In: Chemistry
Given
1) Vapor pressure of liquid Q (P 1) = 0.500 atm
2) Vapor pressure of liquid Q (P 2) = 1 atm
3) T 1 = 20.0 0 C =20.0 + 273 = 293 K
4) T 2 = ?
5) H
vaporization = 26 kJ / mol = 26000 J / mol
We have Clapeyron-Clausius equation, log (P 2 / P
1) = [H
vaporization / 2.303 R ]
[ ( 1 / T 1 ) - ( 1 / T 2 ) ]
log ( 1 atm
/0.5 atm ) = [26000 J / mol / 2.303
8.314 J / mol K ]
[ ( 1 / T 1 ) - ( 1 / T 2 ) ]
0.3010 = [26000 J / mol / 2.303
8.314 J / mol K ]
[ ( 1 / T 1 ) - ( 1 / T 2 ) ]
0.3010 = 1357.9 K
[ ( 1 / T 1 ) - ( 1 / T 2 ) ]
[ ( 1 / T 1 ) - ( 1 / T 2 ) ] = 0.3010 / 1357.9 K
[ ( 1 / T 1 ) - ( 1 / T 2 ) ] =
2.217
10 -04 K -1
( 1 / T 1 ) - 2.217
10 -04 K -1 = ( 1 / T 2 )
( 1/ 293 K ) - 2.217
10 -04 K -1 = ( 1 / T 2 )
3.19
10 -03 K -1 = ( 1 / T
2 )
T 2 = 313.35 K
T 2 = 313.35 K - 273 = 40.35 0 C
Normal boiling point of liquid Q = 40.4 0 C
Boiling point of liquid is greater than normal room temperature. Hence, we can accept that there may be H-Bonding.