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Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
16-07-2019 to 24-10-2019
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
other: Guideline study performed under GLP. All relevant validity criteria were met.
Qualifier:
according to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
Qualifier:
according to guideline
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
Qualifier:
according to guideline
Guideline:
EPA OPPTS 830.7950 (Vapor Pressure)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Remarks:
inspected: August 2018 ; signature: November 2018
Type of method:
effusion method: vapour pressure balance
Temp.:
25 °C
Vapour pressure:
0.149 Pa
Remarks on result:
other: Mean Vp (n=9): extrapolated from log Vp (Pa) vs. 1/T curve of nine (9) individual runs (the first of ten runs was not included due to not having reached equilibrium)
Test no.:
#1
Temp.:
25 °C
Vapour pressure:
0.149 Pa
Test no.:
#2
Temp.:
25 °C
Vapour pressure:
0.15 Pa
Test no.:
#3
Temp.:
25 °C
Vapour pressure:
0.15 Pa
Test no.:
#4
Temp.:
25 °C
Vapour pressure:
0.152 Pa
Test no.:
#5
Temp.:
25 °C
Vapour pressure:
0.151 Pa
Test no.:
#6
Temp.:
25 °C
Vapour pressure:
0.145 Pa
Test no.:
#7
Temp.:
25 °C
Vapour pressure:
0.147 Pa
Test no.:
#8
Temp.:
25 °C
Vapour pressure:
0.152 Pa
Test no.:
#9
Temp.:
25 °C
Vapour pressure:
0.148 Pa
Transition / decomposition:
no
Remarks on result:
other: The test item did not change in appearance under the conditions used in the determination

Table 1.0 – vapour pressure determination of the test item

Run

Log10Vp(25°C)

Vp(25°C) [Pa]

 

 

 

1

-0.828

0.149

2

-0.823

0.150

3

-0.823

0.150

4

-0.818

0.152

5

-0.822

0.151

6

-0.838

0.145

7

-0.832

0.147

8

-0.818

0.152

9

-0.830

0.148

Mean

-0.826

0.149

 

 

 

 

Note the first of ten runs was excluded from assessment as the system had not reached equilibrium (details provided in the full study report).

 

The slope and intercept of the nine (9) individual runs for theplot  ofLog10Vp(Pa) versus reciprocal of temperature (K) is recorded in the full study report. A total of 10 runs were completed for the main sequence. Run 1 was not utilised as the sample was still equilibrating.With regard toequilibrium, this has been assessed to have been reached over runs 1 to 9 (actual #2 to #10) which were used to calculate the vapour pressure. There was no evidence of decomposition during the course of the test (no change in appearance).

Conclusions:
The vapour pressure of the test item has been determined to be 0.149 Pa at 25 °C.
Executive summary:

The vapour pressure was determined using EU Method A.4 guideline using the vapour pressure balance method under GLP. Measurements being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. The temperature of the sample was controlled electronically. The mass and temperature readings were recorded automatically into a computer file. The difference in mass readings with the orifice covered and uncovered is proportional to the vapour pressure at the given oven temperature. Temperature and pressure readings were taken between 25 and 35°C with a one-hour dwell time at 25°C between runs. The vapour pressure of the test item was extrapolated from the vapour pressure curve (log10Vpversus 1/T). The mean vapour pressure based on nine runs was 0.149 Pa at 25°C.

Endpoint:
vapour pressure
Type of information:
experimental study
Adequacy of study:
key study
Study period:
09 April 2021 - 13 April 2021
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
guideline study
Remarks:
Study conducted according to Method A.4 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 and OCSPP 830.7950 of the US EPA Office of Chemical Safety and Pollution Prevention (OCSPP), Series 830: Product Properties Test Guidelines.
Qualifier:
according to guideline
Guideline:
EPA OPPTS 830.7950 (Vapor Pressure)
Qualifier:
according to guideline
Guideline:
OECD Guideline 104 (Vapour Pressure Curve)
Deviations:
no
Qualifier:
according to guideline
Guideline:
EU Method A.4 (Vapour Pressure)
Deviations:
no
GLP compliance:
yes (incl. QA statement)
Remarks:
Signature : August 2021
Type of method:
effusion method: vapour pressure balance
Key result
Temp.:
25 °C
Vapour pressure:
ca. 0.132 Pa
Remarks on result:
other: Mean

Results:


The test item appearance was recorded before and after testing as a clear, pale yellow liquid, thus no change in appearance of the test item occured under the conditions used during the study. 


