A Summary of Different Methods Used to Measure Vaporization Enthalpies

BG Bourdon gaugeC calorimetric determination GC gas chromatographyGCC gas chromatography-calorimetry CGC correlation gas chromatographyDM diaphram manometerDSC differential scanning calorimeterEB ebullometryGS gas saturation, transpirationHG Heise gauge

HSA head space analysisI isoteniscopeIPM inclined piston manometryME Mass effusion-Knudsen effusionMG McLeod gaugeMM mercury manometerOM oil manometerRG Rodebush gaugeSG spoon gaugeSTG strain gaugeT tensimeterTE torsion effusionUV ultraviolet absorption

A Summary of Different Methods Used to Measure Vaporization Enthalpies (continued)

TB thermobalanceTGA thermal gravimetric analysisTPTD temperature programmed thermal

desorption particle beam mass spectrometry

TRM thermoradiometric methodTSGC temperature scanning gas

chromatographyUV ultraviolet absorption HSAV viscosity gaugeVG MKS Baratron Vacuum Gauge

1. Measurement of vapor pressure as a function of temperature - using a manometer

2. Knudsen effusion

P = m(2RT/mw)1/2/ t AKc

Kc = 8r/(3l +8r)

where: P = pressure; m = mass loss from cell;

t = period of time; A = area of opening

mw = molecular weight; T = temperature (K)

r = radius of opening; l = thickness of opening

Measurement of Vaporization Enthalpies

3. Calvet calorimeter

4. Transpiration

5. Head space analysis

6. Correlation gas chromatography

Time (sec)0 100 200 300 400 500

Sig

na

l In

ten

sity

0

50

100

150

200

250

Correlation gas chromatography

T/K 434.3 439.3 444.2 449.1 454.1 459 463.8

t/min

methylene chloride

1.251 1.215 1.246 1.216 1.222 1.228 1.249

tetradecane 3.039 2.695 2.485 2.29 2.145 2.022 1.942

pentadecane 4.107 3.558 3.205 2.887 2.643 2.451 2.288

hexadecane 5.827 4.933 4.344 3.807 3.409 3.084 2.805

heptadecane 8.329 6.907 5.939 5.097 4.47 3.959 3.54

octadecane 12.283 9.994 8.403 7.065 6.071 5.265 4.624

nonadecane 18.549 14.836 12.2 10.075 8.487 7.219 6.211 eicosane 28.345 22.305 17.935 14.57 12.04 10.076 8.522

What is ta?

ta is the adjusted retention time ti - tnrr

ti = retention time of ith componenttnrr = retention time of a non retained reference

What does ta measure?

For a pure component, a plot of ln (vapor pressure) vs 1/T over a narrow temperature range results in a straight line. The slope of the line is equal to - g

lHm(Tm), the

enthalpy of vaporization.

A plot of ln (1/ ta) vs 1/T over a narrow temperature range results in a straight line. What does the slope measure?

Tetradecane

1/T (K)

0.00216 0.00218 0.00220 0.00222 0.00224 0.00226 0.00228 0.00230

ln (

1/ta )

-0.6

-0.4

-0.2

0.0

0.2

0.4

Enthalpy of Transfer Determination for Tetradecane

ln(1/ta) = -gslnHm(Tm)/R + intercept

gslnHm(Tm) * 8.314 J mol-1 = 53.158 kJ mol-1

What is slngHm(Tm) ? What does it measure?

Solute on stationary phase of column gas phase

Thermochemical cycle:

Vapor pure liquid solution on the capillary column

slngHm(Tm) = l

gHm(Tm) + slnHm(Tm)

Characteristics of capillary gas chromatographs with FID detectors

Typical sample sizes ~ microgram quantities

solids or liquids are in “solution” or adsorbed; concentrations are low and too dispersed for crystallization

temperatures are also high for crystals to form

Equations for the temperature dependence of ln(1/ta) for C14 to C20:

Tm = 449 K sln

gHm/R intercept r2

tetradecane -6393.895 14.1610.01 0.9989

pentadecane -6787.973 14.5970.01 0.9994

hexadecane -7251.562 15.1900.01 0.9996

heptadecane -7612.665 15.5870.01 0.9996

octadecane -8014.871 16.0700.01 0.9996

nonadecane -8457.474 16.6400.01 0.9996

eicosane -8919.685 17.2570.01 0.9995

slngHm(449 K)

lgHm (298.15 K) (lit)

lgHm (298.15 K)

