Q12.24P - Chemistry: Molecular Nature Of Matter And Change

Question

Methane $\left({\mathbf{CH}}_4\right)$  has a boiling point of  $\mathbf{-}{\mathbf{164}}^o\mathbf{C}$  at 1atm and a vapour pressure of 42.8atm at  $\mathbf{-}{\mathbf{100}}^o\mathbf{C}$ . What is the heat of vaporization of  ${\mathbf{CH}}_4$?

Step-by-Step Solution

Answer

The enthalpy of vaporization at standard condition,  ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$  is 9.195kJ/mole when the vapour pressure is 1atm at temperature $\mathbf{-}{\mathbf{100}}^o\mathbf{C}$ .

 

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,  ${\mathbf{T}}_1\mathbf{=}\mathbf{-}{\mathbf{164}}^o\mathbf{C}\mathbf{=}\mathbf{109}\mathbf{k}$

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

Temperature,   ${\mathbf{T}}_2\mathbf{=}{\mathbf{100}}^o\mathbf{C}\mathbf{=}\mathbf{173}\mathbf{k}$

Let vapour pressure at temperature,  ${\mathbf{P}}_2\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{8}\mathbf{atm}$

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

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

 

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

  $\begin{array}{c}\mathbf{Log}\left(\frac{P1}{P2}\right)\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{R}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{T}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{T}\mathbf{2}}\mathbf{)}\\ \mathbf{Log}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{42}\mathbf{.}\mathbf{8}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{(}\frac{\mathbf{1}}{\mathbf{109}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{173}}\mathbf{)}\\ \mathbf{Log}\left(\mathbf{1}\right)\mathbf{-}\mathbf{Log}(\mathbf{42}\mathbf{.}\mathbf{8})\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{173}\mathbf{-}\mathbf{109}}{\mathbf{173}\mathbf{\times }\mathbf{173}}\mathbf{\right)}\\ \mathbf{0}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{63}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\left(\frac{64}{18857}\right)\end{array}$

$\begin{array}{c}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{63}\mathbf{=}\frac{\mathbf{-}{{\mathbf{\Delta H}}^o}_{\mathrm{vap}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{J}\mathbf{/}\mathbf{mole}\mathbf{/}\mathbf{k}}\mathbf{\left(}\frac{\mathbf{64}}{\mathbf{132032}\mathbf{.}\mathbf{5}}\mathbf{\right)}\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{63}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{18857}}{64}\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{9195}\mathbf{.}\mathbf{69}\mathbf{Joule}\mathbf{/}\mathbf{mole}\\ {{\mathbf{\Delta H}}^o}_{\mathrm{vap}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{195}\mathbf{KJ}\mathbf{/}\mathbf{mole}\end{array}$

The enthalpy of the vaporization of the liquid  ${{\mathbf{\Delta H}}^o}_{\mathrm{vap}}$ at temperature  $\mathbf{-}\mathbf{100}\mathbf{°}\mathbf{C}$  is 9.195kJ/mole.