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Diss Factsheets

Administrative data

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
Experimental Starting Date: 19 December 2012 Experimental Completion Date: 24 December 2012
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
other: see 'Remark'
Remarks:
Study conducted in compliance with agreed protocols, with no or minor deviations from standard test guidelines and/or minor methodological deficiencies, which do not affect the quality of the relevant results. The study report was conclusive, done to a valid guideline and the study was conducted under GLP conditions.

Data source

Reference
Reference Type:
study report
Title:
Unnamed
Year:
2013
Report date:
2013

Materials and methods

Test guidelineopen allclose all
Qualifier:
according to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
Qualifier:
according to guideline
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
GLP compliance:
yes
Type of method:
effusion method: vapour pressure balance

Test material

Constituent 1
Reference substance name:
A-197
IUPAC Name:
A-197
Details on test material:
Identification: A-197
Description: pale brown liquid
Batch: B12-12212
Purity: 77%
Expiry / Retest Date: 20 September 2013
Storage Conditions: approximately 4 °C in the dark

Results and discussion

Vapour pressure
Temp.:
25 °C
Vapour pressure:
0 Pa
Transition / decomposition
Transition / decomposition:
no

Any other information on results incl. tables

Results

Recorded temperatures, mass differences and the resulting calculated values of vapor pressure are shown in the following tables. Figures 3.2 - 3.5 Graphs of Log10Vapor Pressure vs Reciprocal Temperature are shown in Attachment 2 of this Summary.

Run 4

 

Table 3.1 – Vapor Pressure Data

 

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

90

363.15

0.002753683

53.42

5.342E-08

0.074160646

-1.129826495

91

364.15

0.002746121

54.51

5.451E-08

0.075673845

-1.121054197

92

365.15

0.002738601

57.51

5.751E-08

0.079838614

-1.097787011

93

366.15

0.002731121

54.41

5.441E-08

0.075535020

-1.121851653

94

367.15

0.002723682

72.49

7.249E-08

0.100634692

-0.997252279

95

368.15

0.002716284

73.78

7.378E-08

0.102425542

-0.989591728

97

370.15

0.002701607

84.27

8.427E-08

0.116988350

-0.931857385

98

371.15

0.002694328

88.56

8.856E-08

0.122943969

-0.910292771

99

372.15

0.002687089

82.37

8.237E-08

0.114350663

-0.941761313

100

373.15

0.002679887

114.82

1.148E-07

0.159399577

-0.797512836

 

A plot of log10(vapor pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 gives the following statistical data using an unweighted least squares treatment.

 

Slope  -4.12 x 103
Standard deviation in slope   507

 

Intercept         10.2
Standard deviation in intercept         1.38

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -4.12 x 103/temp(K) + 10.2

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -3.63.

Run 5

 

Table 3.2 – Vapor Pressure Data

 

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

90

363.15

0.002753683

50.02

5.002E-08

0.069440575

-1.158386691

91

364.15

0.002746121

56.21

5.621E-08

0.078033881

-1.107716793

92

365.15

0.002738601

51.42

5.142E-08

0.071384134

-1.146398306

94

367.15

0.002723682

69.89

6.989E-08

0.097025226

-1.013115338

95

368.15

0.002716284

70.69

7.069E-08

0.098135831

-1.008172397

96

369.15

0.002708926

76.68

7.668E-08

0.106451485

-0.972848275

97

370.15

0.002701607

79.58

7.958E-08

0.110477428

-0.956726444

99

372.15

0.002687089

102.04

1.020E-07

0.141657663

-0.848759929

100

373.15

0.002679887

92.65

9.265E-08

0.128621937

-0.890684955

 

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  -4.10 x 103
Standard deviation in slope   404

 

Intercept         10.1
Standard deviation in intercept         1.10

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -4.10 x 103/temp(K) + 10.1

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -3.62.

Run 6

 

Table 3.3 – Vapor Pressure Data

 

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

90

363.15

0.002753683

55.71

5.571E-08

0.077339753

-1.111597220

91

364.15

0.002746121

44.93

4.493E-08

0.062374351

-1.204993960

92

365.15

0.002738601

52.62

5.262E-08

0.073050041

-1.136379535

93

366.15

0.002731121

61.00

6.100E-08

0.084683628

-1.072200544

94

367.15

0.002723682

70.99

7.099E-08

0.098552308

-1.006333203

95

368.15

0.002716284

64.60

6.460E-08

0.089681350

-1.047297861

96

369.15

0.002708926

82.87

8.287E-08

0.115044791

-0.939133040

97

370.15

0.002701607

85.67

8.567E-08

0.118931909

-0.924701612

98

371.15

0.002694328

87.46

8.746E-08

0.121416887

-0.915720906

99

372.15

0.002687089

100.84

1.008E-07

0.139991755

-0.853897542

100

373.15

0.002679887

97.35

9.735E-08

0.135146741

-0.869194423

 

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  -4.48 x 103
Standard deviation in slope   517

 

Intercept         11.2
Standard deviation in intercept         1.40

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -4.48 x 103/temp(K) + 11.2

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -3.87.

Run 7

 

Table 3.4 – Vapor Pressure Data

 

Temperature (ºC)

Temperature (K)

Reciprocal Temperature (K-1)

Mass Difference (µg)

Mass Difference (kg)

Vapor Pressure (Pa)

Log10Vp

91

364.15

0.002746121

54.61

5.461E-08

0.075812671

-1.120258202

92

365.15

0.002738601

55.81

5.581E-08

0.077478578

-1.110818356

93

366.15

0.002731121

54.91

5.491E-08

0.076229148

-1.117878935

94

367.15

0.002723682

65.90

6.590E-08

0.091486083

-1.038644964

95

368.15

0.002716284

63.80

6.380E-08

0.088570745

-1.052709700

96

369.15

0.002708926

72.39

7.239E-08

0.100495866

-0.997851802

97

370.15

0.002701607

85.87

8.587E-08

0.119209560

-0.923688916

98

371.15

0.002694328

76.28

7.628E-08

0.105896183

-0.975119694

99

372.15

0.002687089

89.16

8.916E-08

0.123776923

-0.907360319

100

373.15

0.002679887

103.24

1.032E-07

0.143323570

-0.843682383

 

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  -4.14 x 103
Standard deviation in slope   447

 

Intercept         10.2
Standard deviation in intercept         1.21

 

The results obtained indicate the following vapor pressure relationship:

 

Log10(Vp (Pa)) = -4.14 x 103/temp(K) + 10.2

 

The above yields a vapor pressure (Pa) at 298.15 K with a common logarithm of -3.66.

Summary of Results

 

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

 

Table 3.5 – Summary of Vapor Pressure Data

 

Run

Log10[Vp(25 ºC)]

4

-3.63

5

-3.62

6

-3.87

7

-3.66

Mean

-3.69

Vapor Pressure

2.02 x 10-4Pa

 

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

Applicant's summary and conclusion

Conclusions:
Please see the Executive Summary section below for the conclusion.
Executive summary:

The determination was carried out using a procedure designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008 and Method 104 of the OECD Guidelines for Testing of Chemicals, 23 March 2006.

 

Instrument:                             Vapor pressure balance.

Summary of Results

 

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

 

Table 3.5 – Summary of Vapor Pressure Data

 

Run

Log10[Vp(25 ºC)]

4

-3.63

5

-3.62

6

-3.87

7

-3.66

Mean

-3.69

Vapor Pressure

2.02 x 10-4Pa

 

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

Conclusion

The vapor pressure of the test item has been determined to be 2.0 x 10-4Pa at 25 ºC.