Zumdahl’s Chapter 6

1. Zumdahl’s Chapter 6 Thermochemistry

2. Energy, E Chemical Energy Matter-Energy Chemist’s Enthalpy Enthalpy, H Calorimetry,CV=dE/dT Hess’s Law State Functions Standard Enthalpies of Formation, Hf Elements Compounds Ions in Solution Energy Sources Efficiencies Environmental concern Chapter Contents

3. KINETIC ENERGY POTENTIAL ENERGY —Fixed Total Energy Chemical Energy • Exothermicity • Release of energy (usually as heat) from chemical reactions whose products are at lower potential energy than their reactants. • If total energy is conserved, and it comes as kinetic plus potential energy (E=KE+PE), then the lowered product potential mean raised product kinetic energies, and heat flows out. • Chemists use enthalpy when discussing thermicity.

4. KINETIC ENERGY POTENTIAL ENERGY —Fixed Total Energy Consumption of Energy • Endothermicity • If a reaction’s products are higher in potential energy than its reactants, its kinetic energy must be lower after the reaction. • Heat flows in to equalize temperatures. • Again chemists use enthalpy instead of energy. • Both kinds of chemical energy presume the conservation of total energy.

5. Conservation of Energy • “Energy can be converted from one form to another but can be neither created nor destroyed.” • Empirical observations verify, e.g.: • Count von Rumford, boring out cannon, observed the relationship between the mechanical energy of the drill and the (frictional) heat of the cannon. • True even of atomic energy if mass isE.

6. E = mc2 ? • ActuallyE = [ m02c4 + p2c2 ]½ (Einstein) • Valid for both matter and light. • Light (photon): • m0“rest mass” is strictly zero for light. • So E = pc (?) but light has momentum, p = h / • deBroglie was also correct for both light and matter. • So E = h c /  = h … for photons • Just as we always knew it was!

7. Matter’s Energy Expression • E = { m02c4 + p2c2 }½ • but p = mv for matter. • E = { m02c4 + m2v2c2 }½ • but m = m0/ [ 1 – (v / c)2 ]½(Lorentz) • which explains why you can’t go even as fast as c. • E = { m02c4 + m02v2c2/ [ 1 – (v / c)2]½ }½ • E = m0c2/[ 1 – (v / c)2]½at all v … even v <<c • Indeed, given Lorentz, E mc2 for matter at all v!

8. PE At Garden Variety Velocities • E = m0c2/[ 1 – (v / c)2]½when v<<c or v/c ~  • [ 1 – 2 ]–½ ~ 1 + ½2 + ignorable terms • So E = m0c2 [ 1 + ½(v2 / c2) ] or • E = m0c2 + ½m0v2for matter at low velocity. • E = rest mass energy + kinetic energy • What happened to potential energy?

9. Conservation of Mass, eh? • Potential energy is an algebraic shorthand for changes in mass that occur with the juxtapositions of matter and fields. Use it. • PE = c2m0 implies that atomic weights vary in compounds! Are we worried? • Formation of 1 mole of water yields 286,000 J • m0 = 2.86105 J / (3108 m/s)2 = 3.210–12 kg • With 1.810–2 kg / mol, we won’t miss m0 !

10. E Surroundings Pedestrian Energy System +w –q • Far from Einsteinian Esoterica, we can observe energy changes in systems due to two gross causes: • Heat, q, flowing into the system raises its E. • Work, w, done on the system raises its E. • Together, these macroscopic components imply • E = q + wand when we isolate a system from surroundings, then E = 0.

11. HEAT Thermal energy flow. q = C T By Kinetic Theory, T is proportional to kinetic energy (& q). By Quantum Theory, heat associates with changes in energy level populations. WORK Organized rather than chaotic molecular motion. Comes in many forms. By Quantum Theory, work associates with changes in the energy levels themselves. w q Internal Energy’s Components

12. Work Work Work Work • Electrical work drags a charge Q through an electrical potential difference V, so +Q V • Surface work stretches surface tension  over larger areas, so work is +  A • Newtonian work pushes an object a distance x against a force F, so its work is +F x • Pressure volume work compresses a volume with a pressure P, so it is –PV

13. Pressure Volume Work • Inescapable when doing chemistry under the relentless atmospheric pressure ifVgas changes during a reaction. (gas V  n) • For constant Pext , w = –PextdV = – PextV • But if Pext is always the system’s Pinternal, • w = –PdV = –nRT V–1dV = nRT ln(V1/V2) • which uses the Ideal Gas Law to track Pext • and assumes constant T and is called “reversible work”

14. Calorimetry • We can use E = C T to infer E from observed T if C, heat capacity, is known. • Conceptually, C measures the system’s number of energy modes that can hide thermal energy. • Since E changes with V too, we must fix V in order to measure E by heat capacity, so • E = CV T =qV • and we should work in fixed V “bombs.”

15. At constant P Enthalpy, H, a chemist’s energy • Since PV work is inevitable in reactions open to the omnipresent atmosphere, we’ll be doing a lot of PV calculations … unless • Define H E + PV then H = E + (PV) • But E = qVbomb = qP– PV • So H = qP– PV + PV + VP = qP • Now we get out of the bomb and onto the bench!

