1. Changes of Energy - 1 Change of energy depends only on initial and final states –In general: –Constant volume: Pressure-volume work only No electrical work, etc. –From definition of heat capacity:

2. Energy vs. Enthalpy Define a new state function, enthalpy: –Why? Think about doing an experiment in which you measure heat capacity It is convenient to do this at constant pressure (bench top experiment) You add certain amount of heat, q p, and measure  T For solids and liquids, your measurement of C p is similar to what you measure when the same experiment is carried out at constant volume (C v ) For solids and liquids,  H =  E and C p = C v HOWEVER, when working with a gas, you get different results. You add some heat, q p, raise the temperature,  T But the gas expands,  V = nR  T/P The system did work on the surroundings: w = -P  V Temperature does not go up as much as we expected, lost energy through work We just add this lost work to the energy to get the enthalpy

3. Energy vs. Enthalpy –At constant pressure –At constant volume –What is the difference

4. Measuring Enthalpy Remember  H   E + PV Since both  E and PV are state properties,  H must be a state property. It is therefore an exact differential! Thus basic calculus applies: dH= dE + d(PV)= dE + PdV + V dP Since for reversible pressure-volume systems, dE= dq + dw= dq - PdV dH= dq - VdP So for constant volume pressure processes:  H= q P

5. Another Example This summation property allows us to use information seemingly only tangentially related to our reaction of interest to get to an end point. We want to know: 3 O 2 (g,1atm) + 2 glycine (s)  1 urea (s) + 3 CO 2 (g, 1atm) + 3 H 2 O (l) But we only have: 3 O 2 (g,1atm) + 2 glycine (s)  4 CO 2 (g,1atm) + 2 H 2 O (l) + 2 NH 3 (g)  H= -1163.5 kJ/mol and H 2 O (l) + urea (s)  CO 2 (g) + 2 NH 3 (g)  H= 133 kJ/mol

6. Substracting these last two equations gives the desired result: 3 O 2 (g,1atm) + 2 glycine (s)  4 CO 2 (g,1atm) + 2 H 2 O (l) + 2 NH 3 (g) - H 2 O (l) + urea (s)  CO 2 (g) + 2 NH 3 (g) = 3 O 2 (g,1atm) + 2 glycine (s)  1 urea (s) + 3 CO 2 (g, 1atm) + 3 H 2 O (l)  H= -1163.5 kJ/mol - 133 kJ/mol = -1296.8 kJ/mol mol= mole of reaction! Note: To get a biologically reasonable reaction: glycine(s) +  H 2 O (l)  glycine (aq)

7. Standard Enthalpies We are only ever concerned with  H But we use H to calculate, what is the value of H –“Actual energy” from E = mc 2 Impractical and unnecessary –Arbitrarily agree on a zero –The convention is to assign zero enthalpy to all elements in their most stable states at 1atm pressure. –The standard enthalpy of a compound is defined to be the enthalpy of formation of 1 mol of compund at 1atm pressure from its elements in their standard states. –Although temperature is not specified in the defenition of standard enthalpy, usually use 25 °C = 298 K

8. Standard Enthalpies Enthalpy of reactants and products can be calculated from their heats of formation. 2 C(s) + 3 H 2 (g) + 1/2 O 2 (g)  C 2 H 5 OH (l)  H° 298 =0 is the most stable state at standard T and P. From the table:  H° 298,f = -276.98 kJ/mol But this is not a clean reaction in practice. How do we measure it?

9. Clean Reactions We sum up the enthalpies of “clean” reactions. C(graphite) + O 2 (g)  CO 2 H 2 (g) + 1/2O 2 (g)  H 2 O 2 CO 2 (g)+ 3H 2 O(l)  C 2 H 5 OH(l) + 3O 2 (g)

10. Bond Energies BondD(kJ/mol) C-C 344 C=C 615 C  C 812 C-H 415 C-N 292 C-O 350 C=0 725 C-S 259 N-H 391 O-O 143 O-H 463 S-H 339 H 2 436 N 2 945.4 O 2 498.3 C(graphite) 716.7 So we can calculate the heats of formation for many compounds and the reaction enthalpies for many transformations. From these we can “back-out” average values for how much energy it takes to break a bond. H | | H-C-C-O-H | | H H

11. Errors with Bond Energies OH | C(graphite) + H 2 (g) + O 2 (g)  H-C=O C(graphite)  C(g)716.7 kJ/mol H 2 (g)  2H(g)436.0 kJ/mol O 2 (g)  2O(g)498.3 kJ/mol C(g)+ 2H(g) + 2O(g)1651 kJ/mol C=O + C-O + C-H + O-H1953 kJ/mol Calculated  H f -302 kJ/mol Measured  H f -423.76 kJ/mol

12. Complex Systems Surroundings 100W ~ 9000 kJ/day 1 g protein/carbohydrate= 15kJ= 3.6 Cal 1g fat= 35kJ= 8.4 Cal System

13. Carnot Cycle State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal q=0, w=C V  T q=0, w=-C V  T w= -nRT hot ln(V2/V1) q=-w w= -nRT cold ln(V4/V3) q=-w

14. Carnot Cycle, Schematic View T hot T cold E The engine operates between two reservoirs to and from which heat can be transferred. We put heat into the system from the hot reservoir and heat is expelled into the cold reservoir.

15. Questions about Thermodynamic Cycles How much of the heat put in at high temperature can be converted to work? Can two engines with the same temperature difference drive one another? What does entropy have to do with it? Clausius b. Jan. 2, 1822, Prussia d. Aug. 24, 1888, Bonn "Heat cannot of itself pass from a colder to a hotter body."

16. Carnot Cycle: Step 1 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Isothermal reversible Expansion:  E=0Energy of an ideal gas depends only on temperature q= -w Heat converted to work! T hot

17. Carnot Cycle: Step 2 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Isothermal reversible Expansion: q=0No heat transferred in adiabatic process Energy lost to expansion

18. Carnot Cycle: Steps 3 & 4 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Isothermal reversible Expansion: q=0No heat transferred in adiabatic process Energy lost to expansion

19. Carnot Cycle: Summary Step1Step2Step3Step4 w-nRT hot ln(V 2 /V 1 )-C V  T-nRT cold ln(V 4 /V 3 ) C V  T q-w0-w0  E0w0w w 1 + w 2 + w 3 + w 4 = -nRT hot ln(V 2 /V 1 ) - nRT cold ln(V 4 /V 3 ) q 1 + q 2 + q 3 + q 4 = nRT hot ln(V 2 /V 1 ) + nRT cold ln(V 4 /V 3 ) q = -w

20. Carnot Cycle: PV Diagram Volume Pressure 1 2 3 4 -nRT hot ln(V 2 /V 1 ) -C V  T -nRT cold ln(V 4 /V 3 ) C V  T