1. Copyright © 2014 R. R. Dickerson 1 CHEMICAL THERMODYNAMICS A. ENTHALPY OF FORMATION AND COMBUSTION (In search of the Criterion of Feasibility) 1. First Law of Thermodynamics (Joule 1843 - 48) dU = đQ – đW or đQ = dU + đW The energy of a system is equal to the sum of the heat and the work done on the system. Note dU = dE in some texts. Remember Eq. of state and exact differentials 1. Define Enthalpy (H) dH = dU + d(PV) dH = đQ - đW + PdV + VdP

2. Copyright © 2009 R. R. Dickerson & Z.Q. Li 2 For example, consider the burning of graphitic carbon in oxygen, the change in enthalpy is:  H o C graph + O 2 → CO 2 -94.0 kcal/mole In the combustion of 1 mole (12 g) of pure carbon as graphite (a reasonable approximation of coal) 94 kcal are released. This is enough to raise the temperature of 1.0 L of water 94 °C. The superscript ° stands for standard conditions (25 o C and 1.00 atm). Because we started with elements in their standard state  H° =  H f o the standard heat of formation for CO 2. There is a table of  H f o in Pitts & Pitts, Appendix I, p. 1031.

3. Copyright © 2013 R. R. Dickerson3 At constant pressure and if the only work is done against the atmosphere, i.e. PdV work, then đW = PdV dH p = đQ p and đQ is now an exact differential - that is independent of path. Enthalpy is an especially useful expression of heat.

4. Copyright © 2009 R. R. Dickerson & Z.Q. Li 4 The burning of graphitic carbon might proceed through formation of CO:  H o or  H f o C graph + O 2 → CO 2 -94.0 kcal/mole C graph + 1/2 O 2 → CO -26.4 CO + 1/2 O 2 → CO 2 -67.6 ---------------------------------------------------- NET C graph + O 2 → CO 2 -94.0 kcal/mole This is Hess' law – the enthalpy of a reaction is independent of the mechanism (path). The units of kcal are commonly used because  H f o is usually measured with Dewars and change in water temperature.

5. Copyright © 2009 R. R. Dickerson & Z.Q. Li 5 Heat capacity: The amount of heat required to produce a one degree change in temp in a given substance. C = đ Q/dT C p = (∂Q/∂T) p = (∂H/∂T) p C v = (∂Q/∂T) v = (∂U/∂T) v Because đ Q p = dH and đ Q v = dU For an ideal gas PV = nRT C p = C v + R Where R = 2.0 cal mole -1 K -1 = 287(J kg -1 K -1 )* 0.239 (cal/J) * 29.0E-3 (kg/mole)

6. Copyright © 2009 R. R. Dickerson & Z.Q. Li 6 The heat capacity depends on degrees of freedom in the molecule enjoys. Translation = 1/2 R each (every gas has 3 translational degrees of freedom) Rotation = 1/2 R Vibration = R For a gas with N atoms you see 3N total degrees of freedom and 3N - 3 internal (rot + vib) degrees of freedom. Equipartition principle: As a gas on warming takes up energy in all its available degrees of freedom.

7. Copyright © 2009 R. R. Dickerson & Z.Q. Li 7 Measured Heat Capacities (cal mole -1 K -1 ) C v C p He3.05.0 Ar3.05.0 O 2 5.07.0 N 2 4.956.9 CO5.06.9 CO 2 6.99.0 SO 2 7.39.3 H 2 O6.08.0 C v = R/2 x (T.D.F.) + R/2 x (R.D.F.) + R x (V.D.F.) C p = C v + R Translational degrees of freedom: always 3. Internal degrees of freedom: 3N - 3 Where N is the number of atoms in the molecule.

8. Copyright © 2009 R. R. Dickerson & Z.Q. Li 8 Test Calculation: C v (He): 3R/2 = 3.0 cal/(mole K) good! C v (O 2 ): 3R/2 + 2(R/2) + 1(R) = (7/2)R ≈ 7.0 cal/(mole K)? The table shows 5.0; what's wrong? Not all energy levels are populated at 300 K. Not all the degrees of freedom are active (vibration). O 2 vibration occurs only with high energy; vacuum uv radiation. At 2000K C v (O 2 ) approx 7.0 cal/mole K Students: show that on the primordial Earth the dry adiabatic lapse rate was about 12.6 K/km.

