# Chapter 6 Thermochemistry Chapter 6 Lesson 2 Thermochemistry Prentice Hall UNIT 3 Chapter 5: Energy Changes Section 5.1 The First Law of Thermodynamics: Energy is Conserved The first law of thermodynamics states that: energy can be converted from one form to another but cannot be created or destroyed. Since any change in energy of the universe must be zero, Euniverse = Esystem + Esurroundings = 0 Esystem = Esurroundings if a system gains energy, that energy

comes from the surroundings if a system loses energy, that energy enters the surroundings UNIT 3 Chapter 5: Energy Changes Section 5.1 Enthalpy One way chemists express thermochemical changes is by a variable called enthalpy, H. The change in enthalpy, H, of a system can be measured. It depends only on the initial and final states of the system, and is represented by H = E + (PV) For reactions of solids and liquids in solutions, (PV) = 0 If heat enters a system If heat leaves a system H is positive

H is negative the process is endothermic the process is exothermic UNIT 3 Chapter 5: Energy Changes Section 5.1 The Second Law of Thermodynamics The second law of thermodynamics states that: when two objects are in thermal contact, heat is transferred from the object at a higher temperature to the object at the lower temperature until both objects are the same temperature (in thermal equilibrium) When in thermal contact, energy from hot particles will transfer to cold particles until the energy is equally

distributed and thermal equilibrium is reached. UNIT 3 Chapter 5: Energy Changes Section 5.1 Comparing Categories of Enthalpy Changes: Enthalpy of Solution Three processes occur when a substance dissolves, each with a H value. 1. bonds between solute molecules or ions break 2. bonds between solvent molecules break 3. bonds between solvent molecules and solute molecules or ions form Sum of the enthalpy changes: enthalpy of solution, Hsolution The orange arrow shows the overall H.

UNIT 3 Chapter 5: Energy Changes Section 5.2 Thermochemical Equations and Calorimetry Chemical reactions involve initial breaking of chemical bonds (endothermic) then formation of new bonds (exothermic) Hr is the difference between the total energy required to break bonds and the total energy released when bonds form. UNIT 3 Chapter 5: Energy Changes Section 5.3 Hesss Law

The enthalpy change of nearly any reaction can be determined using collected data and Hesss law. The enthalpy change of any reaction can be determined if: the enthalpy changes of a set of reactions add up to the overall reaction of interest standard enthalpy change, H, values are used UNIT 3 Chapter 5: Energy Changes Section 5.3 Combining Sets of Chemical Equations To find the enthalpy change for formation of SO3 from O2 and S8, you can use UNIT 3 Chapter 5: Energy Changes Section 5.3 Techniques for Manipulating Equations

1.Reverse an equation the products become the reactants, and reactants become the products the sign of the H value must be changed 2.Multiply each coefficient all coefficients in an equation are multiplied by the same integer or fraction the value of H must also be multiplied by the same number Standard Conditions the standard state is the state of a material at a defined set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest

usually 25C substance in a solution with concentration 1 M the standard enthalpy change, H, is the enthalpy change when all reactants and products are in their standard states the standard enthalpy of formation, Hf, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the Hf for a pure element in its standard state = 0 kJ/mol by definition Tro, Chemistry: A Molecular Approach 10 UNIT 3 Chapter 5: Energy Changes

Section 5.3 Standard Molar Enthalpies of Formation Data that are especially useful for calculating standard enthalpy changes: standard molar enthalpy of formation, Hf the change in enthalpy when 1 mol of a compound is synthesized from its elements in their most stable form at SATP conditions enthalpies of formation for elements in their most stable state under SATP conditions are set at zero since formation equations are for 1 mol of compound, many equations include fractions standard molar enthalpies of formation are in Appendix B UNIT 3 Chapter 5: Energy Changes

Section 5.3 Formation Reactions and Thermal Stability The thermal stability of a substance is the ability of the substance to resist decomposition when heated. decomposition is the reverse of formation the opposite sign of an enthalpy change of formation for a compound is the enthalpy change for its decomposition the greater the enthalpy change for the decomposition of a substance, the greater the thermal stability of the substance UNIT 3 Chapter 5: Energy Changes Section 5.3 Using Enthalpies of Formation and Hesss Law

