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Thermochemistry Chapter 5 Energy, Heat, Work Watkins Chemistry 1422, Chapter 5 1 The Nature of Energy Energy is what makes things happen. Matter is what it happens to. Whenever anything happens to matter, Energy is involved. Matter and Energy are the only two things in the universe, and they are related: E = mc2 Watkins Chemistry 1422, Chapter 5 2 The Nature of Energy (Review) Energy cannot be created or destroyed during any process or reaction; it can be transferred from place to place it can be converted from one form to another This fact is called the First Law of Thermodynamics Watkins Chemistry 1422, Chapter 5 3 The Nature of Energy (Review) A few of the many different forms of energy: • • • • • • • • • • Mechanical Work (w = fd = -PDV) Gravitational Energy (-Gm1m2/d = mgh) Kinetic Energy (1/2mv2) 1 J = 1 kg•m2•s-2 Coulombic Energy (kz1z2/d) Internal Energy (DE) 1 cal = 4.184 J Light Energy (hn) (chapter 6) Nuclear Energy (mc2) Electrical Energy (VQ) (chapter 20) Heat flow (q) Chemical (Bond) Energy (DHrxn) Watkins Chemistry 1422, Chapter 5 4 Thermodynamics (Review) DEFINITIONS System (sys) that part of the universe in which we are interested. Surroundings (surr) the rest of the universe. Watkins Chemistry 1422, Chapter 5 5 Thermodynamics (Review) DEFINITIONS H2(g) & O2(g) Internal Energy Total energy of system Esys H2(g) + ½ O2(g) → H2O(g) Esys DEsys < 0 H2O(g) → H2(g) + ½ O2(g) DEsys > 0 H2O(g) Watkins We cannot measure Esys We can measure DEsys when the system changes during a process For example, during a chemical reaction DEsys = Efinal - Einitial Chemistry 1422, Chapter 5 6 First Law of Thermodynamics (Review) Energy cannot be created or destroyed during a process; it can only be converted or transferred. Therefore ... • the total energy in the universe is constant: DEuniv = 0 = DEsys + DEsurr • any energy transferred to or from system must be transferred from or to surroundings Watkins Chemistry 1422, Chapter 5 7 First Law of Thermodynamics (Review) Relating DE to Heat and Work When a system undergoes a physical or chemical change, the change in its internal energy is equal to q, the heat energy flowing into or out of the system, plus w, the work (all other forms of energy) coming into or going out of the system: DEsys = q + w Watkins Chemistry 1422, Chapter 5 8 First Law of Thermodynamics (Review) q=0 adiabatic q>0 endothermic q<0 exothermic System Surroundings q > 0 (+) Heat (–) 0 > q DEsys = q + w w = 0 (?) w>0 w > 0 (+) Work (–) 0 > w done on system w<0 done by system A process occurs in the system Watkins Chemistry 1422, Chapter 5 9 State Function Defined • A State function is an energy term which depends only on the initial and final states of the system, not on how the energy is used. • E and H are state functions, because DE = Efinal – Einitial DH = Hfinal – Hintial • Neither q nor w are state functions (even though their sum DE is a state function) Watkins Chemistry 1422, Chapter 5 10 heating coil State Function E For a given system, DE does not change with the process but q and w do change, depending on how the energy is used. Watkins charged q<0 q<0 w=0 DE w<0 discharged Chemistry 1422, Chapter 5 11 DE = q + w State Function E For most chemical systems, the usual form of work energy is w = -PDV or expansion/compression work. For these systems, when the volume is contant (DV = 0), w = 0 DE = qv DE is measured as heat flow in a constant volume process. Watkins Chemistry 1422, Chapter 5 12 DE = qV However, most chemical reactions are not carried out at constant volume. They are usually carried out at constant pressure (in the open atmosphere). So a new state function was invented for contant pressure processes. Watkins Chemistry 1422, Chapter 5 13 DE = qV For a contant pressure process (DP = 0); if the system expands (DV > 0), it does work on the surroundings: w = -PDV DE = q + w = qP - PDV qP = DE + PDV = DH DH = qP State Function H is called Enthalpy Watkins Chemistry 1422, Chapter 5 14 State Function H DE = qV DH = qP Think of H as the amount of “heat energy” contained in