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Transcript
Chapter 6. Therrnochemistry The Nature & Types of Energy (Section 6.1) Energy Changes in Chemical Reactions (Section 6.2) Introduction to Thermodynamics (Section 6.3) Enthalpy of Chemical Reactions (Section 6.4) Calorimetry (Section 6.5) Standard Enthalpy of Formation and Reaction (Section 6.6) Heat-of Solution and Dilution (Section 6.7) SUMMARY Truly balanced reactions include the energy changes that accompany chemical reactions. This chapter describes these energy changes, techniques used to measure them experimentally and methods used to predict them quantitatively. The Nature & Types of Energy (Section 6.1) Energy. Energy is one of those terms that is difficult to understand because it is an abstract concept that is used in many different ways. The standard scientific definition is a system’s capacity to do work (move an object against an opposing force). There are three principal types of energy, some with a number of sub-classes. The types of energy that are important in our discussion of chemical reactions are kinetic (dynamic) energy: potential (static) energy: radiant (electromagnetic) energy: sub-classes: energy an object possesses by virtue of its motion energy an object possesses by virtue of its position energy a wave possesses by virtue of its frequency chemical energy (kinetic and potential energy E associated with bond breakage & formation) thermal energy (kinetic and potential energy E associated with random particle motion) heat (thermal energy transferred between objects at different temperatures) The law of conservation of energy states that energy is neither destroyed nor created; it is transferred from one part of the universe to another or it is converted from one form to another. Energy Changes in Chemical Reactions (Section 6.2) In order to study energy changes in chemical reactions we divide the universe into the system (where the reaction occurs) and the surroundings. In an open system, mass and energy in the system are exchanged with the surroundings. For example, a burning fire is an open system. The system consists of the reactants (wood and oxygen) and products (CO2, water vapor, ashes and so on). The surroundings are everything else. In a closed system, 108 Chemistry, Ch. 6: Thermochemistry 109 only energy from the system is exchanged with the surroundings. In an isolated system, neither mass nor energy from the system is exchanged with the surroundings. When a reaction in the system produces heat that is released to the surroundings, that is when energy is a product, the reaction is called exothermic. When a reaction in the system consumes heat from the surroundings (heat is a reactant) the reaction is called endothermic. Introduction to Thermodynamics (Section 6.3) The theories that allow us to describe energy changes quantitatively are part of thermodynamics, the study of the interconversion of different types of energy. The first law of thermodynamics (conservation of E) limits the energy absorbed or released by a reaction to two.#arts: work done on or by the system, w, and heat absorbed or released by the system, q. By definition, work done on the system is positive (work done by the system is negative) and heat absorbed by the system is positive (heat released from the system is negative).. For the hypothetical reaction: A~nit~ -) Ann~, the conservation law is AEsys --&Esurr &Esys =q+w where AEsys is the change in the internal energy of the system and AEsu,r is the change in the energy of the surroundings. If we could relate q and w directly to the reactants and products (Ainitiat and A,o~=), we’d have the quantitative description we’re looking for. The problem is that q and w are path dependent properties, in other words their values change as the conditions and procedures used to carry out the reaction change. We need pathindependent properties that