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Reasons for the Development of Estimation Techniques

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If there is no experimental data available, are there any other methods of estimating thermochemical data? Group Additivity ab initio calculations molecular mechanics calculations. Reasons for the Development of Estimation Techniques - PowerPoint PPT Presentation
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If there is no experimental data available, are there any other methods of estimating thermochemical data? Group Additivity ab initio calculations molecular mechanics calculations
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Page 1: Reasons for the Development of Estimation Techniques

If there is no experimental data available, are there any other methods of estimating thermochemical data?

Group Additivity

ab initio calculations

molecular mechanics calculations

Page 2: Reasons for the Development of Estimation Techniques

Reasons for the Development of Estimation Techniques

1. To provide a value for the property when no experimental values are available

2. To provide a means of distinguishing between two or more discordant experimental values.

3. The parameters generated by the estimation technique may provide a handle to measure or understand some molecular property that can not be measured directly.

4. As a diagnostic tool to identify and sometimes quantify unusual interactions.

Page 3: Reasons for the Development of Estimation Techniques

Question: Can thermochemical properties be subdivided into small increments so that the total molecular property in question can be calculated from the sum of its parts?

1. How small can the increments be?

2. Are these group increments interchangable so that they can be used to calculate similar properties of distinctively different molecules?

3. What properties are subject to this treatment?

Page 4: Reasons for the Development of Estimation Techniques

Some Thermochemical properties that have been modeled by group additivity

1. Molecular weight

2. Atomic weight

3. Enthalpy of formation

4. Entropy of formation

4. Enthalpy of vaporization

5. Total phase change entropy

6. Heat capacity

7. Vapor pressure

Page 5: Reasons for the Development of Estimation Techniques

Bond additivity

1. Bond Additivity: An assignment of a value associated with each type of bond.

Group additivity

2. Values are associated with a series of atom combinations is various structural environments.

Page 6: Reasons for the Development of Estimation Techniques
Page 7: Reasons for the Development of Estimation Techniques

Butane

C3H8 Hf(298) = 8 C-H + 2C-C

Hf(298) = [8(-3.83) + 2(2.73)]4.184

= -105.4 kJ mol-1

Hf(298)expt = -104.6 kJ mol-1

Bond Additivity

Page 8: Reasons for the Development of Estimation Techniques

Butadiene

CH2=CH-CH=CH2

Hf(298) = 6 Cd-H + Cd-C

Hf(298) = [6(3.2) + 6.7]*4.184 = 109.4 kJ mol-1

Hf(298)expt = 110 kJ mol-1

Bond Additivity

Page 9: Reasons for the Development of Estimation Techniques

Styrene

Hf(298) = 5 -H + -C +3 Cd-H

Hf(298) = 4.184[5(3.25) + 7.25 + 3(3.2)]

Hf(298) = 138.5 kJ mol-1

Hf(298)expt = 147.9 kJ mol-1

Bond Additivity

Page 10: Reasons for the Development of Estimation Techniques

Benson’s Group Additivity

Values are assigned to groups based on their constitution and on their neighboring environment

Page 11: Reasons for the Development of Estimation Techniques
Page 12: Reasons for the Development of Estimation Techniques

Styrene

Hf(298) = 5 CB-(H) + CB-(Cd) + Cd-(Cd)(H)+ Cd-(H2)

Hf(298) = 4.184[5(3.3) + 5.68 + 6.78 + 6.26]

Hf(298) = 147.4 kJ mol-1

Hf(298)expt = 147.9 kJ mol-1

Group Additivity

Page 13: Reasons for the Development of Estimation Techniques

Group Additivity

Butane

Hf(298) = 2 C-(H)3(C) + 2 C-(H)2(C)2

Hf(298) = 4.184[2(-10.2) +2(-4.93)]

Hf(298) = -126.6

Hf(298)expt = -125.5

Page 14: Reasons for the Development of Estimation Techniques

Group Additivity

CyclobutaneHf(298) = 4 C-(H)2(C)2

Hf(298) = 4.184[4(-4.93)] = -82.5

Hf(298)expt = 28.4

Hf(298) = -82.5-28.4 = -111 kJ mol-1

CH2CH2

-CH2CH2CH2CH2- vs | | CH2CH2

Hf(298) = strain energy in cyclobutane

Page 15: Reasons for the Development of Estimation Techniques
Page 16: Reasons for the Development of Estimation Techniques

Cp(g)

Calculation of heat capacity follows the same rules as enthalpy.

