Q12.140CP - Chemistry: Molecular Nature Of Matter And Change

Question

The $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}\mathbf{°}$ of gaseous dimethyl ether ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{OCH}}_{\mathbf{3}}$  is -185.4 kJ/mol; the vapor pressure is 1.00 atm at  $\mathbf{-}\mathbf{23}\mathbf{.}\mathbf{7}\mathbf{°}\mathbf{C}$ and 0.526 atm at $\mathbf{-}\mathbf{37}\mathbf{.}\mathbf{8}\mathbf{°}\mathbf{C}$. 

(a) Calculate $\mathbf{∆}{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}}^{\mathbf{°}}$  of dimethyl ether. 

(b) Calculate  $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{°}}$ of liquid dimethyl ether.

Step-by-Step Solution

Answer

(a) The enthalpy of vaporization of dimethyl ether is 22.2kJ/mole.

(b) The enthalpy of formation of the liquid dimethyl ether is higher than the gaseous dimethyl ether.

Definition

The enthalpy of vaporization may be defined as the energy required to convert the liquid into the vapor phase. It is also known as the Enthalpy of evaporation.

 

Clausius- Clapeyron equation

 $ln\frac{P2}{P1}=\frac{\Delta H^{o}vap}{R}\left[\frac{1}{T2}-\frac{1}{T1}\right]$

 

P = Pressure

T = Temperature

R = Gas Constant

${\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}$ = Enthalpy of vaporisation

 

Enthalpy of formation can be defined as forming a new bonding that requires energy and releases energy.

Numerical Explanation for (a)

(a) Vapour pressure, P₁ = 1 atm

Temperature, T₁ = $-23.7^{\mathrm{o}}\mathrm{C}$ = 249.3 k

Vapour pressure, P₂ = 0.526 atm

Temperature = $-37.8^{\mathrm{o}}\mathrm{C}$ = 235.2k

 

The Enthalpy of the vaporization:

$$\begin{gathered} \ln \frac{\mathrm{P}2}{\mathrm{P}1}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right] \\ \Rightarrow \ln \frac{0.526}{1}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{1}{235.2}-\frac{1}{249.3}\right] \\ \Rightarrow \ln \frac{0.526}{1}=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{249.3-235.2}{235.2\times 249.3}\right] \\ \Rightarrow -0.642=\frac{-{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{8.314}\left[\frac{14.1}{58635.36}\right] \\ \Rightarrow {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=\frac{0.642\times 8.314\times 58635.36}{14.1} \\ \Rightarrow {\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}=22.2\mathrm{kJ}/\mathrm{mole} \end{gathered}$$ 

The atomic radius of the body-centered unit cell is $5\times {10}^{-8} \mathrm{m}$

Numerical Explanation for (b)

(b) The enthalpy of formation of liquid is higher than the enthalpy of formation of gaseous because the force of attraction in the liquid is greater than the force of attraction in gaseous. It requires more energy to break the bond between the molecule of liquid than a molecule of gaseous.