Properties

## Property Estimation Joback Method

The Joback method is a group contribution method which calculates thermophysical and transport properties as a function of the sum of group parameters. It uses a very simple and easy to assign group scheme. This article shows how to calculate properties using Joback Method in an excel spreadsheet.

Example

Estimate properties for 1-Butanol based on Joback Method

Structure of 1-Butanol consists of following groups

-CH3 Group : 1-CH2 Group : 3-OH (alcohol) Group : 1

Normal Boiling Point

\displaystyle \displaystyle T_{NBP} \left(K\right) = 198 + \sum T_{b,i}

where Tb,i is contribution due to each group. These values for individual group are available in literature and Wikipedia reference below. On combining values for each group normal boiling point comes out to be -

TNBP (K) = 383.10

Critical Temperature

\displaystyle \displaystyle Tc\left(K\right)=T_{NBP}/\left(0.584+0.965*\sum T_{c,i}-\left(\sum T_{c,i}\right)^2\right)

Tc (K) = 545.08

Critical Pressure

\displaystyle \displaystyle Pc \left(bar\right) = \left(0.113+0.0032*N_{A}-\sum P_{c,i}\right)^{-2}

Pc (bar) = 43.86

where NA is number of atoms in the molecular structure.

Critical Volume

\displaystyle \displaystyle Vc \left(cm^{3}/mol\right) = 17.5 + \sum V_{c,i}

Vc (cm³/mol) = 278.50

Critical Compressibility

\displaystyle \displaystyle Zc = (Pc*Vc)/(R*Tc)

Zc = 0.2695

Acentric Factor

Lee-Kesler method can be used to estimate the acentric factor.

\displaystyle \displaystyle \omega= \alpha/\beta

\displaystyle \displaystyle \theta = T_{NBP}/Tc

\displaystyle \displaystyle \alpha = -\ln\left(Pc\right)-5.92714+6.09648/\theta+1.28862*\ln\left(\theta\right)-0.169347*\theta^6

\displaystyle \displaystyle \beta = 15.2518 - 15.6875/\theta - 13.4721*\ln\left(\theta\right)+0.43577*\theta^6

ω = 0.6602

Freezing Point

\displaystyle \displaystyle Tm(K) = 122.5+\sum T_{m,i}

Tm (K) = 195.66

Heat of Formation (Ideal Gas, 298 K)

\displaystyle \displaystyle H_{formation}(kJ/mol) = 68.29+\sum H_{form,i}

Hformation (kJ/mol) = -278.12

Gibbs Energy of Formation (Ideal Gas, 298 K)

\displaystyle \displaystyle G_{formation} (kJ/mol) = 53.58+\sum G_{form,i} H_{form,i}

Gformation (kJ/mol) = -154.02

Heat of Vaporization (at Normal Boiling Point)

Reidel's equation can be used to estimate a liquid's heat of vaporization at its normal boiling point.

\displaystyle \displaystyle \triangle H_{v} (kJ/mol) = \frac{1.092*R*T_{NBP}*\left(\ln(Pc)-1.013\right)}{0.930-T_{NBP}/Tc}

ΔHv (kJ/mol) = 42.38

Heat of Fusion

\displaystyle \displaystyle \triangle H_{fusion}(kJ/mol) = -0.88+\sum H_{fus,i}

ΔHfusion (kJ/mol) = 10.2

Heat Capacity (Ideal Gas)

\displaystyle \displaystyle C_{p} (J/mol.K) = A + B.T + C.T^{2}+D.T^{3}

\displaystyle \displaystyle A = \sum a_{i} - 37.93

\displaystyle \displaystyle B = \sum b_{i} + 0.210

\displaystyle \displaystyle C = \sum c_{i} - 3.91*10^{-4}

\displaystyle \displaystyle D = \sum d_{i} - 2.06*10^{-7}

CP (J/mol.K) = 110.96 (at 300 K)

Heat of Vaporization (@ Temperature T)

Watson equation can be used to estimate heat of vaporization at different temperature as following.

\displaystyle \displaystyle H_{v,2}\left(kJ/mol\right)=H_{v,1}\left(\frac{Tc-T_{2}}{Tc-T_{1}}\right)^{0.38}

ΔHv,2 (kJ/mol) = 49.60 (at 300 K)

Liquid Viscosity

\displaystyle \displaystyle \eta_{L} \left(Pa.s\right)=Mw.\exp\left(A/T+B\right)

\displaystyle \displaystyle A = \sum\eta _{a}-597.82

\displaystyle \displaystyle B = \sum\eta _{b}-11.202

ηL (Pa.s) = 1.936*10-3 (at 300 K)

where Mw is the molecular weight.

Liquid Density

Rackett equation can be used to estimate liquid density from critical properties as following -

\displaystyle \displaystyle \rho_{L}\left(gm/cm^{3}\right)=\frac{Mw}{\left(R.Tc/Pc\right)*Zc^{\left(1+(1-T/Tc)^{2/7}\right)}}

ρL (gm/cm³) = 0.756 (at 300 K)

Liquid Vapor Pressure

Lee-Kesler equation can be used to estimate liquid vapor pressure as following -

\displaystyle \displaystyle Pvp \left(bar\right)=Pc*\exp\left(f0+\omega.f1\right)

\displaystyle \displaystyle f0 = 5.92714 - 6.09648/Tr - 1.28862*ln(Tr) + 0.169347.Tr^{6}

\displaystyle \displaystyle f1 = 15.2518 - 15.6875/Tr - 13.4721*ln(Tr) + 0.43577.Tr^{6}

\displaystyle \displaystyle Tr = T/Tc

Pvp (bar) = 0.018

Resources

References