22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)1 Thermochemistry - Energy of Chemical Reactions Contents Contents: heat, work, forms of energy specific heat and energies of phase changes enthalpy changes in chemical reactions standard enthalpies of formation Hess’s law estimating enthalpies of reaction from Bond Energies

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)2 CHEMICAL ENERGY Chemical bonds are a source of energy BOND BREAKING - requires energy BOND MAKING - releases energy In a chemical reaction : if more energy is released in forming bonds than is used in breaking bonds then EXOTHERMIC... reaction is EXOTHERMIC if more energy is used in breaking bonds than is released in forming bonds then... reaction is ENDOTHERMIC Energy is released as HEAT, LIGHT, SOUND, WORK Energy can be provided by - LIGHT - photochemistry - WORK - electrochemistry - COOLING of surroundings

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)3 Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of a difference in temperature. WORK is the form of energy that results in a macroscopic displacement of matter such as gas expansion or motion of an object (force x distance) Other forms of energy — lightlight electricalelectrical kinetickinetic Chemical Chemical gravitational potential electrostatic potential electrostatic potential

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)4 Specific Heat Capacity Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer. The heat “lost” or “gained” is related to a)sample mass b) change in T, and c) specific heat capacity by Specific heat capacity= heat lost or gained by substance (J) (mass, g) (T change, K)

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)5 Specific Heat Capacity SubstanceSpec. Heat (J/gK) H 2 O4.184 Al0.902 glass0.84 Aluminum Water

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)6 Specific Heat Capacity - an example If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al? where  T = T final - T initial = = -273 K q = (0.902 J/gK)(25.0 g)(-273 K) q = J negative sign of q  heat is “lost by” or transferred from Al J/g.K =  heat gain/lost = q = (specific heat)(mass)(  T)

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)7 Heat Transfer and Changes of State Changes of state involve energy Ice  Water 333 J/g Heat of Fusion (Heat of Fusion) Water  Vapor 2260 J/g (Heat of vaporization)

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)8 Heating/Cooling Curve for Water Heat water Evaporate water Melt ice

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)9 CHEMICAL REACTIVITY What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. In Ch. 4 we saw a number of “driving forces” for reactions that are PRODUCT-FAVORED. formation of a precipitate gas formation H 2 O formation (acid-base reaction)

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)10 Energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are “product-favored”. How do we describe heat transfer in chemical processes ? CHEMICAL REACTIVITY

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)11 Heat Energy Transfer in Physical & Chemical Processes CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C)CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C) Heat flows into the SYSTEM (solid CO 2 ) from the SURROUNDINGS in an ENDOTHERMIC process. heat Surroundings System

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)12 ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 samples because of a difference in temperature. WORK is the form of energy that results in a macroscopic displacement of matter such as gas expansion or motion of an object (force x distance) the same amount In CO 2 sublimation & expansion, the same amount of flows from surroundingssystem of ENERGY flows from surroundings to system If expanding gas is enclosed, part of the energy transfer appears in the form of WORK OF EXPANSION w exp = - P  V(for an ideal gas) If expanding gas is not enclosed, the energy transfer appears HEAT only as HEAT (CO 2 gas gets warm).

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)13 FIRST LAW OF THERMODYNAMICS q = E - w q =  E - w heat energy transferred Energy change work done by the system Energy is conserved! E = q + w OR  E = q + w NB - q and w positive when they are transferred FROM surroundings TO system Surroundings Heat q sys > 0 System Work w sys > 0

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)14 ENTHALPYENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P is called q p with q p =  H =  E - w =  E + P  V =  E+PV) where H = enthalpy H is defined as  E + PV)  H = heat transferred at constant P  H = change in heat content of the system  H = H final - H initial

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)15  H = H final - H initial If H final > H initial then  H is positive Process is ENDOTHERMIC If H final > H initial then  H is positive Process is ENDOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMIC ENTHALPYENTHALPY

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)16 Endo- and Exothermic Surroundings Heat q sys > 0 System ENDOTHERMICENDOTHERMIC Heat q sys < 0 q sys < 0 Surroundings System EXOTHERMIC

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)17 But the reverse reaction, the decomposition of water : H 2 O(g) kJ ---> H 2 (g) + 1/2 O 2 (g) Endothermic reaction — heat is a “reactant”,  H = +242 kJ. This does not occur spontaneously. Consider the combustion of H 2 to form water.. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g)  242 kJ Exothermic reaction — heat is a “product”.  H = -242 kJ. This is spontaneous and proceeds readily once initiated. USING ENTHALPY BUT... Decomposition of water can be made to occur by coupling to another, spontaneous process...

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)18 How can we make H 2 gas ? N. Lewis, American Scientist, Nov. 1995, page 534. Nov. 1995, page 534.

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)19 liquid Making H 2 from liquid H 2 O involves two steps. H 2 O(liq) + 44 kJ  H 2 O(g) H 2 O(g) kJ  H 2 (g) + 1/2 O 2 (g) H 2 O(liq) kJ  H 2 (g) + 1/2 O 2 (g) This is an example of HESS’S LAW — If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns. If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns.

