Question

The enthalpy of vaporization of ethanol is $\mathbf{38}\mathbf{.}\mathbf{7}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$at its boiling point (78°C). Determine$\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{y}\mathbf{s}}\mathbf{,}\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{r}}\mathbf{,}\mathit{\Delta }{\mathit{S}}_{\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{v}}$ when 1.00 mole of ethanol is vaporized at 78°Cand 1.00 atm.

Short answer

$\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{y}\mathbf{s}}\mathbf{=}\mathbf{+}\mathbf{110}\mathbf{.}\mathbf{2}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$,$\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{r}}\mathbf{=}\mathbf{-}\mathbf{110}\mathbf{.}\mathbf{2}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathit{K}$,$\Delta S_{univ}=0$when 1.00 mole of ethanol is vaporized at 78°Cand 1.00 atm

Step-by-step solution

Step 1: Introduction to the Concept

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy, spontaneous change occurs.

$\mathit{\Delta }{\mathit{S}}_{\mathbf{u}\mathbf{n}\mathbf{i}\mathbf{v}}\mathbf{=}\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{y}\mathbf{s}}\mathbf{+}\mathit{\Delta }{\mathit{S}}_{\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{r}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The entropy change of the system and its surroundings equal to the entropy change of the cosmos.

The entropy change ($\Delta S_{sys}$) associated with a process involving a change in a substance's state is provided as:

$\Delta S_{sys}=\Delta H_T-----\left(2\right)$

The following is the relationship betweenentropy change and enthalpy change for the surrounding:

$\Delta S_{surr}=-\Delta H_T-----\left(3\right)$

$\Delta S_{surr}=$Entropy change of the environment

$\Delta H=$Enthalpy change

$T=$Temperature

Step 2: Determination of Entropy change

From the given, the below reaction is thedepiction,

$C_2{H}_{5}OH\left(l\right)\to C_2{H}_{5}OH\left(g\right)\Delta {\text{H}}_{vap}=38.7\text{kJ}/mol$

Let’s compute the entropy change at temperature$T={78}^{\circ }C$

$\Delta S_{sys}=\Delta H_{vap}T$

$=(38.7\text{kJ}/mol)\times (78+273\text{K})$

$=+110.2\text{J}/mol-K$

$\Delta S_{surr}=-\Delta HT$

$=(-38.7\text{kJ}/mol)\times (78+273\text{K})$

$=-110.2\text{J}/mol-K$

$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$

$=+110.2\text{J}/mol-K+(-110.2\text{J}/mol-K)$

=0

Therefore$\Delta S_{sys}=+110.2J/mol-K$,$\Delta S_{surr}=-110.2J/mol-K$, $\Delta S_{univ}=0$.

As a conclusion, the entropy change for the system and its surrounds is equal in size but opposite in direction and the universe's entropy is zero.