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Copyright © 2014 R. R. Dickerson 1 CHEMICAL THERMODYNAMICS A. ENTHALPY OF FORMATION AND COMBUSTION...

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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
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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

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

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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 are released. This is enough to raise the temperature of 1.0 L of water 94 °C.

The superscript ° stands for standard conditions (25 oC and 1.00 atm). Because we started with elements in their standard state H° = Hf

o the standard heat of formation for CO2. There is a table of Hf

o in Pitts & Pitts, Appendix I, p. 1031.

Copyright © 2013 R. R. Dickerson 3

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 © 2009 R. R. Dickerson & Z.Q. Li

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The burning of graphitic carbon might proceed through formation of CO: Ho or Hf

o

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 Hf

o is usually measured with Dewars and change in water temperature.

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

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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

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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 RVibration = RFor 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

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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

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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: show that on the primordial Earth the dry adiabatic lapse rate was about 12.6 K/km.

Copyright © 2011 R. R. Dickerson & Z.Q. Li9

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.

Copyright © 2011 R. R. Dickerson & Z.Q. Li10

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 - VdPdU ≤ đQ - PdV - VdPdU - TdФ + PdV ≤ 0

Because dP and dT are zero we can add VdP and ФdT to the equation.

dU - TdФ - Ф dT + PdV + VdP ≤ 0d(U + PV - TФ) ≤ 0

We define G as (U + PV − TФ) or (H − TФ)

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

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dG = dH − TdФG = H − TФ

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

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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) - Hf

o(reactants)

The change of enthalpy of a reaction is fairly independent of temperature.

EXAMPLE: ENTHALPY CALCULATION

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

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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 waterWe 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

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Another example: What is the C-H bond enthalpy in methane? We want Ho for the reaction: CH4 → Cgas + 4HAny 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

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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

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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

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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 © 2013 R. R. Dickerson & Z.Q. Li

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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

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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 © 2014 R. R. Dickerson & Z.Q. Li

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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 © 2013 R. R. Dickerson & Z.Q. Li

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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.

Gibbs Free Energy and Equilibrium

Copyright © 2013 R. R. Dickerson & Z.Q. Li

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Example: Lightning

Copyright © 2011 R. R. Dickerson23

Example: Lightning

In the absence of industrial processes, lightning is a major source ofodd 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 © 2013 R. R. Dickerson & Z.Q. Li

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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° – Tφ°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

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G(2000) ≈ 31.12 kcal mole-1

Keq = = PNO2/PN2*PO2 = exp (-GT/RT)

= exp (-31.12E3/1.98*2000)

= 3.87E-4If 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

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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.

Copyright © 2013 R. R. Dickerson & Z.Q. Li

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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

)298 - )(1500101/4(1.094x - )298 - )(15008x101/3(26.544

)298 - )(1500x101/2(2.7199 - 298) - 00.7525(150 dT Cp/R

449-337-

2231500

298

-

Copyright © 2013 R. R. Dickerson & Z.Q. Li

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What is H1500 ?

re. temperatuoft independennearly

and positivestrongly is H change; noAlmost

)(kcal/mole 42.506 0.904 - 43.41 0.904 - H H

(cal/mole) 903.81-

cal/moleK) (K R 454.176- |C

2981500

1500298p

Copyright © 2015 R. R. Dickerson29

Mean = 25.2oCStdev = 2.04 oC

Copyright © 2015 R. R. Dickerson 30

Mean = 22.3 oCStdev = 1.14 oC

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What is Your estimated uncertainty?+/- oC

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 oC

= +/- 4.1 oC (~7oF) if we use separate measurements; +/-2.3 together.


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