Chemical Thermodynamics Salby Chapter 4. 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 Copyright © 2017 R. R. Dickerson

Enthalpy (heat) of 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.

Hrxn =  Hfo(products) -  Hfo(reactants) 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. Hrxn =  Hfo(products) -  Hfo(reactants) The change of enthalpy of a reaction is fairly independent of temperature. Example Enthalpy Calculation Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2017 R. R. Dickerson For example, consider the burning of graphitic carbon in oxygen, the change in enthalpy is: Ho Cgraph + O2 → CO2 -94.0 kcal/mole In the combustion of 1 mole (12 g) of pure carbon as graphite (a reasonable approximation of coal) 94 kcal (393.5 kJ) are released. This is enough to raise the temperature of 1.0 L of water by 94 °C. The superscript ° stands for standard conditions (25 oC and 1.00 atm). Because we started with elements in their standard state H° = Hfo the standard heat of formation for CO2. There is a table of Hfo in Pitts & Pitts, Appendix I, p. 1031. Copyright © 2017 R. R. Dickerson

Copyright © 2016 R. R. Dickerson At constant pressure and if the only work is done against the atmosphere, i.e., PdV work, then đW = PdV dHp = đQp and đQ is now an exact differential - that is independent of path. Enthalpy is an especially useful expression of heat. Copyright © 2016 R. R. Dickerson

Copyright © 2016 R. R. Dickerson The burning of graphitic carbon might proceed through formation of CO: Ho or Hfo Cgraph + O2 → CO2 -94.0 kcal/mole Cgraph + 1/2 O2 → CO -26.4 CO + 1/2 O2 → CO2 -67.6 ---------------------------------------------------- NET Cgraph + O2 → CO2 -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 Hfo is usually measured with Dewars and change in water temperature. Copyright © 2016 R. R. Dickerson

Copyright © 2017 R. R. Dickerson Another example: What is the heat of the reaction of methane oxidation? We want Ho for the reaction: CH4 + 2O2→ CO2 + 2H2Ogas Any path will do (equation of state.) Ho = –394 – 2(242) –(–75) = – 803 (kJ/mole) or 192 kcal/mole The change in enthalpy for reaction is also the heat of combustion of methane. Does this tell us anything about the greenhouse forcing of coal vs. natural gas? Copyright © 2017 R. R. Dickerson

Copyright © 2009 R. R. Dickerson & Z.Q. Li Heat capacity: The amount of heat required to produce a one degree change in temp in a given substance. C = đQ/dT Cp = (∂Q/∂T)p = (∂H/∂T)p Cv = (∂Q/∂T)v = (∂U/∂T)v Because đQp = dH and đQv = dU For an ideal gas PV = nRT Cp = Cv + 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) Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li 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. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li Measured Heat Capacities (cal mole-1 K-1) Cv Cp He 3.0 5.0 Ar 3.0 5.0 O2 5.0 7.0 N2 4.95 6.9 CO 5.0 6.9 CO2 6.9 9.0 SO2 7.3 9.3 H2O 6.0 8.0 Cv = R/2 x (T.D.F.) + R/2 x (R.D.F.) + R x (V.D.F.) Cp = Cv + R Translational degrees of freedom: always 3. Internal degrees of freedom: 3N - 3 Where N is the number of atoms in the molecule. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li Test Calculation: Cv(He): 3R/2 = 3.0 cal/(mole K) good! Cv(O2): 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). O2 vibration occurs only with high energy; vacuum uv radiation. At 2000K Cv (O2) approx 7.0 cal/mole K Students: Later you will show that on the primordial Earth the temperature profile was steeper. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2016 R. R. Dickerson Consider entropy; for one mole of an ideal gas, P/T = R/V At constant volume, dU = Cv dT Thus dФ = {Cv dT}/T + {RdV}/V Integrating Ф = Cv ln(T) + R ln(V) + Фo Where Фo is the residual entropy. This equation lets you calculate entropy for an ideal gas at a known T and V. Note, texts by Salby and Wallace and Hobbs use S not Ф. Copyright © 2016 R. R. Dickerson