only depend on Aioitia= and A, na~ and stay the same regardless of how the reaction is carried out. It is a little more work, but we can relate q and w to state functions, which is the technical term for path-independent properties. Work done by gases is -PV (work = force x distance; pressure = force/area; volume = area x distance), so we can define an energy state function especially for reactions in which the pressure is constant. The state function is called the enthalpy, H. The formula for the enthalpy is H=E+PV The following equations show that enthalpy changes (&H) equal the heat transferred from the system by the reaction at constant pressure, q~. AH : H~.., - HinitiaI : (Efina, + PVfinal) - (Ei.,ti~’ + PVinitiaI ) Examples 6.1 -6.3 : (Efinal - EinitiaI) + P (Vfinal - VinitiaI ) = AE + PAV = AE - w = qp Exercises 6.1 -6.5 Enthalpy of Chemical Reactions (Section 6.4) Most reactions are run at constant pressure, so the enthalpy ~s equal to the heat transferred by the reaction. When the enthalpy is included in a chemical reaction it is called a thermochemical reaction. For exothermic reactions, (heat is released by the system) the enthalpy change is negative: CH4(g)+ 2 O2(g)--,’. CO2(g)+ 2 H20(g) AH =-802.4 kJ/mol ~ ~u ~neml~[ry, ~n. o, iRermocnemls[ry For endothermic reactions, (heat is absorbed by the system) the enthalpy change is positive: CaCO~(s) --+ CaO(s) + CO2(g) z~H = 177.8 k J/ When a reaction is reversed, the sign of ,&H is reversed H20(s) --> H20(£) H20(~’) --~ H20(s) &H = 6.01 kJ/mol AH = -6.01 kJ/mel The enthalpy is an extensive property, its value depends on the amount of material in the system. Therefore, scaling the reaction (multiplying by a constant) also scales H20(s) --> H20(~) 2 H20(s) -~ 2 H2Oi#) AH = 6.01 kJ/mol AH = 12.0 kJ/reel The physical states of all reactants and products are crucial to enthalpy values CH4(g)+2 O2(g)--> CO2(g)+2 H20(g) AH =-802.4 k J/reel CH4(g)+ 2 O2(g) -~ CO2(g)+ 2 H20(#) AH =-890.4 kJ/mol Since 80.0 g, of CH4 equals nearly 5 moles, if all the gas was burned under conditions Example that produce gaseous products, approximately 4 MJ (5 mol x 0.8 M J/reel) of energy 6.4 would be released. 80.0gx lm°lCH4 x 802.4kJ released = 4.00 MJ released 16.032 g CH4 moICH4 Exercises 6-6 - 6-7 Calorimetry (Section 6.5) Heat Capacity and Specific Heat. A calorimeter is a device for measuring the heat changes associated with chemical reactions. All calorimetry experiments involve carrying out a reaction in a vessel immersed in a medium such as water and measuring the temperature change (At) of the medium. The heat capacity, C, of a substance is the amount of heat, q, required to raise the temperature of a given mass of substance by 1 °C, The heat capacity of 100 g H20 is 418.4 J/°C. This means that 418.4 J of heat will raise the temperature of 100 g water 1 degree Celsius. The heat capacity is a proportionality constant that relates the amount of heat absorbed or released by a material to its change in temperature: The specific heat, s, is a related term which is the amount of heat required to raise the temperature of one gram of a substance by 1 °C. In terms of the specific heat, the heat absorbed or released is q = msAt. Chemistry, Ch. 6: Thermochemistry 111 Constant-Volume Calorimetry. Bomb calorimeters, as shown in Figure 6.8 in the textbook, are used to measure the heat evolved in combustion reactions. A high-pressure steel vessel, called a bomb, is loaded with a small amount of a combustible substance and 02 at 30 atm of pressure. The loaded bomb is immersed in a known amount of water. The heat evolved during combustion is absorbed by the water and the calorimeter: q~, : -(qcal + qwater ) The change in the temperature of the wate~ and calorimeter are related to the heat evolved by the reaction, q, and to the enthalpy, AH. AH = q~xn = -(CcalAt + mswaterAt) Reactions in a bomb calorimeter occur under constant