Question

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?

Answer 1

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.