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 PresentationTRANSCRIPT
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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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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.
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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?
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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
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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.
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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
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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
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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
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Benson’s Group Additivity
Values are assigned to groups based on their constitution and on their neighboring environment
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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
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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
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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
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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
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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.
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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)
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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
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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
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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
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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
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H CH3
CH3
CH2
H2C
CH3
H H
1 gauche interaction
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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