(calc)

tetradecane 53.2 71.7 71.81.0

pentadecane 56.4 76.8 76.51.0

hexadecane 60.3 81.4 821.1

heptadecane 63.3 86.5 86.31.2

octadecane 66.6 91.4 91.11.3

nonadecane 70.3 96.4 96.41.4

eicosane 74.2 101.8 101.91.4

lgHm (298.15 K) = (1.4360.019) sln

gHm(Tm) – (4.540.35); r2 = 0.9991

g

slnHm (298.15 K) kJ mol-1

50 55 60 65 70 75 80

glH

m (

298.

15 K

) / k

J m

ol-1

70

75

80

85

90

95

100

105

Why does lgHm (298.15 K) correlate with sln

gHm(Tm) in a linear fashion?

gslnHm(Tm) = g

lHm(Tm) + slnHm(Tm)

We know that glHm(298.15 K) 4.69 (nC -nQ) + 3.0

However T = 298.15 K is an arbitrary temperature

glHm(Tm) = AT (nC) + BT where A is some constant and B is a variable

but small in magnitude

Lets assume for the moment that

slnHm(Tm) = Asln(nC) + Bsln where B is a variable but small in magnitude

The slope of the line from the correlation is given by:

slope = lgHm (298.15 K) / sln

gHm(Tm)

slope = lgHm (298.15 K)/sln

gHm(Tm)

slope = [A298 (nC) + B298]/{[AT (nC) + BT]+ Asln(nC) + Bsln}

slope = [A298 (nC) + B298]/{(AT + Asln)(nC) + (BT + Bsln)}

let A’ = (AT + Asln); B’= (BT + Bsln)

slope =/[A298 (nC) + B298]/{(A’)(nC) + (B’)}

if = (A’)(nC) > B’ and A298 > B298

then slope = (A298)/(A’) = constant

Table 4. Parameters of the Cox Equation. Tb Ao 103A1 106A2

tetradecane 526.691 3.13624 -2.063853 1.54151pentadecane 543.797 3.16774 -2.062348 1.48726hexadecane 559.978 3.18271 -2.002545 1.38448heptadecane 575.375 3.21826 -2.04 1.38octadecane 590.023 3.24741 -2.048039 1.36245nonadecane 603.989 3.27626 -2.06 1.35eicosane 617.415 3.31181 -1.02218 1.34878

Cox Equation

ln (p/po) = (1-Tb/T)exp(Ao +A1T +A2T 2)

Ruzicka, K.; Majer, V. “Simultaneous treatment of vapor pressures and related thermal data between the triple point and normal boiling temperatures for n-alkanes C5-C20,” J. Phys. Chem. Ref. Data 1994, 23, 1-39.

Hv(298) Hv(lit) (449) Hslnv Hsln

tetradecane 71.7 56.909 53.2 -3.709pentadecane 76.8 60.701 56.4 -4.301hexadecane 81.4 64.485 60.3 -4.185heptadecane 86.5 68.171 63.3 -4.871octadecane 91.4 72.092 66.6 -5.492nonadecane 96.4 75.998 70.3 -5.698eicosane 101.8 79.793 74.2 -5.593

Number of carbons, nC

13 14 15 16 17 18 19 20 21

lg Hm (

298.

15 K

), (

sl

n Hm (

T),

O /

kJ m

ol-1

-2e+4

0e+0

2e+4

4e+4

6e+4

8e+4

1e+5

lgHm(T)/ kJ mol-1= 3.816nC+3.43

slnHm(T)/ kJ mol-1= 0.34816nC+1.08

lgHm(449 K) / kJ mol-1= 3.82nC+3.43

slnHm (449 K) / kJ mol-1= -0.35nC+1.08

lgHm(298.15 K) / kJ mol-1= 4.98nC+1.88

lgHm(298.15 K)/sln

gHm(449 K) = (4.98nC+1.88)/(3.82nC+3.43- 0.35nC+1.08)

lgHm(298.15 K)/sln

gHm(449 K)= 4.98/(3.47) = 1.435

lgHm (298.15 K) / sln

gHm(Tm) = (1.4360.019)