Q12.21P - Chemistry: Molecular Nature Of Matter And Change

Question

A liquid has a ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$ of 35.5 kJ/mole and a boiling point of  $\mathbf{122}\mathbf{°}\mathbf{C}$ at 1.00 atm. What is its vapour pressure at $\mathbf{113}\mathbf{°}\mathbf{C}$ ?

Step-by-Step Solution

Answer

The vapour pressure of the liquid is 0.8atm at temperature, $113°C$  as the enthalpy of vaporization at standard condition, ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  is 35.5kJ/mole. 

 

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

${\mathbf{P}}_{\mathbf{1}}$  = Vapour pressure at temperature  ${\mathbf{T}}_{\mathbf{1}}$

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

R = Gas constant at a universal level.

Step 2: Numerical Explanation

Temperature,  ${\mathbf{T}}_1\mathbf{=}{\mathbf{122}}^o\mathbf{C}\mathbf{=}\mathbf{395}\mathbf{k}$

Vapour pressure at temperature,  $\mathbf{P}\mathbf{1}\mathbf{=}\mathbf{1}\mathbf{atm}$

Temperature,  ${\mathbf{T}}_2\mathbf{=}{\mathbf{113}}^o\mathbf{C}\mathbf{=}\mathbf{386}\mathbf{k}$ 

Let vapour pressure at temperature,  ${\mathbf{P}}_2\mathbf{=}\mathbf{‘}\mathbf{x}\mathbf{’}\mathbf{atm}$

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

Enthalpy of vaporization,  ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{5}\mathbf{kJ}\mathbf{/}\mathbf{mole}$

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

  $\begin{array}{c}\mathbf{Log}\left(\frac{P1}{P2}\right)\mathbf{=}\frac{\mathbf{-}\Delta {{\mathbf{H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}\\ \mathbf{Log}\left(\frac{1}{x}\right)\mathbf{=}\frac{\mathbf{-}\mathbf{35}\mathbf{.}\mathbf{5}\mathbf{KJ}\mathbf{/}\mathbf{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{395}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{386}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\frac{-35500J/\mathrm{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{386}\mathbf{-}\mathbf{395}}{\mathbf{395}\mathbf{\times }\mathbf{386}}\mathbf{\right)}\\ \mathbf{Log}\left(\frac{1}{x}\right)\mathbf{=}\mathbf{-}\mathbf{4300}\mathbf{\left(}\frac{\mathbf{-}\mathbf{9}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{152470}}\mathbf{\right)}\end{array}$

$\begin{array}{c}\mathbf{Log}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\\ \left(\frac{1}{x}\right)\mathbf{=}\mathbf{Antilog}(\mathbf{0}\mathbf{.}\mathbf{11})\\ \left(\frac{1}{x}\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\\ \mathbf{x}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{1}\mathbf{.}\mathbf{3}\\ \mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{atm}\end{array}$

The Pressure at temperature ${\mathbf{113}}^o\mathbf{C}$  is 0.8atm.