For example: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) UNIT 3 Chapter 5: Energy Changes Section 5.3 LEARNING CHECK Determine Hr for the following reaction using the enthalpies of formation that are provided. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) Hf of C2H5OH(l): 277.6 kJ/mol Hf of CO2(g): 393.5 kJ/mol Hf of H2O(l): 285.8 kJ/mol Answer on the next slide

UNIT 3 Chapter 5: Energy Changes Section 5.3 LEARNING CHECK Hr = [(2 mol)(Hf CO2(g)) + (3 mol)(Hf H2O(l))] [(1 mol)(Hf C2H5OH(l)) + (3 mol)(HfO2(g)] Hr = [(2 mol)(393.5 kJ/mol) + (3 mol)(285.8 kJ/mol)] [(1 mol)(277.6 kJ/mol) + (3 mol)(0 kJ/mol)] Hr = (1644.4 kJ) (277.6 kJ) Hr = 1366.8 kJ Relationships Involving Hrxn when reaction is multiplied by a factor, Hrxn is multiplied by that factor because Hrxn is extensive C(s) + O2(g) CO2(g) H = -393.5 kJ

2 C(s) + 2 O2(g) 2 CO2(g) H = 2(-393.5 kJ) = 787.0 kJ if a reaction is reversed, then the sign of H is reversed CO2(g) C(s) + O2(g) Tro, Chemistry: A Molecular Approach H = +393.5 kJ 16 Sample Hesss Law Given the following information: 2 NO(g) + O2(g) 2 NO2(g) 2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq) N2(g) + O2(g) 2 NO(g) H = -173 kJ H = -255 kJ H = +181 kJ Calculate the H for the reaction below:

3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) H = ? [[32 NO O2(g)] H = (+259.5 NO22(g) (g) 32 NO(g) NO(g) ++ 1.5 O2(g)] x 1.5 H = 1.5(+173 kJ) kJ) [1 HNO [2 N N22(g) (g) ++ 2.5 5 OO ++2 1HH

4 2HNO x 0.5 2(g) 2O(l) 3(aq)] 2(g) 2O(l) 3(aq)] [[22 NO(g) NO(g) N N22(g) (g) ++ O O22(g)] (g)] 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) Tro, Chemistry: A Molecular Approach H kJ) kJ) H == (-128

0.5(-255 H H == -181 -181 kJ kJ H = - 49 kJ 17 The Combustion of CH4 Tro, Chemistry: A Molecular Approach 18 Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 1. Write formation reactions for each compound and determine the Hf for each

2 C(s, gr) + H2(g) C2H2(g) Hf = +227.4 kJ/mol C(s, gr) + O2(g) CO2(g) Hf = -393.5 kJ/mol H2(g) + O2(g) H2O(l) Hf = -285.8 kJ/mol Tro, Chemistry: A Molecular Approach 19 Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 2.

Arrange equations so they add up to desired reaction 2 C2H2(g) 4 C(s) + 2 H2(g) H = 2(-227.4) kJ 4 C(s) + 4 O2(g) 4CO2(g) H = 4(-393.5) kJ 2 H2(g) + O2(g) 2 H2O(l) H = 2(-285.8) kJ 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) Tro, Chemistry: A Molecular Approach H = -2600.4 kJ

20 Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) Hreaction = n Hf(products) - n Hf(reactants) Hrxn = [(4HCO2 + 2HH2O) (2HC2H2 + 5HO2)] Hrxn = [(4(-393.5) + 2(-285.8)) (2(+227.4) + 5(0))] Hrxn = -2600.4 kJ Tro, Chemistry: A Molecular Approach 21 Example How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? Given: Find: Concept Plan: 1.0 x 1011 kJ

mass octane, kg Write the balanced equation per mole of octane Hfs kJ Relationships: Solution: H rxn nH f products - nH f reactants Hrxn from above mol C8H18 MMoctane = 114.2 g/mol, 1 kg = 1000 g g C8H18

