Can some please help me part a? 2. (30 points) Later you find in the CRC Handboo
ID: 228320 • Letter: C
Question
Can some please help me part a?
2. (30 points) Later you find in the CRC Handbook of Chemistry and Physics the following information on 2-chlorophenol. TCC) Po (mm Hg) 12.1 51.210 82.040 106.0 100 149.8400 174.5760 a) Calculate the vapor pressure, p, in atmosphere and in pascal of 2-chlorophenol at 40 °C using the vapor pressure-temperature data given in the table above. b) Compare the value for p'iu with the value obtained in 1a) c) From these data in the table, calculate AvapHi (in kJ/mol) for 2-chlorophenol d) From these data, also calculate AvapSi (in J/molK) for 2-dichlorophenol (Hint: there is a relationship between vapH and AvapS at Tb.Explanation / Answer
you have to use the Clausius-Claperyon equation to solve this
it gives the relationship between vapor pressure and temperature
the equation is ln (P1/P2) = (del Hvap/R) * (1/T2 - 1/T1)
to find del Hvap first we will use any two set of values for vapour pressure and temperature from the given table
lets take P1= 10 mm hg= 0.0131atm T1= 51.2 +273= 324.2K
P2= 40mm hg= 0.0526 atm T2= 82 + 273= 355K and R= 8.314 J/ mol.K
putting all the values in equation
ln (P1/P2) = (del Hvap/R) * (1/T2 - 1/T1)
ln (0.0131/0.0526) = (del Hvap/ 8.314) * ( 1/355- 1/324.2)
del Hvap = 43.185 KJ
now to find vapour pressure at 40oC put all the values in the equation and find the unknown
ln (P1/P2) = (del Hvap/R) * (1/T2 - 1/T1)
ln (P1/0.0526) = (43185/8.314) * (1/355- 1/313)
ln (P1/0.0526) = 5194.25 * (-3.77 * 10-4)
eln(P1/0.0526)= e-1.963
(P1/0.0526) = e-1.963
(P1/0.0526) = 0.1404
P1= 0.1404*0.0526 = 0.007386 atm
P1= 5.61 mm Hg at 40oC
note : for atm to mmhg multiply by 760 and for mm hg to atm multiply by 0.00131