Question

Calculate the entropy change for the vaporization ofliquid methane and hexane using the following data:

	Boiling point (1 atm)	$\mathit{\Delta }{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}}$
Methane	role="math" localid="1658770380495" $\mathbf{112} \mathbf{K}$	$\mathbf{8}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{KJ}\mathbf{/}\mathbf{mol}$
Hexane	$\mathbf{342} \mathbf{K}$	$\mathbf{28}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{KJ}\mathbf{/}\mathbf{mol}$

Compare the molar volume of gaseous methane at 112 K with that of gaseous hexane at 342 K. How do the differences in molar volume affect the values of $\mathit{\Delta }{\mathit{S}}_{\mathbf{v}\mathbf{a}\mathbf{p}}$ for these liquids?

Short answer

$\mathit{\Delta }{\mathit{S}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{n}\mathbf{e}}\mathbf{=}\mathbf{+}\mathbf{73}\mathbf{.}\mathbf{2}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$, Molar volume $\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{19}\mathit{L}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$

$\mathit{\Delta }{\mathit{S}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{h}\mathbf{e}\mathbf{x}\mathbf{a}\mathbf{n}\mathbf{e}}\mathbf{=}\mathbf{+}\mathbf{84}\mathbf{.}\mathbf{5}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$, Molar volume$\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{1}\mathit{L}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$

Higher molar volume will have higher vaporisation entropy.

Step-by-step solution

Step 1: Introduction to the Concept

• Entropy (S) is a measure of disorder or unpredictability; the higher the degree of disorder, the higher the entropy.

• For a process involving a change in the state of a substance, the entropy change ($\Delta S$) is given as:

$\mathit{\Delta }\mathit{S}\mathbf{=}\frac{\mathbf{\Delta }\mathbf{H}}{\mathbf{T}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

$\mathit{\Delta }\mathit{H}\mathbf{=}$enthalpy change.

$T=$Temperature

Step 2: Given information

Boiling point of methane$=112K$

$\Delta H_{vap}=8.20KJ/mol$

Boiling point of hexane$=342\text{K}$

$\Delta H_{vap}=28.9KJ/mol$

Step 3: Determination of ΔSvap for methane and hexane

From equation (1),

$\mathit{\Delta }{\mathit{S}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{n}\mathbf{e}}\mathbf{=}\frac{\mathbf{\Delta }{\mathbf{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{n}\mathbf{e}}}{{\mathbf{T}}_{\mathbf{b}\mathbf{·}\mathbf{p}\mathbf{t}}}$

$=\frac{{\displaystyle 8.20KJ/mol}}{{\displaystyle 112\text{K}}}$

$=73.2J/mol-K$

$\mathit{\Delta }{\mathit{S}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{h}\mathbf{e}\mathbf{x}\mathbf{a}\mathbf{n}\mathbf{e}}\mathbf{=}\frac{\mathbf{\Delta }{\mathbf{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{,}\mathbf{h}\mathbf{e}\mathbf{x}\mathbf{a}\mathbf{n}\mathbf{e}}}{{\mathbf{T}}_{\mathbf{b}\mathbf{·}\mathbf{p}\mathbf{t}}}$

localid="1658772753783" width="120">=28.9KJ/mol342K

$=84.5J/molK$

Step 4: Determination of molar volume for methane and hexane

A gas's molar volume is the volume (V) occupied by one mole (n) of the gas.Using the ideal gas equation as a guide:

$PV=nRT$

$\frac{V}{n}=\frac{RT}{P}$

For methane:

$\frac{V}{n}=\frac{0.0821\text{Latm/molK}\times 112K}{1\text{atm}}$

$=9.19\text{L/mol}$

For hexane:

$\frac{V}{n}=\frac{0.0821\text{Latm/molK}{\displaystyle \times }{\displaystyle 342}{\displaystyle }{\displaystyle K}}{{\displaystyle 1\text{atm}}}$

$=28.1\text{L/mol}$

The entropy changes for liquid methane and hexane vaporisation are$\mathbf{73}\mathbf{.}\mathbf{2}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$and$\mathbf{84}\mathbf{.}\mathbf{5}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$, respectively. Hexane has a greater molar volume (localid="1658774162633" $\mathbf{28}\mathbf{.}\mathbf{1}\mathit{L}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$) than methane ($\mathbf{9}\mathbf{.}\mathbf{19}\mathit{L}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$). Because an increase in volumeincreases entropy, the gas with a higher molar volume will have higher vaporisation entropy.