volume rather than constant pressure conditiops, and so the heat released does not equal AH exactly. For most reactions, the difference is small and can be neglected. For instance, for the combustion of 1 mole of pentane, the difference is only 7 kJ out of 3500 kJ. Constant-Pressure Calorimeter. The heat evolved in non-combustion reactions is measured in a constant-pressure calorimeter. Coffee cup calorimeters, shown in Figure 6.9 in the textbook, are very inexpensive examples of these devices. Water-soluble reactants are added to water in the calorimeter. The reaction occurs in the solution, absorbing or liberating heat that changes the temperature of the solution (mostly water), and the calorimeter: qrxn - -(q~.l + qwater ) Tl~ese calorimeters are constructed of low heat capacity materials, such as Styrofoam, so the heat absorbed by the calorimeter can be neglected: qrxn =-qwater ’ Examples 6.5 - 6.7 Reactions in a coffee cup calorimeter occur under constant pressure conditions, Exercises and so the heat released is equal to AH. 6-8 - 6-13 AH = qrxn = -mSwaterz~t Standard Enthalpy of Formation and Reaction (Section 6.6) Recapping the information presented so far, we can calculate AH for any reaction for which areactants and Hproducts are known and measure AH for most reactions. The problem is that there is no way to measure Hreactants and aproducts. Absolute energy values, such as Hreactants and Hproducts, can’t be measured because there, is no way to define absolute zero energy. Relative energy values can be measured because the reference energy is arbitrary. The reference point for enthalpy values is the standard of enthalpy of formation of the elements which is defined as zero at 1 atm (g) or 1 M and 25 °C (in their standard state). The standard enthalpy of atlotropes and single atoms of diatomic elements have non-zero values. For example, 112 Chemistry, Ch. 6: Thermochemistry AH~(O2 ) = 0 kJ / mol ° AHf(O3)=142 kJ/ mol AH~ (Cgraphite ) = 0 kJ / mol ° AHf(Cdiamond)=l.9 kJ/mol AH~ (H2 ) = 0 kJ / mol AH~(H)=218.2 kJ/mol Direct Calculation ofz~H,~,. Now we can calculate AH for any reaction using A/-!) of the reactants and products. The enthaipy change for any reaction is the difference of the SUM of the standard enthalpies of formation of the products multiplied by their stoichiometric coefficients and the SUM of the standard enthaipies of formation of the reactants multiplied by their stoichiometric coefficients. For e_xample, the enthaipy of the decomposition of CaCO3, CaCO3(s) --> CaO(s) + CO2(g), is AH~xn = ~produotsnproduclsAH~(products)-’~reactantsnreactantsAH~(reactants) = AHf(CaO) + AHf (CO2)- AHf(CaCO3) = (1 mol)( 635.6 kJ / mol) + (1 mol)(-393.5 kJ / mol) - (1 mol)(-1206.9 kJ / tool) = 177.8 kJ / tool Indirect Calculation ofz~H~,. Some reactions don’t proceed as written. For example, the reaction may proceed too slowly or generate side-products. In other cases, the enthalpy of formation of one or more reactant or products may not be available. The alternative comes from Hess’s law of heat summation. When a reaction is the sum of several reaction steps, then the enthalpy change for the reaction is equal to the sum of the enthalpy changes of the reaction steps. For example, it’s not convenient to measure the enthalpy of the conversion of graphite to diamond. This reaction takes thousands of years at high pressure in the earth. However, it is easy to measure the enthalpies of combustion of graphite and diamond: Cgraphite -}- 02 ~ CO2 AHoor~b :-393.51 kJ/mol Cdiamond + 02 --~ 002 AHoomb :-395.40 kJ/mol Notice that when we reverse the diamond thermochemical reaction, the sum of the Examples combustion reactions is the graphite to diamond conversion: 6.8-6.9 Cgraphite ÷ 02 --)- CO2 AHcomb =-393.51 kJ/mol 002 --). Cdiamond + 02 AHcomb = 395.40 kJ/mol AH~,n = 1.89 kJ / mol Cgraphite ~)" Cdiarnond Exercises 6-14-6-17 Heat of Solution and Dilution (Section 6.7) Heat of Solution. Dissolving ionic solutes in a solvent often produces temperature changes as the extent of solute hydration changes. We can think of solute dissolution as requiring 2 steps: solute disassociation to gaseous ions followed by ion hydration to form the solution. This is not what actually happens, but these are steps for which AH values are available. The energy required to separate 1 mole of an ionic compound to gaseous ions (step 1) is called the lattice energy, Etattice. The enthalpy change of the hydration of 1 mole of gaseous Chemistry, Ch. 6: Thermochemistry 113 ions (step 2) is the enthalpy of hydration, AHhydr. The enthalpy of solution is the sum of these two terms. For the dissolution of LiCI: LiCl(s)+energy-->Li+(g)+CI (g) Elattice =853 kJ/mol Li+(g) + Cl-(g) --> Li+(aq) + CI (aq) AH,ydr = -890.1 kJ / mol LiCl(s) --> Li+(aq) + CI (aq) AHso~n -- Elattice + AH~dr = -37.1 kJ/mol Heat of Dilution. When Elat~ice, which is ~lways positive, is smaller than AHhydr, the solvation reaction is exothermic. If more water is added to such a solution, more heat will be released by the system. (This is why you never add water to acid, but always add acid to water with vigorous stirring. If the water boils it is much less dangerous than if concentrated acid boils!) Using the same reasoning, if the solvation reaction is endothermic, dilution will cause more heat to be absorbed from the.surroundings. Example 6.10 Exercises 6-18-6-19 Chemistry, Ch, 6: Thermochemistry 115 GLOSSARY LIST kinetic energy potential energy chemical energy thermal energy heat radiant energy energy conservation law work heat state function enthalpy endothermic exothermic thermochemical equation thermodynamics system surroundings open system closed system Isolated system enthalpy of reaction standard state heat capacity specific heat calorimeter standard enthalpy of formation Hess’ law lattice energy heat of hydration heat of solution enthalpy of solution heat of dilution EQUATIONS Algebraic Equation AE-q+w w -- -PAV H=E+PV AH - AE + PAV AE = AH - RTAn C=ms q = CAt q = msAt AH~o = ~nAH~(products)- ~mAH~(reactants) English Translation The energy of a system consists of heat and work 118 Chemistry, Ch. 6: Thermochemistry WORKED EXAMPLES EXAMPLE 6.1 Work Done on the System Calculate the work done on the system when 6.0 L of a gas is compressed to 1.0 L by a constant external pressure of 2.0 atm, ¯ Solution The work done is: W = -P AV = -P(V2 - Vl) = -2.0 atm (I.0 L - 6.0 L) = + 10 L.atm The answer can be converted to joules. w = 1.01 × 103 J ¯ Comment Obtaining a positive value for work means that work is done on the system by the surroundings in a compression. A positive work value means the system gains energy. EXAMPLE 6.2 Gas Expansion Work A gas, initially at a pressure of 10.0 arm and having a volume of 5.0 L, is al{owed to expand at constant temperature against a constant external pressure of 4,0 arm until the new volume is 12.5 L. Calculate the work done by the gas on the surroundings. o Solution In this problem the system does work on the surroundings as it expands, and by convention (Table 6.5 in the text) the sign is negative: w = -Pz~V = -P(V2 - V1) P is the pressure opposing the expansion, ~V is the change in volume of the system, w = -4,0 atm (12.5 L - 5.0 L) = -30 L.atm This quantity can be expressed in units of joules. w = -30 L,atm x 101.3 J = -3.0 x 103 J t L. atm Chemistry, Ch. 6: Thermochemistry 119 EXAMPLE 6.3 The First Law of Thermodynamics A gas is allowed to expand at constant temperature from a volume of 10.0 L to 20.0 L against an external pressure of 1.0 atm. If the gas also absorbs 250 J of heat from the surroundings, what are the values of q, w, and AE? Solution The work done by the system is w = -P AV = -P(V2 - V1) =-1.0 atm (20.0 L- 10.0 L) =- 10 L.atm w=-10L.atmx 101.3J =_1.0x103j 1 L ¯ atm The amdunt of heat absorbed was 250 J, and