CHEMICAL THERMODYNAMICS A ENTHALPY OF FORMATION AND COMBUSTION

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

For example, consider the burning of graphitic carbon in oxygen, the change in enthalpy is: Cgraph + O 2 → CO 2 Ho -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° = Hfo the standard heat of formation for CO 2. There is a table of Hfo in Pitts & Pitts, Appendix I, p. 1031. Copyright © 2009 R. R. Dickerson & Z. Q. Li 2

At constant pressure and if the only work is done against the atmosphere, i. e. Pd. V work, then đW = Pd. V d. Hp = đQp and đQ is now an exact differential - that is independent of path. Enthalpy is an especially useful expression of heat. Copyright © 2013 R. R. Dickerson 3

The burning of graphitic carbon might proceed through formation of CO: Ho or Hfo Cgraph + O 2 → CO 2 -94. 0 kcal/mole Cgraph + 1/2 O 2 → CO -26. 4 CO + 1/2 O 2 → CO 2 -67. 6 --------------------------NET Cgraph + 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 Hfo is usually measured with Dewars and change in water temperature. Copyright © 2009 R. R. Dickerson & Z. Q. Li 4

Heat capacity: The amount of heat required to produce a one degree change in temp in a given substance. C = đQ/d. T Cp = (∂Q/∂T)p = (∂H/∂T)p Cv = (∂Q/∂T)v = (∂U/∂T)v Because đQp = d. H and đQv = d. U For an ideal gas PV = n. RT Cp = C v + R Where R = 2. 0 cal mole-1 K-1 = 287(J kg-1 K-1)* 0. 239 (cal/J) * 29. 0 E-3 (kg/mole) Copyright © 2009 R. R. Dickerson & Z. Q. Li 5

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 3 N total degrees of freedom and 3 N - 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 6

Measured Heat Capacities (cal mole-1 K-1) Cv Cp He 3. 0 5. 0 Ar 3. 0 5. 0 O 2 5. 0 7. 0 N 2 4. 95 6. 9 CO 5. 0 6. 9 CO 2 6. 9 9. 0 SO 2 7. 3 9. 3 H 2 O 6. 0 8. 0 Cv = R/2 x (T. D. F. ) + R/2 x (R. D. F. ) + R x (V. D. F. ) Cp = C v + R Translational degrees of freedom: always 3. Internal degrees of freedom: 3 N - 3 Where N is the number of atoms in the molecule. Copyright © 2009 R. R. Dickerson & Z. Q. Li 7

Test Calculation: Cv(He): 3 R/2 = 3. 0 cal/(mole K) good! Cv(O 2): 3 R/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 2000 K Cv (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. Copyright © 2009 R. R. Dickerson & Z. Q. Li 8

For one mole of an ideal gas, P/T = R/V At constant volume, d. U = Cv d. T Thus d. Ф = {Cv d. T}/T + {Rd. V}/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. Copyright © 2011 R. R. Dickerson & Z. Q. Li 9

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

d. G = d. H − 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. Grxn° = Gf°(products) - Gf°(reactants) Copyright © 2009 R. R. Dickerson & Z. Q. Li 11

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 12

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 → 2 H + 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 Copyright © 2009 R. R. 13 the b. d. e. for either O-H bond. Dickerson & Z. Q. Li

Another example: What is the C-H bond enthalpy in methane? We want Ho for the reaction: CH 4 → Cgas + 4 H Any path will do (equation of state. ) Ho (kcal/mole) CH 4 + 2 O 2 → CO 2 + 2 H 2 O -193 CO 2 → Cgraph + O 2 +94 2 H 2 O → 2 H 2 + O 2 +116 2 H 2 → 4 H +208 Cgraph → Cgas +171 -----------------------NET CH 4 → Cgas + 4 H + 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 14

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 15

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 Pd. V is too small to have an impact. Copyright © 2009 R. R. Dickerson & Z. Q. Li 16

Example 2. The formation of nitric oxide from nitrogen and oxygen occurs at combustion temperature. We know that H >>0 at room temperature, but what about at combustion temperature? We can calculate H as a function of temperature with heat capacities, Cp, found in tables. Remember that R = 1. 99 cal/mole. K and d. H = Cp d. T N 2 + O 2 2 NO Copyright © 2009 R. R. Dickerson & Z. Q. Li 17

N 2 + O 2 2 NO At room temperature: H 298 = +

Gibbs Free Energy, G, and Equilibrium Constants, Keq Consider the isothermal expansion of an ideal gas. d. G = Vd. P From the ideal gas law, d. G = (n. RT/P)d. P Integrating both sides, Copyright © 2013 R. R. Dickerson & Z. Q. Li 19

Consider the following reaction, a. A + b. B ↔ c. C + d. D Where small case letters represent coefficients. For each gas individually: G = n. RT ln(P 2/P 1) Remember what an equilibrium constant is: thus G = -n. RT ln (Keq) Copyright © 2013 R. R. Dickerson & Z. Q. Li 20

Gibbs Free Energy and Equilibrium a. A + b. B ↔ c. C + d. D G° = -n. RT 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. Copyright © 2013 R. R. Dickerson & Z. Q. Li 21

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

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 2 NO How can this happen? Let’s calculate the Gibbs Free Energy for the reaction for 298 K and again for 2000 K. G° = – n. RT ln (Keq) Gf° = Hf° 0. 0 for N 2 and O 2 Hf° (NO) = 21. 6 kcal mole-1 Gf° (NO) = 20. 7 kcal mole-1 R = 1. 98 cal mole-1 K-1 G° = 2*(20. 7) = 41. 4 kcal mole-1 Keq = exp (– 41. 4 E 3/1. 98*298) = 3. 4 E-31 Copyright © 2011 R. R. Dickerson 23

G° = 2*(20. 7) = 41. 4 kcal mole-1 = exp (- G°/RT) = 3. 4 E-31 Assume PN 2 = 0. 8 atm; PO 2 = 0. 2 atm = 2. 3 E-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. 04 E-3 kcal mole-1 K-1 GT ≈ 43. 2 – 2000* 6. 04 E-3 = 31. 12 kcal mole-1 Copyright © 2013 R. R. Dickerson & Z. Q. Li 24

G(2000) ≈ 31. 12 kcal mole-1 Keq = = PNO 2/PN 2*PO 2 = exp (- GT/RT) = exp (-31. 12 E 3/1. 98*2000) = 3. 87 E-4 If the total pressure is 1. 00 atm PNO = 7. 9 E-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. 4 E-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 25

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, 143149, 1977. Rakov V. , and M. A. Uman, Lightning: Physics and Effects, University Press, Cambridge, 2003. Copyright © 2013 R. R. Dickerson & Z. Q. Li 26

What is ∆H at 1500 K? If d. H = Cpd. T then H 1500 = H 298 + Integral from 298 to 1500 of Cp d. T Cp = 2 Cp (NO) - Cp (N 2) - Cp(O 2) We can approximate Cp with a Taylor expansion. Cp (O 2)/R = 3. 0673 + 1. 6371 x 10 -3 T - 5. 118 x 10 -7 T 2 Cp (N 2)/R = 3. 2454 + 0. 7108 x 10 -3 T - 0. 406 -7 T 2 Cp (NO)/R = 3. 5326 - 0. 186 x 10 -3 T - 12. 81 x 10 -7 T 2 - 0. 547 x 10 -9 T 3 Cp /R = 0. 7525 - 2. 7199 x 10 -3 T + 26. 5448 x 10 -7 T 2 - 1. 094 x 10 -9 T 3 Copyright © 2013 R. R. Dickerson & Z. Q. Li 27