Chapter 12, Problem 12A.1ST

Interpretation Introduction

Interpretation:

At which temperature the relative proportion of molecules in axial and equatorial conformations in a sample of methylcyclohexane is $0.30$ or $30$ percentage has to be determined.

Concept Introduction:

Conformations of methylcyclohexane:

There are two different conformations for methylcyclohexane molecules which are axial and equatorial conformations. The equatorial conformation is the lower energy conformation. Hence, it is the most stable conformation whereas the axial conformation is the higher energy conformation. Hence, it is the less stable conformation. The energy difference between these two conformations is $6.0\text{kJ}\text{mol}{}^{\text{-1}}$.

Answer to Problem 12A.1ST

At $599.4\text{K}$ the relative proportion of molecules in axial and equatorial conformations in a sample of methylcyclohexane is $0.30$ or $30$ percentage.

Explanation of Solution

At which temperature the relative proportion of molecules in axial and equatorial conformations in a sample of methylcyclohexane is $0.30$ or $30$ percentage can be determined as shown here:

$\frac{{\text{N}}_{\text{a}}}{{\text{N}}_{\text{e}}}\text{=}\text{e}{}^{\text{-}\left({\text{E}}_{\text{a}}{\text{-E}}_{\text{e}}\right)\text{/RT}}\Rightarrow \text{Boltzman}\text{distribution}$

$\begin{array}{c}{\text{N}}_{\text{a}}\text{=}\text{Population}\text{of}\text{axial}\text{confomer}\\ {\text{N}}_{\text{e}}\text{=}\text{Population}\text{of}\text{equatorial}\text{confomer}\\ {\text{E}}_{\text{a}}\text{=}\text{Energy}\text{of}\text{axial}\text{conformer}\\ {\text{E}}_{\text{e}}\text{=}\text{Energy}\text{of}\text{equatorial}\text{conformer}\end{array}$

$\begin{array}{c}\text{R}\text{=}\text{Gas}\text{constant}\text{=}\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature}\\ {\text{E}}_{\text{a}}{\text{-E}}_{\text{e}}\text{=}\text{6}\text{.0}\text{×}{\text{10}}^{\text{3}}\text{J}{\text{mol}}^{\text{-1}}\\ \text{T}\text{=}\text{?}\end{array}$

$\begin{array}{c}\frac{{\text{N}}_{\text{a}}}{{\text{N}}_{\text{e}}}\text{=}\text{e}{}^{\text{-}\left({\text{E}}_{\text{a}}{\text{-E}}_{\text{e}}\right)\text{/RT}}=0.30\\ 0.30=\exp \frac{-\left(6.0\times {10}^3\text{J}{\text{mol}}^{-1}\right)}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{T}\right)}\\ \ln 0.30=\frac{-\left(6.0\times {10}^3\text{J}{\text{mol}}^{-1}\right)}{\left(\text{8}\text{.3145}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{T}\right)}\\ -1.2040=-721.6309\times \frac{\text{1}}{\text{T}}\end{array}$

$\begin{array}{c}\text{T}=\frac{\bcancel{-}721.6309}{\bcancel{-}1.2040}\\ =599.3612\\ \simeq 599.4\text{K}\end{array}$

Therefore, at $599.4\text{K}$ the relative proportion of molecules in axial and equatorial conformations in a sample of methylcyclohexane is $0.30$ or $30$ percentage.

Conclusion

At which temperature the relative proportion of molecules in axial and equatorial conformations in a sample of methylcyclohexane is $0.30$ or $30$ percentage has been determined.