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

Grxn° =  Gf°(products) -  Gf°(reactants) dG = dH − TdФ 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. Grxn° =  Gf°(products) -  Gf°(reactants) Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li 3. Bond Energies & Bond Dissociation 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 H2O → 2H + O +221 kcal/mole We add together the two steps: H2O → 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. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li Another example: What is the C-H bond enthalpy in methane? We want Ho for the reaction: CH4 → Cgas + 4H Any path will do (equation of state.) Ho (kcal/mole) CH4 + 2 O2 → CO2 + 2H2O -193 CO2 → Cgraph + O2 +94 2H2O → 2 H2 + O2 +116 2H2 → 4H +208 Cgraph → Cgas +171 ------------ ----------------------------------- NET CH4 → Cgas + 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. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li 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. Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2009 R. R. Dickerson & Z.Q. Li Example 1 H2O(l)  H2O(g) P = 10 torr, T = 25oC U = + 9.9 kcal/mole The enthalpy, H, is also positive, about 10 kcal/mole, and PdV is too small to have an impact. Copyright © 2009 R. R. Dickerson & Z.Q. Li

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, Cp, found in tables; but this is more detail than is necessary. Remember that R = 1.99 cal/moleK and dH = CpdT N2 + O2  2NO Copyright © 2009 R. R. Dickerson & Z.Q. Li

Copyright © 2013 R. R. Dickerson & Z.Q. Li N2 + O2  2NO At room temperature: H298 = + 43.14 kcal/mole The reaction is not favored, but combustion and lightning heat the air, and Cp ≡ (∂H/∂T)p. Copyright © 2013 R. R. Dickerson & Z.Q. Li

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

Copyright © 2014 R. R. Dickerson & Z.Q. Li Consider the following reaction, aA + bB ↔ cC + dD Where small case letters represent coefficients. For each gas individually: G = nRT ln(P2/P1) Remember what an equilibrium constant is: thus G = -nRT ln (Keq) Copyright © 2014 R. R. Dickerson & Z.Q. Li

Copyright © 2013 R. R. Dickerson & Z.Q. Li Gibbs Free Energy and Equilibrium aA + bB ↔ cC + dD G° = -nRT ln (Keq) 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. Equilibrium constants are always dimensionless. Copyright © 2013 R. R. Dickerson & Z.Q. Li

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

Example: Lightning In the absence of industrial processes, lightning is a major source of odd nitrogen (NOx) and thus nitrate (NO3-) in the atmosphere. Even today, lightning is a major source of NOx in the upper free troposphere. N2 + O2  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 (Keq) Gf° = Hf° 0.0 for N2 and O2 Gf° (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 Keq = exp (– 41.4E3/1.98*298) = 3.4 E-31 Copyright © 2011 R. R. Dickerson

Copyright © 2013 R. R. Dickerson & Z.Q. Li G° = 2*(20.7) = 41.4 kcal mole-1 = exp (-G°/RT) = 3.4 E-31 Assume PN2 = 0.8 atm; PO2 = 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. GT ≈ H° – Tφ° 41.4 = 43.2 – 298 φ° φ° = +6.04E-3 kcal mole-1 K-1 GT ≈ 43.2 – 2000* 6.04E-3 = 31.12 kcal mole-1 Copyright © 2013 R. R. Dickerson & Z.Q. Li

Keq = = PNO2/PN2*PO2 = exp (-GT/RT) = exp (-31.12E3/1.98*2000) G(2000) ≈ 31.12 kcal mole-1 Keq = = PNO2/PN2*PO2 = exp (-GT/RT) = exp (-31.12E3/1.98*2000) = 3.87E-4 If the total pressure is 1.00 atm PNO = 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 Keq = 3.4E-3 and [NO] = 2.3%. In high temperature combustion, such as a car engine or power plant, NO arises from similar conditions. Copyright © 2013 R. R. Dickerson & Z.Q. Li

Copyright © 2013 R. R. Dickerson & Z.Q. Li References Allen, D. J., and K. E. Pickering, Evaluation of lightning flash rate parameterization 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. Copyright © 2013 R. R. Dickerson & Z.Q. Li

Cp = 2Cp (NO) - Cp (N2) - Cp(O2) What is ∆H at 1500K? If dH = CpdT then H1500 = H298 + Integral from 298 to 1500 of Cp dT Cp = 2Cp (NO) - Cp (N2) - Cp(O2) We can approximate Cp with a Taylor expansion. Cp (O2)/R = 3.0673 + 1.6371x10-3 T - 5.118x10-7 T 2 Cp (N2)/R = 3.2454 + 0.7108x10-3 T - 0.406-7 T 2 Cp (NO)/R = 3.5326 - 0.186x10-3 T - 12.81x10-7 T 2 - 0.547x10-9 T 3 Cp /R = 0.7525 - 2.7199x10-3 T + 26.5448x10-7 T 2 - 1.094x10-9 T 3 Copyright © 2013 R. R. Dickerson & Z.Q. Li

Copyright © 2016 R. R. Dickerson What is H1500 ? Copyright © 2016 R. R. Dickerson

Copyright © 2009 R. R. Dickerson & Z.Q. Li