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Diss Factsheets

Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
The study was conducted between 07 August 2015 and 11 August 2015
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
other: 1= Reliable without restriction o GLP guideline study

Data source

Reference
Reference Type:
study report
Title:
Unnamed
Year:
2015
Report date:
2015

Materials and methods

Test guideline
Qualifier:
according to guideline
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Type of method:
effusion method: vapour pressure balance

Test material

Constituent 1
Chemical structure
Reference substance name:
2-{2-[(2-ethylhexyl)oxy]ethoxy}ethyl prop-2-enoate
Cas Number:
117646-83-0
Molecular formula:
C15H28O4
IUPAC Name:
2-{2-[(2-ethylhexyl)oxy]ethoxy}ethyl prop-2-enoate
Test material form:
other: liquid
Details on test material:
Identification: M
120

Appearance/Physical State:
Clear colorless liquid

Batch:
NC-5635-01

Purity:
99.4%

Expiry Date:
05 October 2015

Storage Conditions:
Room temperature in the dark

Results and discussion

Vapour pressure
Temp.:
25 °C
Vapour pressure:
0.044 Pa

Any other information on results incl. tables

Results Recorded temperatures, mass differences and the resulting calculated values of vapour pressure are shown in the following tables:

Run 5

Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

30

303.15

0.003298697

65.49

6.549E-08

0.090916898

-1.041355388

31

304.15

0.003287851

75.60

7.560E-08

0.104952169

-0.979008583

32

305.15

0.003277077

88.00

8.800E-08

0.122166545

-0.913047707

33

306.15

0.003266373

102.31

1.023E-07

0.142032492

-0.847612294

34

307.15

0.003255738

116.43

1.164E-07

0.161634669

-0.791465481

35

308.15

0.003245173

135.01

1.350E-07

0.187428469

-0.727164441

36

309.15

0.003234676

154.41

1.544E-07

0.214360640

-0.668854956

37

310.15

0.003224246

181.94

1.819E-07

0.252579333

-0.597602188

38

311.15

0.003213884

203.80

2.038E-07

0.282926613

-0.548326199

39

312.15

0.003203588

234.11

2.341E-07

0.325004659

-0.488110414

40

313.15

0.003193358

270.96

2.710E-07

0.376161900

-0.424625195

 

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 gives the following statistical data using an unweighted least squares treatment.

 

Slope:

-5.84 x 103

Standard error in slope:

32.6

Intercept:

18.2

Standard error in intercept:

0.106

 

The results obtained indicate the following vapor pressure relationship:

 Log10(Vp (Pa)) = -5.84 x 103/temp(K) + 18.2

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -1.36.

Run 6

Vapor Pressure Data

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

30

303.15

0.003298697

65.80

6.580E-08

0.091347258

-1.039304485

31

304.15

0.003287851

75.04

7.504E-08

0.104174745

-0.982237553

32

305.15

0.003277077

87.43

8.743E-08

0.121375239

-0.915869900

33

306.15

0.003266373

102.46

1.025E-07

0.142240730

-0.846976027

34

307.15

0.003255738

114.63

1.146E-07

0.159135808

-0.798232086

35

308.15

0.003245173

129.37

1.294E-07

0.179598704

-0.745696801

36

309.15

0.003234676

153.85

1.539E-07

0.213583216

-0.670432878

37

310.15

0.003224246

176.24

1.762E-07

0.244666272

-0.611425895

38

311.15

0.003213884

200.77

2.008E-07

0.278720197

-0.554831560

39

312.15

0.003203588

223.43

2.234E-07

0.310178082

-0.508388893

40

313.15

0.003193358

260.31

2.603E-07

0.361376971

-0.442039527

 

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 6 gives the following statistical data using an unweighted least squares treatment.

 

Slope:

-5.67 x 103

Standard error in slope:

56.9

Intercept:

17.6

Standard error in intercept:

0.185

 

The results obtained indicate the following vapor pressure relationship:

 Log10(Vp (Pa)) = -5.67 x 103/temp(K) + 17.6

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -1.35.

Run 7

Vapor Pressure Data

 

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

30

303.15

0.003298697

63.24

6.324E-08

0.087793322

-1.056538517

31

304.15

0.003287851

74.27

7.427E-08

0.103105788

-0.986716955

32

305.15

0.003277077

85.38

8.538E-08

0.118529314

-0.926174228

33

306.15

0.003266373

100.89

1.009E-07

0.140061168

-0.853682257

34

307.15

0.003255738

117.19

1.172E-07

0.162689744

-0.788639824

35

308.15

0.003245173

130.58

1.306E-07

0.181278494

-0.741653714

36

309.15

0.003234676

151.69

1.517E-07

0.210584583

-0.676573427

37

310.15

0.003224246

168.12

1.681E-07

0.233393632

-0.631910997

38

311.15

0.003213884

198.88

1.989E-07

0.276096393

-0.558939267

39

312.15

0.003203588

224.62

2.246E-07

0.311830107

-0.506081956

40

313.15

0.003193358

257.51

2.575E-07

0.357489854

-0.446736280

 

A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 7 gives the following statistical data using an unweighted least squares treatment.

 

Slope:

-5.72 x 103

Standard error in slope:

66.8

Intercept:

17.8

Standard error in intercept:

0.217

 

The results obtained indicate the following vapor pressure relationship:

 Log10(Vp (Pa)) = -5.72 x 103/temp(K) + 17.8

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -1.36.

Summary of Results

The values of vapor pressure at 25 °C extrapolated from each graph are summarized in the following table:

 

Table3.4 – Summary of Vapor Pressure Data

 

Run

Log10[Vp(25 ºC)]

5

-1.36

6

-1.35

7

-1.36

Mean

-1.36

Vapor Pressure

4.37 x 10-2Pa

 

The test item did not change in appearance under the conditions used in the determination.

Discussion

A total of seven runs were completed for the main sequence. Equilibrium with regard to vapor pressure was assessed to have been reached over the final three runs. Thus the final three runs have been used to calculate the definitive vapor pressure value for the test item.

Applicant's summary and conclusion

Conclusions:
The vapor pressure of the test item has been determined to be 4.4 x 10-2 Pa at 25 ºC.