the system. (H can be called the “heat content”) When a reaction occurs at constant pressure, the heat that flows (qP) between system and surroundings is exactly the enthalpy change DH Watkins Chemistry 1422, Chapter 5 15 State Function H H ("heat content") • We cannot measure H, only DH = Hfinal - Hinitial = qp • H is the total heat energy in the system • qp is heat added to or subtracted from the system at constant pressure. Watkins initial final DH < 0 heat out exo DH > 0 heat in endo final initial Chemistry 1422, Chapter 5 H is heat energy just sitting in the system ("latent" heat) q is heat flow (heat enrgy being transferred!) 16 Enthalpies of Reaction How can you find qp = DHrxn? Two basically different procedures: Calculate it Bond Energies Hess’s Law Measure it Calorimetry Watkins Chemistry 1422, Chapter 5 17 Enthalpies of Reaction (Review: Chapter 8, section 8, page 328) Bond Dissociation Energy is the energy absorbed (+D) to break a chemical bond. Bond breaking is always endothermic When a chemical bond is formed, “Bond Energy” is the energy emitted (-D). Bond making is always exothermic DHrxn is the sum of all bond breaking and bond making energies in a reaction. Watkins Chemistry 1422, Chapter 5 18 Enthalpies of Reaction Hess’s Law I For a chemical reaction at constant pressure: DHrxn = SH(products) - SH(reactants) H is an extensive property - the magnitude of H is directly proportional to amount: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH1 = -803 kJ DH1 = [H(CO2(g)) + H(2H2O(g))] - [H(CH4(g)) + H(2O2(g))] = [H(CO2(g)) + 2H(H2O(g))] - [H(CH4(g)) + 2H(O2(g))] [Note that H (kJ) is extensive, but H (kJ/mol) is intensive] 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) DH2 = -1606 kJ Watkins Chemistry 1422, Chapter 5 19 Enthalpies of Reaction Hess’s Law I For a chemical reaction at constant pressure: DHrxn = SH(products) - SH (reactants) When a reaction is reversed, the sign of DH changes: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -803 kJ CO2(g) + 2H2O(g) CH4(g) + 2O2(g) DH = +803 kJ Watkins Chemistry 1422, Chapter 5 20 Enthalpies of Reaction Hess’s Law I For a chemical reaction at constant pressure: DHrxn = SH(products) - SH (reactants) Enthalpy depends on state: H2O(l) H2O(g) DHrxn = +88 kJ/mol DHrxn = H[H2O(g)] - H[H2O(l)] > 0 H[H2O(g)] > H[H2O(l)] Watkins Chemistry 1422, Chapter 5 21 Enthalpies of Reaction Hess’s Law I For a chemical reaction at constant pressure: DHrxn = SH(products) - SH (reactants) Any reaction can be written as a sum of two or more other reactions; DH for the overall reaction is the sum of the DH's of the individual reactions. Watkins Chemistry 1422, Chapter 5 22 Enthalpies of Reaction Hess’s Law I DHrxn = SH(products) - SH (reactants) CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 2H2O(g) 2H2O(l) DH1 DH2 CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH3 [H(CO2(g)) + 2H(H2O(g))] – [H(CH4(g)) + 2H(O2(g))] = -803 kJ [2H(H2O(l))] – [2H(H2O(g))] = -176 kJ DH3 = DH1 + DH2 = -979 kJ To add these two reactions ... • Cancel common reactant-product pairs • Write remaining reactants and products • Add Enthalpies (watch signs!) Watkins Chemistry 1422, Chapter 5 23 Enthalpies of Reaction Hess’s Law II DHrxn = SH(products) - SH (reactants) Any reaction can be written as a sum of two or more other reactions; DH for the overall reaction is the sum of the DH's of the individual reactions. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 2H2O(g) 2H2O(l) DH1 DH2 CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH3 = DH1 + DH2 DH1 = [1H(CO2(g)) + 2H(H2O(g))] – [1H(CH4(g)) + 2H(O2(g))] DH2 = [2H(H2O(l))] – [2H(H2O(g))] DH3 = [1H(CO2(g)) + 2H(H2O(l))] – [1H(CH4(g)) + 2H(O2(g))] Is there a way to find H? Watkins Chemistry 1422, Chapter 5 24 Enthalpies of Reaction Hess’s Law II DHrxn = SH(products) - SH (reactants) We can measure qp = DHrxn. We cannot measure H(prod), H(react), or H(x), the molar heat content of any substance x. But we can measure something that is exactly equivalent to H(x). We can measure DHof (x), the Standard Heat of Formation of substance x. Watkins Chemistry 