114.2 g 1 mol 1 kg 1000 g kg C8H18 C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) 1 mol C8H18 H rxn nH f products 114.22 - ngH f reactants1Material kg Hf, kJ/mol - 1.0 10 kJ

25 Look up Hthe H H 1000 f H-250.1 8H f kJ CO 2 1 9mol H O C H

- 5074.1 C H g C H (l) f 2 8 18 f 8 18 f O 2 8 18 2 11 for each material 00 kJ 2(g) 8 393.5 kJ 9 241.84 kJ O250 .1 kJ 25

in Appendix 2.3 106 kg C8 H18 2 CO2(g) -393.5 5074.1 kJ H2O(g) -241.8 Check: the units and sign are correct the large value is expected 22 0 The standard enthalpy of reaction (Hrxn ) is the enthalpy of a reaction carried out at 1 atm.

aA + bB cC + dD 0 Hrxn = [ cH0f (C) + dH0f (D) ] - [ aH0f (A) + bH0f (B) ] 0 Hrxn = nH0f (products) - mHf0 (reactants) Hesss 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 doesnt matter how you get there, only where you start and end.) 6.5 6.5

Calculate the standard enthalpy of formation of CS2 (l) given that: 0 C(graphite) + O2 (g) CO2 (g) Hrxn = -393.5 kJ S(rhombic) + O2 (g) CS2(l) + 3O2 (g) SO2 (g) 0 Hrxn = -296.1 kJ CO2 (g) + 2SO2 (g) H0rxn = -1072 kJ 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. C(graphite) + O2 (g) 2S(rhombic) + 2O2 (g) + CO2(g) + 2SO2 (g) 0 CO2 (g) Hrxn = -393.5 kJ 0 2SO2 (g) Hrxn = -296.1x2 kJ CS2 (l) + 3O2 (g) H0rxn = +1072 kJ C(graphite) + 2S(rhombic) CS2 (l)

H0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ 6.5 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) 0 Hrxn = nH0f (products) - mHf0 (reactants) 0 Hrxn = [ 12H0f (CO2) + 6H0f (H2O)] - [ 2Hf0 (C6H6) ] 0 Hrxn = [ 12x393.5 + 6x187.6 ] [ 2x49.04 ] = -5946 kJ

-5946 kJ = - 2973 kJ/mol C6H6 2 mol 6.5 The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. Hsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.6 The Solution Process for NaCl

Hsoln = Step 1 + Step 2 = 788 784 = 4 kJ/mol 6.6 Energy Use and the Environment in the U.S., each person uses over 105 kWh of energy per year most comes from the combustion of fossil fuels combustible materials that originate from ancient life C(s) + O2(g) CO2(g) Hrxn = -393.5 kJ CH4(g) +2 O2(g) CO2(g) + 2 H2O(g) Hrxn = -802.3 kJ C8H18(g) +12.5 O2(g) 8 CO2(g) + 9 H2O(g) Hrxn = -5074.1 kJ

fossil fuels cannot be replenished at current rates of consumption, oil and natural gas supplies will be depleted in 50 100 yrs. Tro, Chemistry: A Molecular Approach 29 Energy Consumption the increase in energy consumption in the US Tro, Chemistry: A Molecular Approach 30 The Effect of Combustion Products on Our Environment because of additives and impurities in the fossil fuel, incomplete combustion and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy

therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming Tro, Chemistry: A Molecular Approach 31 Global Warming CO2 is a greenhouse gas it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the earth to escape into outer space it acts like a blanket CO2 levels in the atmosphere have been steadily increasing current observations suggest that the average global air

temperature has risen 0.6C in the past 100 yrs. atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats Tro, Chemistry: A Molecular Approach 32 CO2 Levels Tro, Chemistry: A Molecular Approach 33 Renewable Energy our greatest unlimited supply of energy is the sun new technologies are being developed to capture

the energy of sunlight parabolic troughs, solar power towers, and dish engines concentrate the suns light to generate electricity solar energy used to decompose water into H 2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity H2(g) + O2(g) H2O(l) Hrxn = -285.8 kJ hydroelectric power wind power Tro, Chemistry: A Molecular Approach 34