so q -- 250 J. Substituting into the first law of thermodynamics gives the energy change, AE. AE=q+w = 250 J - 1000 J AE = -750 J o Comment tn this example, the system did more work than the energy absorbed as heat, therefore the internal energy E decreased. EXAMPLE 6.4 Thermochemical Equations The thermochemical equation for the combustion of propane is: C3H8(g) + 5 O2(g) -~ 3 CO2(g) + 4 H20 AH~n = -2220 kJ/mol a. b. How many kJ of heat are released when 0.50 mole of propane reacts? How much heat is released when 88.2 g of propane reacts? o Solution for a Let q = the heat absorbed or released by the reaction. The heat released by an exothermic reaction is an extensive property. This means that q depends on the amount of propane consumed. The equation indicates that 2220 kJ of heat is released per mole of propane burned. q = 0.50 mol C3H8 x q =-1.1 x 103 kJ -2220 kJ 1 mol C3H8 120 Chemistry, Ch. 6: Thermochemistry o Solution for b Since we know the he~t.of reaction per mole, we convert the number of grams of C3H8 to moles. q = 88.2 g C3H8 × 1 mol C3H8 -2220 kJ x 44.1 g C3H8 1 mol C3H8 q = -4440 kJ EXAMPLE 6.5 Specific Heat a.- What is the heat capacity of a block of lead if the temperature of a 425-g block increases 2.31 °C when it absorbs 492 J of heat? b. What is the specific heat of lead? ¯ Solution a. The heat capacity of the block of lead is the heat absorbed divided by the temperature rise: C - q - 492J =213J/°C At (2,31 °C) b. In terms of the specific heat (s), the amount of heat absorbed when an object of mass m is heated from T~ to Tr is: q = ms At Substituting into the equation the given quantities for q, m, and/~t gives: 492 J = (425 g) s (2.31 °C) Rearranging to solve for the specific heat s: s- 492 J = 0.501 J/g.°C (425 g)(2.31 °C) ¯ Comment Since C = ms we could have solved part b from s = C/m using our answer from part a. Chemistry, Ch. 6: Thermochemistry 121 EXAMPLE 6.6 Determining The Calorimeter Constant The combustion of benzoi~ acid is often used as a standard source of heat for calibrating combustion bomb calorimeters. The heat of combustion of benzoic acid has been accurately determined to be 26.42 kJ/g. When 0.8000 g of benzoic acid was burned in a calorimeter containing 950 g of water, a temperature rise of 4.08 °C was observed. What is the heat capacity of the bomb calorimeter (the calorimeter constant)? ¯ Solution The combustion of 0.8000 g of benzoic acid produces a known amount of heat. q~o = 0.8000 g × 26.41 kJ =-21.14 kJ =-2.114 × 104j lg And since At and the amount of water are known, Cc,~, the heat capacity of the calorimeter. can be calculated. All the heat from the combustion reaction is absorbed by the bomb calorimeter and water. q~n = -(q~,l + qwater) qrxn = -(Cca~ At + ms At) The heat absorbed by the water was qwater = ms At, where m = 950 g water s = 4.184 J/g-°C At = 4.08 °C qwater = 950 g (4.184 J/g’°C)(4.08 °C) = 1.620 x 104 J The bomb calorimeter must have absorbed the difference between the 21.14 × 103 J released, and the 16.20 x 103 J absorbed by the water. qr~o = -(qo~l + qc,I = -qr~o - q~,te~ = --(--2.114 X 104 J) - 1.620 x 104 J qoal = 4.94 x 103 J The heat capacity of the calorimeter is Cc~- qbomb _ 4.94x10~ J At 4.08 °C Cc~ = 1210 J/°C 122 Chemistry, Ch. 6: Thermochemistry EXAMPLE 6.7 Determining the Heat of Combustion The thermochemical equation for the combustion of pentane (C5H12) is: C5H12(~) + 8 O2(g) --> 5 CO2(g) + 6 H20(~) AH" : -3509 k J/mot From the following information calculate the heat of combustion, AH°, and compare your result to the value given above. 