9. Copyright © 2011 R. R. Dickerson & Z.Q. Li 9 For one mole of an ideal gas, P/T = R/V At constant volume, dU = C v dT Thus dФ = {Cv dT}/T + {RdV}/V Integrating Ф = C v ln(T) + R ln(V) + Ф o Where Ф o is the residual entropy. This equation lets you calculate entropy for an ideal gas at a known T and V.

10. Copyright © 2011 R. R. Dickerson & Z.Q. Li 10 GIBBS FREE ENERGY The Second Law states that for a reversible reaction: d Ф = đ Q/T For an irreversible reaction, d Ф > đ Q/T At constant temperature and pressure for reversible and irreversible reactions: dU = đ Q - PdV - VdP dU ≤ đ Q - PdV - VdP dU - Td Ф + PdV ≤ 0 Because dP and dT are zero we can add VdP and Ф dT to the equation. dU - Td Ф - Ф dT + PdV + VdP ≤ 0 d(U + PV - T Ф ) ≤ 0 We define G as (U + PV − T Ф ) or (H − T Ф )

11. Copyright © 2009 R. R. Dickerson & Z.Q. Li 11 dG = dH − Td Ф  G =  H − T  Ф  G is the Gibbs free energy that stands as the criterion of feasibility. G tends toward the lowest values, and if  G for a reaction is positive, the reaction cannot proceed! The Gibbs free energy of a reaction is the sum of the Gibbs free energy of formation of the products minus the sum of the Gibbs free energy of formation of the reactants.  G rxn ° =   G f °(products) -   G f °(reactants)

12. Copyright © 2009 R. R. Dickerson & Z.Q. Li 12 2. ENTHALPY OF REACTIONS The heat of a reaction is the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.  H rxn =   H f o (products) -   H f o (reactants) The change of enthalpy of a reaction is fairly independent of temperature. EXAMPLE: ENTHALPY CALCULATION

13. Copyright © 2009 R. R. Dickerson & Z.Q. Li 13 3. BOND ENERGIES See Appendix III of Pitts for a table of bond energies. The quantity is actually heat not energy. Definitions: Bond Dissociation Energy - The amount of energy required to break a specific bond in a specific molecule. Bond Energy - The average value for the amount of energy required to break a certain type of bond in a number of species. EXAMPLE: O-H in water We want H 2 O → 2H + O +221 kcal/mole We add together the two steps: H 2 O → OH + H +120 OH → O + H +101 --------------------------------- NET +221 Bond energy (enthalpy) for the O-H bond is 110.5 kcal/mole, but this is not the b.d.e. for either O-H bond.

14. Copyright © 2009 R. R. Dickerson & Z.Q. Li 14 Another example: What is the C-H bond enthalpy in methane? We want  H o for the reaction: CH 4 → C gas + 4H Any path will do (equation of state.)  H o (kcal/mole) CH 4 + 2 O 2 → CO 2 + 2H 2 O -193 CO 2 → C graph + O 2 +94 2H 2 O → 2 H 2 + O 2 +116 2H 2 → 4H +208 C graph → C gas +171 ------------ ----------------------------------- NET CH 4 → C gas + 4H + 396 kcal/mole The bond energy for C-H in methane is +396/4 = +99 kcal/mole. Bond energies are useful for "new" compounds and substances for which b.d.e. can't be directly measured such as radical.

15. Copyright © 2009 R. R. Dickerson & Z.Q. Li 15 FREE ENERGY We have a problem, neither internal energy (E or U) nor enthalpy (H) is the "criterion of feasibility". Chemical systems generally tend toward the minimum in E and H, but not always. Everyday experience tells us that water evaporates at room temperature, but this is uphill in terms of the total energy.

16. Copyright © 2009 R. R. Dickerson & Z.Q. Li 16 Example 1 H 2 O( l )  H 2 O(g) P = 10 torr, T = 25 o C  U = + 9.9 kcal/mole The enthalpy,  H, is also positive, about 10 kcal/mole, and PdV is too small to have an impact.

17. Copyright © 2009 R. R. Dickerson & Z.Q. Li 17 Example 2. The formation of nitric oxide from nitrogen and oxygen occurs at combustion temperature. We know that  H >>0 at room temperature, also at combustion temperatures. We can calculate  H as a function of temperature with heat capacities, C p, found in tables; but this is more detail than is necessary. Remember that R = 1.99 cal/moleK and dH = C p dT N 2 + O 2  2NO

18. Copyright © 2013 R. R. Dickerson & Z.Q. Li 18 N 2 + O 2  2NO At room temperature:  H 298 = + 43.14 kcal/mole The reaction is not favored, but combustion and lightning heat the air, and C p ≡ (∂H/∂T) p.

