Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

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 6.1

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. Temperature = Thermal Energy 400C 900C greater thermal energy 6.2

Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. SURROUNDINGS SYSTEM open closed isolated Exchange: mass & energy energy nothing 6.2

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) 6.2

DH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. 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 6.3

Thermochemical Equations 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. H2O (s) H2O (l) DH = 6.01 kJ 6.3

Thermochemical Equations 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. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ 6.3

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 6.3

Thermochemical Equations 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) DH = -3013 kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.3

Heat (q) absorbed or released: The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. 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. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.4

Dt = tfinal – tinitial = 50C – 940C = -890C How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J 6.4

Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves! 6.4

Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.4

6.4

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f 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 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 6.5

6.5

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.) 6.5

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 6.5

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–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C6H6 6.5

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? Which substance(s) could be used for a cold pack? 6.6

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.6

Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

Thermodynamics DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDV when a gas expands against a constant external pressure 6.7

Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure, q = DH and w = -PDV DE = DH - PDV DH = DE + PDV 6.7

Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — light electrical kinetic and potential

Energy & Chemistry Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

Energy & Chemistry These reactions are PRODUCT FAVORED They proceed almost completely from reactants to products, perhaps with some outside assistance.

Energy & Chemistry 2 H2(g) + O2(g) --> 2 H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 ---> 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O ---> 4 OH- CCR, page 845

Potential & Kinetic Energy Potential energy — energy a motionless body has by virtue of its position.

Potential Energy on the Atomic Scale Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy NaCl — composed of Na+ and Cl- ions.

Potential Energy on the Atomic Scale Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy

Potential & Kinetic Energy Kinetic energy — energy of motion • Translation

Potential & Kinetic Energy Kinetic energy — energy of motion.

Internal Energy (E) PE + KE = Internal energy (E or U) Int. E of a chemical system depends on number of particles type of particles temperature

Internal Energy (E) PE + KE = Internal energy (E or U)

Internal Energy (E) The higher the T the higher the internal energy So, use changes in T (∆T) to monitor changes in E (∆E).

Heat transfers until thermal equilibrium is established. Thermodynamics Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions. Heat transfers until thermal equilibrium is established.

Directionality of Heat Transfer Heat always transfer from hotter object to cooler one. EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down T(surr) goes up

Directionality of Heat Transfer Heat always transfer from hotter object to cooler one. ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T (surr) goes down

Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE.

Energy Change in Chemical Processes PE of system dropped. KE increased. Therefore, you often feel a T increase.

UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = 4.184 joules James Joule 1818-1889

HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?

Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity

Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = Tfinal - Tinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

Heat Transfer No Change in State q transferred = (sp. ht.)(mass)(∆T)

Heat Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass)

Heat Transfer and Changes of State Liquid ---> Vapor Requires energy (heat). This is the reason a) you cool down after swimming you use water to put out a fire. + energy

Heating/Cooling Curve for Water Evaporate water Heat water Note that T is constant as ice melts Melt ice

Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g +333 J/g +2260 J/g

Heat & Changes of State How much heat is required to melt 500. g of ice and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total heat energy = 1.51 x 106 J = 1510 kJ

Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery

Chemical Reactivity But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are product-favored. So, let us consider heat transfer in chemical processes.

Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Heat transfers from surroundings to system in endothermic process.

Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) A regular array of molecules in a solid -----> gas phase molecules. Gas molecules have higher kinetic energy.

∆E = E(final) - E(initial) Energy Level Diagram for Heat Energy Transfer CO2 gas ∆E = E(final) - E(initial) = E(gas) - E(solid) CO2 solid

Heat Energy Transfer in Physical Change CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Two things have happened! Gas molecules have higher kinetic energy. Also, WORK is done by the system in pushing aside the atmosphere.

FIRST LAW OF THERMODYNAMICS heat energy transferred ∆E = q + w work done by the system energy change Energy is conserved!

∆E = q + w SYSTEM heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM ∆E = q + w w transfer in (+w) w transfer out (-w)

ENTHALPY ∆H = Hfinal - Hinitial Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal - Hinitial

ENTHALPY ∆H = Hfinal - Hinitial Process is ENDOTHERMIC If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC

USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

USING ENTHALPY Making liquid H2O from H2 + O2 involves two exothermic steps. H2 + O2 gas H2O vapor Liquid H2O

USING ENTHALPY Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ H2O(g) ---> H2O(liq) + 44 kJ ----------------------------------------------------------------------- H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

∆H along one path = ∆H along another path This equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not on how the system got there. V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

Standard Enthalpy Values Most ∆H values are labeled ∆Ho Measured under standard conditions P = 1 bar = 105 Pa = 1 atm /1.01325 Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas

Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O2(g) --> H2O(liquid) ∆H˚ = -286 kJ

Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆Hfo = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2

∆Hfo, standard molar enthalpy of formation Enthalpy change when 1 mol of compound is formed from the corresponding elements under standard conditions H2(g) + 1/2 O2(g) --> H2O(g) ∆Hfo (H2O, g)= -241.8 kJ/mol By definition, ∆Hfo = 0 for elements in their standard states.

Using Standard Enthalpy Values Use ∆H˚’s to calculate enthalpy change for H2O(g) + C(graphite) --> H2(g) + CO(g) (product is called “water gas”)

Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find H2(g) + 1/2 O2(g) --> H2O(g) ∆Hf˚ = - 242 kJ/mol C(s) + 1/2 O2(g) --> CO(g) ∆Hf˚ = - 111 kJ/mol

Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ C(s) + 1/2 O2(g) --> CO(g) ∆Ho = -111 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) ∆Honet = +131 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic.

Using Standard Enthalpy Values Calculate ∆H of reaction? In general, when ALL enthalpies of formation are known: ∆Horxn =  ∆Hfo (products) -  ∆Hfo (reactants) Remember that ∆ always = final – initial

Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Horxn for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn =  ∆Hfo (prod) -  ∆Hfo (react)

Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn =  ∆Hfo (prod) -  ∆Hfo (react) ∆Horxn = ∆Hfo (CO2) + 2 ∆Hfo (H2O) - {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} ∆Horxn = -675.6 kJ per mol of methanol

Measuring Heats of Reaction CALORIMETRY Measuring Heats of Reaction Constant Volume “Bomb” Calorimeter Burn combustible sample. Measure heat evolved in a reaction. Derive ∆E for reaction.

Calorimetry Total heat evolved = qtotal = qwater + qbomb Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb

Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane Temp rises from 25.00 to 33.20 oC Calorimeter contains 1200 g water Heat capacity of bomb = 837 J/K

Measuring Heats of Reaction CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