0.5521 g of C5H12 was burned in the presence of excess 02 in a bomb calorimeter. The heat capacity of the calorimeter was 1.800 x 103 J/°C, the temperature of the calorimeter and 1.000 x 103 g of water rose from 21.22 °C to 25.70 °C. ¯ Solution The heat evolved by the combustion reaction is absorbed by the water and the calorimeter assembly. qrxn = -[qoal + where qrxn = -(Cca=At + msAt) At = 25.70 °C - 21.22 °C = 4.48 °C Substituting: 1"800x103 qca~ =--x 4.48 J 1°C qwater : 1.000 x 103 g x °C = 8.06 x 103 J 4.184 J x 4.48 oC = 1.87 x 104 J qrxn = -[8,060 J + 18,700 J] qr~n = --26,800 J = - 26,8 kJ (three significant figures) AH° refers to the reaction of 1 mole of pentane. The 26,800 J was evolved by 0.5521 g C5H12. The molar mass of pentane is 72.15 g. AH° = q 26.8 kJ 0.5521 g 72.15 g C5H12 1 mol AH° = -3500 kJ/mol = -3.50 x 103 kJ/mol ¯ Comment To three significant figures the result compares well with the value given in the thermochemical equation. Small differences will arise because the heat evolved at constant volume is not quite the same as the heat evolved at constant pressure (which is equal to AE). Only the heat evolved at constant pressure exactly equals AH, the enthalpy change. Chemistry, Ch. 5: Thermochemistry 123 EXAMPLE 6.8 Using Enthalpies of Formation Using AH~ values in Appendix 3 of the textbook calculate the standard enthalpy change for the incomplete combustion of ethane (C2H6). 5 C2H6(g) + ~O2(g)-~ 2 CO(g) + 3 H20(~) ¯ Solution The enthalpy change for this chemical reaction in terms of enthalpies of formation is: - AH~xn = ~’nAH~ (products) - ~ m AH~ (reactants) AH~n = [2 x AH~ (CO)+ 3 x AH~ (H20)]- [1 x AH~ (C2H6)- ~- x AH~ (O2)] To avoid cumbersome notation, the physical states of the reactants and products were omitted from this equation. Be careful to obtain from Appendix 3 the appropriate value ofAHf. Note that the coefficients from the chemical equation are equal to the number of moles of each substance and that al~l the terms involving AH~ of the reactants are preceded by a negative sign. Substituting the values from Appendix 3: AH~o = (2 mol)(-110.5 kJ/mol) + (3 mol)(-285.8 kJ/mol)] - [(1 mol)(-84.68 kJ/mol) + (5/2 mol)(0 kJ/mol)] = (-221.0 k J) + (-857.4 k J) - (-84.68 k J) = -221.0 kJ - 857.4 kJ + 84.68 kJ AH~n = -993,7 kJ EXAMPLE 6.9 Hess’s Law The standard enthalpy change for the combustion of 1 mole of ethanol is: C2HsOH(~) + 3 O2(g) --~ 2 CO2(g) + 3 H20([) AH~xn = -1367 kJ/mol What is AH~n for the following reaction in which H20 is formed as a gas, rather than as a liquid? C2H5OH(~) + 3 O2(g) :~ 2 CO2(g) + 3 H20(g) Given the heat of vaporization of water H20(~) -+ H20(g) AHvap = 44 kJ/mol 124 Chemistry, Ch. 6: Thermochemistry ¯ Solution We can imagine a two-step path for this reaction. In the first step, ethanol undergoes combustion to form liquid H20, followed by the second step in which H20(I) is vaporized. The sum of the two steps gives the desired overall reaction, and according to Hess’s law the sum of the two AH’s gives the overall L~H. C2HsOH(I) + 3 O2(g) -~ 2 CO2(g) + 3 H20(I) 3 H2O(I) -+ 3 H20(g) C2HsOH(I) + 3 O2(g) --~ 2 C~)2(g) + 3 H20(g) z~H~ = -1367 k J/tool AH = 132 kJ/mol AH=o = -1235 kJ/mol EXAMPLE 6.10 Heat of Hydration The heat of solution of NaBr is -1.0 k J/tool. The lattice energy of NaBr is 735 kJ/mol, Determine the heat of hydration of NaBr and write the equation for the hydration reaction. Solution Recall that the heat of solution is the sum of the lattice energy (E~.~) and the heat of hydration ~Hsoln --" Elattice + AHhydr. Rearrranging for the heat of hydration: = -1.0 k J/tool - 735 k J/tool = -736 kJ/mol ~,Hhydr refers to the heat liberated or absorbed when the gas phase ions are dissolved in water: The equation is: H20 Na*(g) + Br-(g) ~ Na+(aq) + Br-(aq) ¯ Comment For NaBr, I::latt~ce and AH.y~ have essentially the same values, but are opposite in sign, Therefore, when NaBr dissolves in water, the energy E~,~e required to break up the crystal lattice is "paid back" by the hydration of the Na÷ and Br- ions. Also, NaBr does not dissolve in nonpolar solvents. Molecules of nonpolar solvents can only interact weakly with ions in the crystal lattice, and so "solvation" is not sufficient to compensate for the lattice energy.