A Summary of Different Methods Used to Measure Vaporization Enthalpies
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Transcript of A Summary of Different Methods Used to Measure Vaporization Enthalpies
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
nal I
nten
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
sln
gHm(449 K) lgHm (298.15 K) (lit) l
gHm (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
l
gHm (298.15 K) = (1.4360.019) slngHm(Tm) – (4.540.35); r2 = 0.9991
g
slnHm (298.15 K) kJ mol-1
50 55 60 65 70 75 80
glH
m (2
98.1
5 K
) / k
J mol
-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 (2
98.1
5 K
), (
slnH
m (T
), O
/ kJ
mol
-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)