Question

The enthalpy of vaporization of chloroform (${\text{CHCI}}_3$) is 31.4KJ/moleat its boiling point (61.7°C).Determine$\Delta S_{sys},\Delta S_{surr},and\Delta S_{univ}$ when 1.00 mole of chloroform isvaporized at 61.7°C and 1.00 atm.

Short answer

The entropy change for the system and its surroundings is identical in magnitudebut opposite in direction, resulting in a zero entropy universe.

Step-by-step solution

Introduction to the Concept

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

According to the second rule of thermodynamics, the entropy of the cosmos always rises. The entropy change of the universe $\Delta S_{univ}$is calculated as the sum of the entropy change of the system $\Delta S_{sys}$ and the environment $\Delta S_{surr}$.

$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}-----\left(1\right)$

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

$\Delta S_{sys}=\frac{\Delta H}{T}-----\left(2\right)$

$∆H$=enthalpy change.

T = temperature.

The entropy change of the environment ($\Delta S_{surr}$) is calculated as follows:

$\Delta S_{surr}=-\frac{\Delta H}{T}-----\left(3\right)$

Solution Explanation

From the given reaction,

$CHC{l}_3(l)\to CHC{l}_3(g)∆H=31.4kJ/mol$

Let’s compute the entropy change of the systemat $T=61.7°C$,

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

$=\frac{31.4\text{kJ}}{61.7+272\text{K}}$

$=+93.8J/K$

$\Delta S_{surr}=-\frac{\Delta H_{vap}}{T}$

$=-\frac{31.4\text{kJ}}{334.7\text{K}}$

$=-93.8J/K$

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

$\Delta S_{univ}=0$