19. Copyright © 2013 R. R. Dickerson & Z.Q. Li 19 Gibbs Free Energy,  G, and Equilibrium Constants, K eq Consider the isothermal expansion of an ideal gas. dG = VdP From the ideal gas law, dG = (nRT/P)dP Integrating both sides,

20. Copyright © 2014 R. R. Dickerson & Z.Q. Li 20 Consider the following reaction, aA + bB ↔ cC + dD Where small case letters represent coefficients. For each gas individually:  G = nRT ln(P 2 /P 1 ) Remember what an equilibrium constant is: thus  G = -nRT ln (K eq )

21. Copyright © 2013 R. R. Dickerson & Z.Q. Li 21 IIA. ENTHALPY (HEAT) 1. FORMATION Definition: The enthalpy of formation.  Hfo is the amount of heat produced or required to form a substance from its elemental constituents. The standard conditions, represented by a super "o", are a little different from those for the Ideal Gas Law: 25oC (not 0oC), 1.0 atm. and the most stable form of elements. The standard heat of formation is zero for elements. This quantity is very useful for calculating the temperature dependence of equilibrium constants and maximum allowed rate constants. It was thought for a long time that  H was the criterion of feasibility. Although  H tends toward a minimum, it is not the criterion. Things usually tend toward minimum in  H, but not always. Examples are the expansion of a gas into a vacuum, and the mixing of two fluids. Which is hotter, an oxygen-acetylene flame or an oxygen-methane flame? REACTIONS C2H2 + 2.5O2 → 2CO2 + H2O CH4 + 2O2 → CO2 + 2H2O Note: melting point iron = 1535 C. IIB. EXAMPLE 2. The formation of nitric oxide from nitrogen and oxygen occurs at combustion temperature. N2 + O2  2NO We know that  H >>0 at room temperature, but we know that combustion produces large quantities of NO. Therefore, these reactions must be driven by some force other than internal energy or heat. Note: There is little change in  H with T. You can prove it by integrating  Cp (Cp of the products - Cp of the reactants) from 298 to 1500 K. At room temperature  H298 = + 43.14 kcal/mole;  H1500 = + 42.506 kcal/mole ENTROPY AND THE SECOND LAW OF THERMODYNAMICS DFN: dS = DQ/T For a reversible reaction the change in entropy,  S, is a function of state of system only. To get a feel for what entropy is, let us derive an expression for the entropy of an ideal gas. GENERAL RELATIONS: dE = DQ - PdV dS = DQ/T dS = dE/T + PdV/T or, for one mole Keq = exp ({-  G}/RT ) NOTE: Keq = 1 if and only if  G = 0.00 If a + b is not the same as c + d, we can get into trouble trying to take the log of an expression with units. For this type of reaction, the Gibbs free energy is the sum of the  G for the chemical reaction and the  G for the change in pressure. Assuming that the reactants start at 1.0 atm and go to an equilibrium pressure and assuming that the products finish at 1.0 atm. aA(PA = 1)  aA(PA)  G A = aRTln (PA/1) Similarly for B. For the products: cC(PC)  cC(PC = 1)  G C = cRTln (1/{ PC}) Similarly for D.  Go =  Grxn +  GA +  GB +  GC +  GDI =  Grxn + aRTln(PA /1) + bRTln(PB /1) + cRTln(1/PC) + dRTln(1/PD) Combining the log terms,  Go =  Grxn + RT ln {(PA)a (PB)b}/{(PC)c (PD)d} For the equilibrium partial pressures where  Grxn = 0,  Go = RT ln [{(PA)a (PB)b}/{(PC)c (PD)d}] = -RT ln [{(PC)c (PD)d }/{(PA)a (PB)b }] If you remember that each of the partial pressures was a ratio with the initial or final pressure taken as 1.0 and that the ln(1) = 0 are left out you can see that Keq is always dimensionless. aA + bB ↔ cC + dD  G° = -nRT ln (K eq ) This holds only for reactants that start (state 1) at standard conditions and products that finish (state 2) at standard conditions (1.00 atm). Standard conditions are 25 °C and 1.00 atm pressure. Watch out for units – Gibbs Free Energy of formation is tabulated for these standard conditions. Gibbs Free Energy and Equilibrium

