Given: for H2(g): nH–H = 1 mol; DH–H = 432 kJ/mol; for F2(g): nF–F = 1 mol; DF–F = 154 kJ/mol; Required:ΔH Analysis: ∆H = Σn x D bonds broken –Σn x D bonds formed ΔH = (nH-FDH-F + nF-F DF-F) – (nH-FDH-F 5.3

Solution: 1 mol each of H–H and F–F bonds are broken The bonds formed are 2 mol of H–F bonds ∆H= (nH-HDH-H + nF-FDF-F) – nH-FDH-F (1 mol x 432KJ) + (1 mol x 154 KJ) - (2 mol x 565 KJ mol mol mol ∆H = -544 KJ The enthalpy change for the reaction of 1 mol hydrogen gas and 1 mol fluorine gas to ptoduce 2 mol. Hydrogen fluoride is -544 KJ 5.3

Bond Energies Review Video Presentation

Hess’s Law

5.5: Standard enthalpy of formation Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 DH0 (C, graphite) = 0 f f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 5.5

5.5

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 5.5

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 5.5

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ - 6534.9 kJ 2 mol = - 3267.44 kJ/mol C6H6 5.5

DHsoln = Hsoln - Hcomponents The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? NaCl, KCl, NH4Cl, NH4NO3 Which substance(s) could be used for a cold pack? LiCl & CaCl2 6.6