1422, Chapter 5 25 Formation Reactions Hess’s Law II DHrxn = SH(products) - SH (reactants) A formation reaction is a chemical reaction, at some temperature T, which forms substance x from its elements under the following conditions: • All reactants are elements • The product is one mole of substance x • All reactants and products are at temperature T and are in their standard states (s, l, g, aq) • Solids and liquids are pure, gases are at 1 atm, aqueous solutions are at 1 M. Watkins Chemistry 1422, Chapter 5 26 Formation Reactions Hess’s Law II An example of a formation reaction is: 2C(graphite) + 2H2(g) + O2(g) → HC2H3O2(l) At 25 oC, reactant element carbon is in the form of solid graphite, and elements hydrogen and oxygen are gases, each at 1 atm; the product, one mole of pure acetic acid, is a liquid. The heat of this reaction, qp = DHrxn, is called the “standard heat of formation” and is given the special symbol DH°f (kJ/mol). Note that this is a molar quantity just like H! Watkins Chemistry 1422, Chapter 5 27 Formation Reactions Hess’s Law II Examples (at 25 oC; listed in Appendix C) DH°f = - 1670 kJ/mol H2(g) + S(s) + 2O2(g) H2SO4(l) DH°f = - 814 kJ/mol H2(g) + 1/2O2(g) H2O(l) DH°f = - 286 kJ/mol 2H2(g) + O2(g) 2H2O(l) DH°rxn = - 572 kJ H2O(l) H2(g) + 1/2O2(g) DH°rxn = + 286 kJ C(s) C(diamond) DH°f = + 1.9 kJ/mol C(s) C(graphite) DH°f = 0 kJ/mol 2Al(s) + 3/2O2(g) Al2O3(s) DH°f = 0 for an element in its standard state. Watkins Chemistry 1422, Chapter 5 28 Hess’s Law II Formation Reactions Why are formation reactions important? Any chemical reaction can be written as a sum of reactions derived from formation reactions! CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) DHrxn = ? (from Appendix C) • CH4(g) → C(s) + 2H2(g) DH1 = -DH°f = + 75 kJ • 2O2(g) → 2O2(g) DH2 = -2DH°f = -0 kJ •C(s) + O2(g) → CO2(g) DH3 = DH°f = - 394 kJ •2H2(g) + O2(g) → 2H2O(g) DH4 = 2DH°f = - 484 kJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) DHrxn = DH1 + DH2 + DH3 + DH4 = -803 kJ Watkins Chemistry 1422, Chapter 5 29 Hess’s Law II Formation Reactions Why are formation reactions important? Any chemical reaction can be written as a sum of reactions derived from formation reactions! CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) DHrxn = [H(CO2) + 2H(H2O)] DHrxn = -803 kJ - [H(CH4) + 2H(O2)] DHrxn=[DHof(CO2)+2DHof(H2O)]- [DHof(CH4)+2DHof(O2)] H DHof Watkins Chemistry 1422, Chapter 5 30 Enthalpies of Formation Hess’s Law II Hess's Law with DH°f will calculate any DHrxn For any reaction: aA + bB + ... cC + dD + ... DHrxn = [cDHof(C) + dDHof(D)+...] - [aDHof(A) + bDHof(B)+...] DHrxn = SpDH°f (products) - SrDH°f (reactants) p = c, d, ... (stoichiometric coefficients of products) r = a, b, ... (stoichiometric coefficients of reactants) Watkins Chemistry 1422, Chapter 5 31 Enthalpies of Reaction Hess’s Law II 2 KOH(s) + CO2(g) → K2CO3(s) + H2O(g) DHfo kJ/mol -424.7 -393.5 -1150.2 -136.1 (all values from Appendix C) DHrxn = [1(-1150.2)+1(-136.1)]-[2(-424.7)+1(-393.5)] DHrxn = -43.4 kJ Pay close attention to signs!!! Pay close attention to states!!! H2O(g) and H2O(l) have different DHof Watkins Chemistry 1422, Chapter 5 32 Calorimetry Calorimetry = measurement of heat flow q. A calorimeter is an apparatus that measures heat flow q. The 1st rule of heat flow: heat always flows from a hot place (high temperature source) to a cold place (low temperature sink). If Tsource = Tsink, q = 0. Tsource usually decreases and Tsink usually increases (but not always) Watkins Chemistry 1422, Chapter 5 33 Calorimetry Calorimetry = measurement of heat flow q. A calorimeter is an apparatus that measures heat flow q. Heat capacity = the amount of heat required to change the temperature of an object by 1 centigrade degree (°C or K). The units of heat capacity are J/deg. The increase in T measures the increased kinetic energy of the molecules. Watkins Chemistry 1422, Chapter 5 34 Heat capacity (J/deg) The heat capacity of an object depends on how big the object is. Example: H2O(l ) Calorimetry (1 g) (1 mol) (1 L) (1 m3 = 1,000 L) 1 cm3 18 cm3 1,000 cm3 1,000,000 cm3 4.18 J/deg 75.3 