22. Copyright © 2013 R. R. Dickerson & Z.Q. Li 22 Example: Lightning

23. Copyright © 2011 R. R. Dickerson 23 Example: Lightning In the absence of industrial processes, lightning is a major source of odd nitrogen (NOx) and thus nitrate (NO 3 - ) in the atmosphere. Even today, lightning is a major source of NOx in the upper free troposphere. N 2 + O 2  2NO How can this happen? Let ’ s calculate the Gibbs Free Energy for the reaction for 298 K and again for 2000 K.  G° = – nRT ln (K eq )  G f ° =  H f ° 0.0 for N 2 and O 2  G f ° (NO) = 20.7 kcal mole -1  Hf° (NO) = 21.6 kcal mole -1 R = 1.98 cal mole -1 K -1  G° = 2*(20.7) = 41.4 kcal mole -1 K eq = exp (– 41.4E3/1.98*298) = 3.4 E-31

24. Copyright © 2013 R. R. Dickerson & Z.Q. Li 24  G° = 2*(20.7) = 41.4 kcal mole -1 = exp (-  G°/RT) = 3.4 E-31 Assume P N2 = 0.8 atm; P O2 = 0.2 atm = 2.3E-16 atm (pretty small) Let ’ s try again at a higher temperature (2000 K). Remember  H and  φ are independent of temperature.  G T ≈  H° – T  φ° 41.4 = 43.2 – 298  φ°  φ° = +6.04E-3 kcal mole -1 K -1  G T ≈ 43.2 – 2000* 6.04E-3 = 31.12 kcal mole -1

25. Copyright © 2013 R. R. Dickerson & Z.Q. Li 25  G (2000) ≈ 31.12 kcal mole -1 K eq = = P NO 2 /P N2 *P O2 = exp (-  G T /RT) = exp (-31.12E3/1.98*2000) = 3.87E-4 If the total pressure is 1.00 atm P NO = 7.9E-3 atm = [0.79% by volume] You can show that the mole fraction of NO at equilibrium is nearly independent of pressure. Try repeating this calculation for 2500 K; you should obtain K eq = 3.4E-3 and [NO] = 2.3%. In high temperature combustion, such as a car engine or power plant, NO arises from similar conditions.

26. Copyright © 2013 R. R. Dickerson & Z.Q. Li 26 References Allen, D. J., and K. E. Pickering, Evaluation of lightning flash rate parmaterization For use in global chemical transport models, J. Geophys. Res., 107(23), Art. No. 4711, 2002. Chameides W. L., et al., NOx production in lightning, J. Atmos. Sci., 34, 143- 149, 1977. Rakov V., and M. A. Uman, Lightning: Physics and Effects, University Press, Cambridge, 2003.

27. Copyright © 2013 R. R. Dickerson & Z.Q. Li 27 What is ∆H at 1500K? If dH = C p dT then  H 1500 =  H 298 + Integral from 298 to 1500 of  C p dT  C p = 2C p (NO) - C p (N 2 ) - C p (O 2 ) We can approximate C p with a Taylor expansion. C p (O 2 )/R = 3.0673 + 1.6371x10 -3 T - 5.118x10 -7 T 2 C p (N 2 )/R = 3.2454 + 0.7108x10 -3 T - 0.406 -7 T 2 C p (NO)/R = 3.5326 - 0.186x10 -3 T - 12.81x10 -7 T 2 - 0.547x10 -9 T 3  C p /R = 0.7525 - 2.7199x10 -3 T + 26.5448x10 -7 T 2 - 1.094x10 -9 T 3

28. Copyright © 2013 R. R. Dickerson & Z.Q. Li 28 What is  H 1500 ?

29. Copyright © 2015 R. R. Dickerson 29 Mean = 25.2 o C Stdev = 2.04 o C

30. Copyright © 2015 R. R. Dickerson30 Mean = 22.3 o C Stdev = 1.14 o C

31. 31 What is Your estimated uncertainty? +/- o C 1.0, 1, 1, 0.2, 0.5, 2.0, 0.5, 0.1, 0.5, 0.6, 5.0, 1.0, “not” with median 1.0K. If the data are Gaussian then 95% CI = +/- 2  = +/- 2.2 o C = +/- 4.1 o C (~7 o F) if we use separate measurements; +/-2.3 together.