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


Run 1:


Table 1 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321115.281.153e-070.16004-0.79577
29302.150.003310118.641.186e-070.16470-0.78331
30303.150.003299125.841.258e-070.17470-0.75771
31304.150.003288135.541.355e-070.18816-0.72547
32305.150.003277141.891.419e-070.19698-0.70558
33306.150.003266153.471.535e-070.21306-0.67150
34307.150.003256166.371.664e-070.23096-0.63646
35308.150.003245178.611.786e-070.24796-0.60562
36309.150.003235194.481.945e-070.26999-0.56865
37310.150.003224213.802.138e-070.29681-0.52752
38311.150.003214226.082.261e-070.31386-0.50326


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


 


Slope:                                  -2.85 x 103 


Standard error in slope:       98.3


Intercept:                              8.65


Standard error in intercept: 0.321


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.85 x 103 / temp (K) + 8.65


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.910.


 


 


Run 2:


Table 2 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321126.151.262e-070.17513-0.75664
29302.150.003310125.591.256e-070.17435-0.75858
30303.150.003299131.641.316e-070.18275-0.73814
31304.150.003288138.241.382e-070.19191-0.71690
32305.150.003277149.421.494e-070.20743-0.68313
33306.150.003266160.331.603e-070.22258-0.65251
34307.150.003256173.711.737e-070.24115-0.61771
35308.150.003245188.521.885e-070.26171-0.58218
36309.150.003235203.222.032e-070.28212-0.54957
37310.150.003224218.112.181e-070.30279-0.51886
38311.150.003214242.312.423e-070.33639-0.47316

 


A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T (K)) for Run 2 gives the following statistical data using an unweighted least square treatment.


 


Slope:                                  -2.78 x 103 


Standard error in slope:       147


Intercept:                              8.43


Standard error in intercept: 0.482


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.78 x 103 / temp (K) + 8.43


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.883.


 


 


Run 3:


Table 3 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321126.951.270e-070.17624-0.75390
29302.150.003310127.851.279e-070.17749-0.75083
30303.150.003299132.231.322e-070.18357-0.73620
31304.150.003288140.261.403e-070.19472-0.71059
32305.150.003277149.921.499e-070.20813-0.68167
33306.150.003266162.251.623e-070.22524-0.64735
34307.150.003256172.801.728e-070.23989-0.61999
35308.150.003245186.521.865e-070.25894-0.58680
36309.150.003235206.652.067e-070.28688-0.54230
37310.150.003224223.452.235e-070.31021-0.50834
38311.150.003214239.442.394e-070.33240-0.47834

 


A plot of Log10(vapor pressure (Pa)) versus reciprocal temperature (1/T (K)) for Run 3 gives the following statistical data using an unweighted least square treatment.


 


Slope:                                  -2.74 x 103 


Standard error in slope:       147


Intercept:                              8.33


Standard error in intercept: 0.479


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.74 x 103 / temp (K) + 8.33


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.877.


 


Run 4:


Table 4 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321130.141.301e-070.18067-0.74311
29302.150.003310128.881.289e-070.17892-0.74734
30303.150.003299134.971.350e-070.18737-0.72730
31304.150.003288142.261.423e-070.19749-0.70445
32305.150.003277150.541.505e-070.20899-0.67987
33306.150.003266164.951.650e-070.22899-0.64018
34307.150.003256176.901.769e-070.24558-0.60981
35308.150.003245188.961.890e-070.26232-0.58117
36309.150.003235206.932.069e-070.28727-0.54171
37310.150.003224224.882.249e-070.31219-0.50558
38311.150.003214242.892.429e-070.33719-0.47213

 


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 square treatment.


 


Slope:                                  -2.71 x 103 


Standard error in slope:       152


Intercept:                              8.21


Standard error in intercept: 0.496


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.71 x 103 / temp (K) + 8.21


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.868.


 


Run 5:


Table 5 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321127.431.274e-070.17691-0.75225
29302.150.003310128.821.288e-070.17884-0.74754
30303.150.003299134.101.341e-070.18617-0.73009
31304.150.003288141.651.417e-070.19665-0.70631
32305.150.003277151.161.512e-070.20985-0.67809
33306.150.003266165.031.650e-070.22910-0.63997
34307.150.003256175.221.752e-070.24325-0.61395
35308.150.003245188.751.888e-070.26203-0.58165
36309.150.003235205.232.052e-070.28491-0.54529
37310.150.003224220.272.203e-070.30579-0.51458
38311.150.003214239.602.396e-070.33263-0.47804

 


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 square treatment.


Slope:                                  -2.69 x 103 


Standard error in slope:       124


Intercept:                              8.14


Standard error in intercept: 0.407


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.69 x 103 / temp (K) + 8.14


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.870.


