Q12.23P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  of a liquid that has a vapour pressure of 621 torr at  $\mathbf{85}\mathbf{.}\mathbf{2}\mathbf{°}\mathbf{C}$ and a boiling point of  $\mathbf{95}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{C}$ at 1 atm?

Step-by-Step Solution

Answer

The enthalpy of vaporization at standard condition,  ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$ is 12.15kJ/mole when the vapour pressure is 1atm at temperature  $\mathbf{95}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{C}$.

 

Step 1: Enthalpy of Vaporization

As Enthalpy means “heat” so enthalpy of the heat of vaporization can be defined as the heat which is a form of energy required to change the liquid phase into the vapour phase. 

  $\mathbf{Log}\frac{P1}{P2}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}$

 

Where,

  ${\mathrm{P}}_1$= Vapour pressure at temperature  ${\mathrm{T}}_1$

 ${\mathrm{P}}_2$ = Vapour pressure at temperature  ${\mathrm{T}}_2$

R = Gas constant at the universal level.

 

The relationship between pressure and temperature is directly proportional, if the temperature increases the pressure also increases and vice-versa.

 

Step 2: Numerical Explanation

Temperature,  ${\mathbf{T}}_1\mathbf{=}\mathbf{85}\mathbf{.}{\mathbf{2}}^o\mathbf{C}\mathbf{=}\mathbf{358}\mathbf{.}\mathbf{2}\mathbf{k}$

Vapour pressure at temperature,  ${\mathbf{P}}_{\mathbf{1}}\mathbf{ }\mathbf{=}\mathbf{621}\mathbf{torr}$

Temperature, ${\mathbf{T}}_2\mathbf{=}\mathbf{95}\mathbf{.}{\mathbf{6}}^o\mathbf{C}\mathbf{=}\mathbf{368}\mathbf{.}\mathbf{6}\mathbf{k}$  

Let vapour pressure at temperature,  ${\mathbf{P}}_2\mathbf{=}\mathbf{1}\mathbf{atm}\mathbf{=}\mathbf{760}\mathbf{torr}$

Universal gas constant, $\mathbf{R}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mol}\mathbf{/}\mathbf{k}$ .

Enthalpy of vaporization, ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$ =?

 

Put the values in the above equation of enthalpy of vaporization:

$\begin{array}{c}\mathbf{Log}\left(\frac{P1}{P2}\right)\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{621}}{\mathbf{700}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{358}\mathbf{.2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{368}\mathbf{.6}}\mathbf{)}\\ \mathbf{Log}\left(\mathbf{621}\right)\mathbf{-}\mathbf{Log}\left(\mathbf{700}\right)\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{368}\mathbf{.}\mathbf{6}\mathbf{-}\mathbf{358}\mathbf{.}\mathbf{2}}{\mathbf{358}\mathbf{.}\mathbf{2}\mathbf{\times }\mathbf{368}\mathbf{.}\mathbf{6}}\mathbf{\right)}\\ \mathbf{2}\mathbf{.}\mathbf{79}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{84}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{10}\mathbf{.}\mathbf{4}}{\mathbf{132032}\mathbf{.}\mathbf{5}}\mathbf{\right)}\end{array}$

$\begin{array}{c}\\ \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\left(\frac{10.4}{132032.5}\right)\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{132032}\mathbf{.}\mathbf{5}}{\mathbf{10}\mathbf{.}\mathbf{4}}\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{12154}\mathbf{.}\mathbf{06}\mathbf{Joule}\mathbf{/}\mathbf{mole}\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{15}\mathbf{KJ}\mathbf{/}\mathbf{mole}\end{array}$

The enthalpy of the vaporization of the liquid ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  at temperature ${\mathbf{95}}^o\mathbf{C}$  is 12.15kJ/mole.