What is the heat capacity of 2-methylhexane?

Cp(g)(298) = 3 C-(H)3C+3 C-(H)2(C)2 +C-(H)(C)3

Cp(g)(298) = 4.184[3(6.19) + 3(5.5) +4.54]

Cp(g)(298) = 165.7 J mol-1K-1

Page 17: Reasons for the Development of Estimation Techniques

Entropy: (S(298)(g)

The calculation of entropy follows the same rules as for enthalpy and heat capacity but has two additional corrections.

Correction terms: (1) symmetry (-Rln)

(2) mixing (-Riniln(ni))

where ni refers to the mol fraction and is the symmetry number. The symmetry number is the number of identical structures that can be obtained by a symmetry operation such as rotation.

Page 18: Reasons for the Development of Estimation Techniques

What is the entropy of 2-methylhexane?

S(298) (g) = 3 C-(H)3C+3 C-(H)2(C)2 +C-(H)(C)3 + corrections

S(298) (g) = 4.184[3(30.41)+3(9.42)-12.07] + corrections

S(298) (g) = 449.4

corrections = -3Rln(3) =-27.4

S(298) (g) = 449.4 - 27.4 = 422 J mol-1 K-1

S(298) (l)expt = 314.6, 315.1, 323.3

S(298) (g)

Page 19: Reasons for the Development of Estimation Techniques

S(298)(g) = S(298)(l) + Cp(l)lnTB/298) + Hv/TB +Cp(g)ln(298/TB)

S(298)(l) = S(298)(g) - [Cp(l)lnTB/298) + Hv/TB+Cp(g)ln(298/TB)]

S(298)(l) = 422 -[222.9ln(363/298)+34900/363+165.7ln(298/363)]

S(298)(l) = 314.6 J mol-1K-1

S(298) (l)expt = 314.6, 315.1, 323.3

S(298) (l) TB=363 K; Cp(l)=222.9J mol-1K-1; Hv= 34.9 J mol-1K-1

Page 20: Reasons for the Development of Estimation Techniques

What is the entropy of 3-methylhexane?

S(298) (g) = 3 C-(H)3C+3 C-(H)2(C)2 +C-(H)(C)3 + corrections

S(298) (g) = 4.184[3(30.41)+3(9.42)-12.07] + corrections

S(298) (g) = 449.4

corrections = -3Rln(3) =-27.4; +Rln(2)

S(298) (g) = 449.4 - 27.4 +5.76 = 427 J mol-1 K-1

S(298) (l)expt = 309.6

Page 21: Reasons for the Development of Estimation Techniques

S(298) (l) TB=364 K; Cp(l)=216.7J mol-1K-1; Hv= 35.1 J mol-1K-1

S(298)(g) = S(298)(l) + Cp(l)lnTB/298) + Hv/TB +Cp(g)ln(298/TB)

S(298)(l) = S(298)(g) - [Cp(l)lnTB/298) + Hv/TB+Cp(g)ln(298/TB)]

S(298)(l) = 427 -[216.7ln(364/298)+35100/364+165.7ln(298/364)]

S(298)(l) = 320.4 J mol-1K-1

S(298) (l)expt = 309.6 J mol-1K-1

Page 22: Reasons for the Development of Estimation Techniques

What is the enthalpy of formation of 3-methylhexane?

Hf(298)(g) = 3 C-(H)3C+3 C-(H)2(C)2 +C-(H)(C)3 + corrections

Hf(298)(g) = 4.184[3(-10.2)+3(-4.93)-1.9] + corrections

Hf(298)(g) = -197.9 + corrections

correction = 0.8*4.184

Hf(298)(g) = -194.5 kJ mol-1

Hf(298)(g)expt = -191.3 1.3

Page 23: Reasons for the Development of Estimation Techniques

H CH3

CH3

CH2

H2C

CH3

H H

1 gauche interaction

Page 24: Reasons for the Development of Estimation Techniques

Summary on Using Benson Group Values

1. Identify the groups: Hf, Cp, S

2. Add non-bonded interactions: Hf, Cp, S

3. Add symmetry and chirality considerations: S


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