 


Run 6:


Table 6 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321127.311.273e-070.17674-0.75267
29302.150.003310128.221.282e-070.17800-0.74958
30303.150.003299134.361.344e-070.18653-0.72925
31304.150.003288138.321.383e-070.19202-0.71665
32305.150.003277150.081.501e-070.20835-0.68121
33306.150.003266163.591.636e-070.22710-0.64378
34307.150.003256177.341.773e-070.24619-0.60873
35308.150.003245189.021.890e-070.26241-0.58102
36309.150.003235204.222.042e-070.28351-0.54743
37310.150.003224221.422.214e-070.30739-0.51231
38311.150.003214239.952.400e-070.33311-0.47741

 


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 square treatment.


Slope:                                  -2.72 x 103 


Standard error in slope:       140


Intercept:                              8.26


Standard error in intercept: 0.458


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.72 x 103 / temp (K) + 8.26


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.874.


 


Run 7:


Table 7 - Vapor Pressure Data

















































































































Temperature (ºC)Temperature (K)Reciprocal Temperature (K-1)Mass Difference (µg)Mass Difference (kg)Vapor Pressure (Pa)Log10 Vp
28301.150.003321126.181.262e-070.17517-0.75654
29302.150.003310128.451.285e-070.17832-0.74880
30303.150.003299132.991.330e-070.18462-0.73372
31304.150.003288144.791.448e-070.20101-0.69678
32305.150.003277153.131.531e-070.21258-0.67248
33306.150.003266163.671.637e-070.22722-0.64355
34307.150.003256175.491.755e-070.24363-0.61327
35308.150.003245188.251.883e-070.26134-0.58279
36309.150.003235205.052.051e-070.28466-0.54567
37310.150.003224222.612.226e-070.30904-0.50999
38311.150.003214241.462.415e-070.33521-0.47468

 


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 square treatment.


Slope:                                  -2.73 x 103 


Standard error in slope:       115


Intercept:                              8.27


Standard error in intercept: 0.375


 


The results obtained indicate the following vapour pressure relationship: 


Log10(Vp (Pa)) = -2.73 x 103 / temp (K) + 8.27


 


The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.872.


 


Table 8 - Summary of Vapor Pressure Data


 









































RunLog10 [Vp(25 ºC)]
1-0.910
2-0.883
3-0.877
4-0.868
5-0.870
6-0.874
7-0.872
Mean-0.879

 

Conclusions:
The vapor pressure of the test item ST 04 C 21 has been determined to be 0.132 Pa at 25 ºC.
Executive summary:

The vapor pressure of the test item has been determined to be 0.132 Pa at 25 °C, using the vapor pressure balance method, designed to be compatible with Method A.4 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 and OCSPP 830.7950 of the US EPA Office of Chemicals Safety and Pollution Prevention (OCSPP), Serie 830: Product Properties Test Guidelines. 

Description of key information

Weight of evidence : The substance vapour pressure is : 0.132 Pa at 25 °C (2021)


1. Vapour Pressure: 0.149 Pa at 25 °C, EU Method A.4 – vapour pressure balance method, 2019


2. Vapour Pressure: 0.132 Pa at 25 °C, EU Method A.4 – vapour pressure balance method, 2021

Key value for chemical safety assessment

Vapour pressure:
0.132 Pa
at the temperature of:
25 °C

Additional information

Key study 1: EU Method A.4, 2019 : The vapour pressure was determined using EU Method A.4 guideline using the vapour pressure balance method under GLP. Measurements being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. The temperature of the sample was controlled electronically. The mass and temperature readings were recorded automatically into a computer file. The difference in mass readings with the orifice covered and uncovered is proportional to the vapour pressure at the given oven temperature. Temperature and pressure readings were taken between 25 and 35°C with a one-hour dwell time at 25°C between runs. The vapour pressure of the test item was extrapolated from the vapour pressure curve (log10Vpversus 1/T). The mean vapour pressure based on nine runs was 0.149 Pa at 25°C.


 


Key study 2: EU Method A.4, 2021 : The vapour pressure was determined using EU Method A.4 guideline using the vapour pressure balance method under GLP. Measurements being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. The temperature of the sample was controlled electronically. The mass and temperature readings were recorded automatically into a computer file. The difference in mass readings with the orifice covered and uncovered is proportional to the vapour pressure at the given oven temperature. Temperature and pressure readings were taken between 28 and 38°C with a one-hour dwell time at 25°C between runs. The vapour pressure of the test item was extrapolated from the vapour pressure curve (log10Vpversus 1/T). The mean vapour pressure based on seven runs was 0.132 Pa at 25°C


Conclusion : the vapour pressure of the substance is 0.132 Pa at 25°C (2021)