16. Practical E vs. H Relation • Chemical energies, E, are best measured in bomb calorimeters, but enthalpies, H, are most conveniently used. So relate them: • H = E + (PV) = qVbomb + RT(ngas) • Which makes the (quite defensible) assumption that all gases are ideal (enough). • Since their non-idealities can be determined, we can spruce this up at will.

17. Temperature Dependence of H • Just as E = CVT (for constant CV) … • H = CPT (for constant CP) • So we can extrapolate H at T other than at 25°C (from standard tables) if we know CP. • But neither C is truly independent of T, so • H =  CP(T) dT and it’s so common … • We find tables of CP~a + bT + c / T 2 • Molar C (J/mol K) vs.“Specific Heat”C (J/g K)

18. Hess’s Law • “State functions” are thermodynamic variables (like energy or enthalpy) that have the same value when you return the system to the same state (same P,V,T,n). • Hess: “Enthalpy changes between reactants and products are not dependent upon how the reaction is brought about.” • Otherwise return to reactants wouldn’t undo H!

19. Hess’s Joyful Consequences • C2H6 + 3.5 O2 2 CO2 + 3 H2O H1 • Then Hess’s Law guarantees that • 2 CO2 + 3 H2O  C2H6 + 3.5 O2 has –H1 • Even though the latter’s not a feasible reaction. • Since H is extensive (scales with # of moles), • 4 CO2 + 6 H2O  2 C2H6 + 7 O2 is –2H1 • Which permits us to do algebra with reactions.

20. Chemical Algebra • Suppose calorimetry gave us the following: • C2H6 + 3.5 O2 2 CO2 + 3 H2O H1 • C2H4 + 3 O2 2 CO2 + 2 H2O H2 • H2 + ½ O2 H2O H3 • If we reverse the first reaction and add, • C2H4 + H2  C2H6 results and has a H • Where H = H2 + H3– H1 • We find ethene’s hydrogenation enthalpy without having to hydrogenate ethene! Just torch it.

21. Chemical Reference Points • We just used the fully oxidized forms of the compounds to do thermochemistry on a reaction that didn’t even involve oxygen! • And we did it because calorimetry is easy. • But Hess lets us use any consistent reference point, and standard elemental states are best. • We turn calorimetric results into Standard Enthalpies of Formation from Elements.

22. Std. Enthalpy of Formation, Hf • Elements will be as their most stable allotrope at 1 bar and 25°C, symbolized ⊖. • These are standard conditions for thermo. • Hf of all stable allotropes is ZERO. • Each element is its own reference point. • This works because reactions don’t destroy atoms. • H2 + ½O2 H2O(l) Hf = –286 kJ/mol • The Standard Enthalpy of Formation of liquid water.

23. Compounds in Solution • While standard enthalpies apply to pure solids and gases at 1 bar (25°C), solution concentrations will scale their enthalpies. • So choose 1 M as a standard concentration (activity, to be precise) for a solute. • Ions present a special case (because of counterions): • So only [H+(aq)] = 1M is defined to have 0 enthalpy of formation, and all other ions measure against it.

24. Reaction Enthalpy from Hf • Hrxn =  nproduct Hf–  nreactant Hf • where n are stoichiometric coefficients from the reaction and the sums are over all products and all reactants. (Not bothering with the 0 elements.) • Reactants are subtracted because they are being destroyed, notformed, by the reaction! • If some Tother than 25°C is needed, use CP obtained just like Hrxn above, a difference of stoichiometrically-scaled molar heat capacities.

25. Two Examples • 2 Ag(s) + S(s)  Ag2S(s) • Hrxn = Hf(Ag2S) – 2 Hf(Ag) – Hf(S) • Hrxn = (–32 kJ/mol) – 2(0) – 0 = – 32 kJ/mol • but • 2 Ag+(aq) + S2–(aq)  Ag2S(s) • Hrxn = Hf(Ag2S) – 2 Hf(Ag+) – Hf(S2–) • Hrxn = –32 – 2(+105) – (+33) = –275 kJ/mol • Why so different? (Think “lattice energies.”)

26. Plundering the Legacy • H2SO4 is our highest volume chemical commodity unless you consider petroleum. • While we do use petroleum in syntheses (feedstock of plastics), mostly we torch it. • For the Joules. • Fossil ferns are the cheapest source of mobile energy; the costs don’t take into consideration the socioeconomic disaster of Global Warming.

27. Solar Energy Conversions • We don’t photosynthesize. • We can’t yet match the efficiency of things that do. They reverse carbohydrate combustion. • But when we can split water into its elements photolytically, we’ll have solved The Energy Problem. H2O is as mobile as gasoline. • And resultant H2 will be our mobile fuel. • It won’t even upset Earth’s solar E budget!

28. Energy Efficiencies • What’s a tank of water worth as gasoline? • For automobile fuel, we consider weight. • Each kg of water holds 55.5 moles of H2. • Our solar cars extract it while sitting in parking lots. • That’s worth 55.5(286 kJ) ~ 16 MJ • But a kg of isooctane has 8.77moles of C8H18. • That’s worth 8.77(5461 kJ) ~ 48 MJ • A recycling water car needs 3 the fuel wt. • But only 1/3 the fuel weight when the O2 is ejected!