Question

In: Chemistry

Calculate the normal boiling point of liquid Q if the vapor pressure is .500 atm at...

Calculate the normal boiling point of liquid Q if the vapor pressure is .500 atm at 20 degrees C. The change/ delta of H of vaporization for liquid Q is 26 kJ/ mole. Do you except liquid Q to have H bonds?

Solutions

Expert Solution

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 = 1 / 3.19 10 -03  K -1

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.


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