J/deg 4.18 kJ/deg 4.18 MJ/deg Watkins Chemistry 1422, Chapter 5 35 Calorimetry Heat capacity (J/deg) The heat capacity of an object also depends on what the object is. 1 g H2O(l) 1 g Air(g) 1 g NaCl(s) 1 g S(s) 1 g U(s) 4.18 J/deg 1.01 J/deg 0.86 J/deg 0.71 J/deg 0.12 J/deg Non-metals generally have HIGH heat capacities. Metals generally have LOW heat capacities. Watkins Chemistry 1422, Chapter 5 36 Calorimetry Heat capacity (J•deg-1) = the amount of heat required to change the temperature of an object by 1 degree centigrade (°C or K). Insulator – an object in which the rate of heat flow is slow (in part due to a large heat capacity) Conductor – an object in which the rate of heat flow is fast (in part due to a small heat capacity) Watkins Chemistry 1422, Chapter 5 37 Calorimetry Molar heat capacity C (J•mol-1•deg-1) The heat required to change the temperature of 1 mole of a substance by 1 degree; if n moles change by DT degrees, then: q = n•C•DT Gram heat capacity (“specific heat”) s (J•g-1•deg-1) The heat required to change the temperature of 1 gram of a substance by 1 degree; if m grams change by DT degrees, then: q = m•s•DT Watkins Chemistry 1422, Chapter 5 38 Calorimetry q = n•C•DT q = m•s•DT DT = Tfinal - Tinitial To raise the temperature (DT > 0) of a system, add heat to it (q > 0, endothermic). To lower the temperature (DT < 0) of a system, subtract heat from it (q < 0, exothermic). When does transfering heat to/from a system not change its temperature? During a phase change, when the added heat energy breaks intermolecular bonds instead of speeding up the molecules. Watkins Chemistry 1422, Chapter 5 39 Calorimetry How much heat must be added to m grams of ice at -20 oC to convert it into m grams of steam at 120 oC? (m grams of steam) Heat added (kJ) 120 vaporization 100 80 60 ~ ~ oC 40 20 0 -20 fusion Heating Curve for H2O (m grams of ice) Watkins Chemistry 1422, Chapter 5 40 s1 = 2.09 J/g.deg DHfus = 6.01 kJ/mol s3 = 4.18 J/g.deg DHvap = 40.67 kJ/mol s5 = 1.84 J/g.deg (Water data: Appendix B) q1=ms1DT: heat added to raise 120 Heat added (kJ) H2O(s) from -20 to 0 oC vaporization Calorimetry 100 q2=(m/18)DHfus (only breaks intermolecular bonds) q3=ms3DT: heat added to raise H2O(l) from 0 to 100 oC 80 60 ~ ~ oC 40 q4=(m/18)DHvap (only breaks intermolecular bonds) q5=ms5DT: heat added to raise H2O(g) from 100 to 120 oC Watkins 20 fusion 0 -20 q1 q2 Chemistry 1422, Chapter 5 q3 q5 q4 41 Calorimetry How much heat must be added to 18.02 grams of ice at -20 oC to convert it into 18.02 grams of steam at 120 oC? H2O(s, -20oC) → H2O(s, 0oC) q1 = DH1 = +0.75 kJ H2O(s, 0oC) → H2O(l, 0oC) q2 = DH2 = +6.01 kJ H2O(l, 0oC) → H2O(l, 100oC) q3 = DH3 = +7.53 kJ H2O(l, 100oC) → H2O(g, 100oC) q4 = DH4 = +40.67 kJ H2O(g, 100oC) → H2O(g, 120oC) q5 = DH5 = +0.66 kJ H2O(s, -20oC) → H2O(g, 120oC) DH = +55.63 kJ Hess’s Law: Any reaction can be written as the sum of two or more other reactions. Watkins Chemistry 1422, Chapter 5 42 Constant Pressure Calorimetry Enthalpy of a Solution Reaction Atmospheric pressure is constant for an open container! • Find specific heat of the solution ssoln • Weigh solution msoln • Measure DTrxn during reaction • qp = msoln× ssoln× DTrxn = DHrxn Watkins Chemistry 1422, Chapter 5 43 Calorimetry In a constant pressure calorimeter, 25 g of ice at -10 oC is added to 100 g of water at 25 oC. After thermal equilibrium is reached, what is the final temperature in the calorimeter? s(s) = 2.09 J/g.deg s(l) = 4.18 J/g.deg DHfus = 6.01 kJ/mol H2O(25g, s,-10) → H2O(25g, s,0) q1 = (25)(2.09)(10) J H2O(25g,s,0) → H2O(25g,l,0) q2 = (25/18)(6010) J (DT = 0) H2O(25g,l,0) → H2O(25g,l,T) q3 = (25)(4.18)(T – 0) J H2O(100g,l,25) → H2O(100g,l,T) q4 = (100)(4.18)(T – 25) J q1 + q2 + q3 + q4 = 0 T=? Watkins Chemistry 1422, Chapter 5 44 Constant Volume Calorimetry Bomb Calorimeter (DV = 0) • Used for combustion reactions. • Measure the heat capacity of the calorimeter (K) with a known reaction. • Measure qV = K•DT = DE for the unknown reaction. • Convert DE(comb) to DH(comb) (not in this class) DHof is derived from DH(comb) Watkins Chemistry 1422, Chapter 5 45