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)20 Calc.  H rxn for S(s) + 3/2 O 2 (g) --> SO 3 (g) knowing that S(s) + O 2 (g) --> SO 2 (g)  H 1 = kJ SO 2 (g) + 1/2 O 2 (g) --> SO 3 (g)  H 2 = kJ Hess’s Law - a second example : The two rxns. add to give the desired rxn., S(s) + 3/2 O 2 (g) --> SO 3 (g) so  H rxn =  H 1 +  H 2 = kJ  H 3 = kJ

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)21 energy 2 S solid SO 3 gas direct path + 3/2 O  H 3 = kJ SO 2 gas +O 2 H1H1 = kJ + 1/2 O 2 H2H2 = kJ  H 3 =  H (2+3) = =   H along one path =   H along another path  H 3 =  H (2+3) = =   H along one path =   H along another path

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)22 This equation is valid because  H is a STATE FUNCTIONThis equation is valid because  H is a STATE FUNCTION These depend only on the state of the system and not how it got there.These depend only on the state of the system and not how it got there. Other state functions include:Other state functions include: V, T, P, energy..   H along one path =   H along another path   H along another path   H along one path =   H along another path   H along another path — and your bank account! Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot measure absolute H. Can only measure  H. measure absolute H. Can only measure  H.

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)23 Standard Enthalpy Values Most  H values are labeled  H o P = 1 atmosphere ( = 760 torr = kPa) Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas o means measured under standard conditions

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)24 - the enthalpy change when 1 mol of compound is formed from elements under standard conditions. Values: Kotz, Table 6.2 and Appendix K By definition,  H o f = 0 for elements in their standard states.  H o f = standard molar enthalpy of formation H 2 (g) + 1/2 O 2 (g) --> H 2 O(g)  H o f = kJ/mol H 2 (g) + 1/2 O 2 (g) --> H 2 O(g)  H o f = kJ/mol

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)25 Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known,  H o rxn =   H o f (products) -  H o f (reactants) Calculate  H of reaction?

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)26  H o rxn =   H o f (prod) -   H o f (react) Example: Calculate the heat of combustion of ethanol, i.e.,  H o rxn for C 2 H 5 OH(g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O(g)  H o rxn = { 2  H o f (CO 2 ) + 3  H o f (H 2 O) } - {7/2  H o f (O 2 ) +  H o f (C 2 H 5 OH)} = { 2 ( kJ) + 3 ( kJ) } - {7/2 (0 kJ) + ( kJ)}  H o rxn = kJ per mol of ethanol

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)27 Given by D - the bond dissociation energy D = energy required to break a bond in a gas phase molecule under standard conditions e.g. CH 4 (g)  C (g) + 4 H (g)  H rxn = kJ = 4 * D(C-H) D(C-H) = 416 kJ per mole of C-H bonds D (C-H) (kJ/mol) varies slightly among compounds : CH 4 416C 2 H 6 392C 3 H C 2 H 4 432C 2 H 2 445C 6 H Bond Energies (Kotz, sect. 9.4, pp ) D can be derived from  H rxn for atomization...

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)28 The GREATER the number of bonds (bond order) the HIGHER the bond dissociation energy The GREATER the number of bonds (bond order) the HIGHER the bond dissociation energy BOND D (kJ/mol) (Bond Energy) H—H436 C—C347 C=C611 C  C 837 N—N159 N  N946 see table 9.5 for Dissociation Energies of other bonds. D is similar for same bond in different molecules Average values over many compounds are tabulated Bond energy depends on bond order Bond Energies

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)29 Using Bond Energies Estimate the energy of the reaction H—H + Cl—Cl ----> 2 H—Cl Net energy =  H rxn = energy required to break bonds - energy evolved when bonds are made

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)30 H—H = 436 kJ/mol H—H = 436 kJ/mol Cl—Cl = 243 kJ/mol H—Cl = 431 kJ/mol H—Cl = 431 kJ/mol H—H = 436 kJ/mol H—H = 436 kJ/mol Cl—Cl = 243 kJ/mol H—Cl = 431 kJ/mol H—Cl = 431 kJ/mol Sum of H-H + Cl-Cl bond energies = 436 kJ kJ = +679 kJ 2 mol H-Cl bond energies = 862 kJ Net =  H = +679 kJ kJ = -183 kJ, ? ? ? THEREFORE,  H f for H-Cl is ? ? ? Estimating  H rxn for H—H + Cl—Cl   2 H—Cl

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)31 Is the reaction exo- or endothermic? Energy for bond breaking: 4 mol O—H bonds = 4 (464 kJ) 2 mol O—O bonds = 2 (138 kJ) TOTAL = 2132 kJ Energy from bond making : 1 mol O=O bonds = 498 kJ 4 mol O—H bonds = 4 (464 kJ) TOTAL = 2354 kJ EXAMPLE 2: Estimate the energy of the reaction 2 H—O—O—H ----> O=O + 2 H—O—H Which is larger: energy req’d to break bonds... or energy evolved on making bonds?

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)32 2 H—O—O—H ----> O=O + 2 H—O—H More energy is evolved on making bonds than is expended in breaking bonds. The reaction is exothermic! Net energy = kJ kJ = kJ

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)33 Enthalpies of Reaction from Bond Energies REACTANTS Gaseous Atoms PRODUCTS ENDOTHERMIC Bond Breaking costs more than is gained by Bond Making Bond Making releases more E than required for Bond Breaking PRODUCTS REACTANTS EXOTHERMIC Gaseous Atoms

22 September, 1997Chem 1A03E/1E03E THERMOCHEMISTRY (Ch. 6)34 Key Concepts from Chapter 6: Thermochemistry heat transfer - specific heat phase transitions - heats of fusion, vaporization, etc First law of thermodynamics  E = q - w endothermic versus exothermic reactions enthalpy change in chemical reactions Hess’s law standard molar enthalpies of formation  H rxn =   H f (products) -   H f (reactants) bond energies  H rxn =  D(bonds broken) -  D(bonds made)