Q12.22P - Chemistry: Molecular Nature Of Matter And Change

Question

Diethyl ether has a ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$ of 29.1 kJ/mole and a vapour pressure of 0.703 atm at  $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$ . What is its vapour pressure at $\mathbf{95}\mathbf{°}\mathbf{C}$  ?

Step-by-Step Solution

Answer

The vapour pressure of the liquid is 6.4atm at temperature, $\mathbf{95}\mathbf{°}\mathbf{C}$  as the enthalpy of vaporization at standard conditions, ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  is 29.1kJ/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

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

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

R = Gas constant at a 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,  ${\mathbf{T}}_1\mathbf{=}{\mathbf{25}}^o\mathbf{C}\mathbf{=}\mathbf{298}\mathbf{k}$

Vapour pressure at temperature,  ${\mathbf{P}}_1\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{703}\mathbf{atm}$

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

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,  ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{1}\mathbf{kJ}\mathbf{/}\mathbf{mole}$

 

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

$\begin{array}{c}\mathbf{Log}\mathbf{\left(}\frac{\mathbf{P}\mathbf{1}}{\mathbf{P}\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{29}\mathbf{.}\mathbf{1}\mathbf{KJ}\mathbf{/}\mathbf{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{298}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{366}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\frac{-29100J/\mathrm{mole}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{368}\mathbf{-}\mathbf{298}}{\mathbf{298}\mathbf{\times }\mathbf{368}}\mathbf{\right)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3500}\mathbf{.}\mathbf{12}\mathbf{\left(}\frac{\mathbf{70}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{109664}}\mathbf{\right)}\end{array}$

$\begin{array}{c}\mathbf{Log}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{97}\\ \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{Antilog}(\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{97})\\ \mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\\ \mathbf{x}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{703}}{\mathbf{0}\mathbf{.}\mathbf{11}}\\ \mathbf{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{atm}\end{array}$

The Vapour pressure at temperature  $95°C$ is 6.4atm.