Chapter 18, Problem 18.1P

Which of the following reactions has a decrease in entropy $\text{(}\Delta S<0)?$
a. $C{O}_2(s)\to C{O}_2(g)$
b. $I_2(g)\to \text{2 I}(g)$
c. $CaC{O}_3(s)\to CaO(s)+C{O}_2(g)$
d. $A{g}^{+}(aq)+B{r}^{-}(aq)\to AgBr(s)$

Interpretation Introduction

(a)

Interpretation:

Whether the reaction given below represents a decrease in entropy in the forward direction should be determined.

${\text{CO}}_{\text{2}}\left(s\right)\to {\text{CO}}_{\text{2}}\left(g\right)$

Concept introduction:

For phase transition such as solid to gas or liquid to gas the entropy change is positive in the direction that leads to more randomness. The gases are associated with the highest degree of randomness as they have molecules far apart. The least randomness is found in a solid-state that has a closely packed structure of atoms. Liquids have randomness intermediate to its solid-gas counterparts.

Mathematically, when randomness increases, this physically means that $\text{ΔS}<\text{0}$, whereas when randomness decreases $\text{ΔS}>\text{0}$.

Answer to Problem 18.1P

No, $\text{ΔS}$ is not negative for the given reaction.

Explanation of Solution

Given information:

The reaction is given as follows:

${\text{CO}}_{\text{2}}\left(s\right)\to {\text{CO}}_{\text{2}}\left(g\right)$

The reactant is dry ice that is in solid sate while the product represents gaseous ${\text{CO}}_{\text{2}}$. Since the extent of randomness is more in gaseous state than in rigid solid state so the forward reaction shows increase in entropy.

Interpretation Introduction

(b)

Interpretation:

Whether the reaction given below represents a decrease in entropy in the forward direction should be determined.

${\text{I}}_{\text{2}}\left(g\right)\to 2\text{I}\left(g\right)$

Concept introduction:

For phase transition such as solid to gas or liquid to gas the entropy change is positive in the direction that leads to more randomness. The gases are associated with the highest degree of randomness as they have molecules far apart. The least randomness is found in a solid-state that has a closely packed structure of atoms. Liquids have randomness intermediate to its solid-gas counterparts.

Mathematically, when randomness increases, this physically means that $\text{ΔS}<\text{0}$, whereas when randomness decreases $\text{ΔS}>\text{0}$.

Answer to Problem 18.1P

No, $\text{ΔS}$ is not negative for the given reaction.

Explanation of Solution

Given information:

The reaction is given as follows:

${\text{I}}_{\text{2}}\left(g\right)\to 2\text{I}\left(g\right)$

Here the phase transition does not occur however the number of atoms is more in the product side so there is more randomness in the product side so $\text{ΔS}>\text{0}$ for the forward reaction.

Interpretation Introduction

(c)

Interpretation:

Whether the reaction given below represents a decrease in entropy in the forward direction should be determined.

${\text{CaCO}}_3\left(s\right)\to C\text{aO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Concept introduction:

For phase transition such as solid to gas or liquid to gas the entropy change is positive in the direction that leads to more randomness. The gases are associated with the highest degree of randomness as they have molecules far apart. The least randomness is found in a solid-state that has a closely packed structure of atoms. Liquids have randomness intermediate to its solid-gas counterparts.

Mathematically, when randomness increases this physically means that $\text{ΔS}<\text{0}$, whereas when randomness decreases $\text{ΔS}>\text{0}$.

Answer to Problem 18.1P

No, $\text{ΔS}$ is not negative for the given reaction in the forward direction.

Explanation of Solution

Given information:

The reaction is given as follows:

${\text{CaCO}}_3\left(s\right)\to \text{CaO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

It can be observed that the reactant includes only they solid ${\text{CaCO}}_3$ while the products have gaseous ${\text{CO}}_{\text{2}}$ in addition to solid $\text{CaO}$. So the randomness is more on the product side and thus $\text{ΔS}>\text{0}$ in the forward direction.

Interpretation Introduction

(c)

Interpretation:

Whether the reaction given below represents a decrease in entropy in the forward direction should be determined.

${\text{Ag}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\to \text{AgBr}\left(s\right)$

Concept introduction:

Dissolution is phenomenon where the ionic lattice is disrupted and the ion gets solvated. Usually such solvation results in increases in randomness and thus, $\text{ΔS}>\text{0}$. For instance the lattice of ${\text{HgCl}}_{\text{2}}$, $\text{KCl}$, $\text{KBr}$ show $\text{ΔS}>\text{0}$ for their dissolution. However for lattice such as ${\text{CaSO}}_4$ dissolution accompanies a negative change in entropy.

Answer to Problem 18.1P

Yes, $\text{ΔS}<\text{0}$ for the given reaction in the forward direction.

Explanation of Solution

Given information:

The reaction is given as follows:

${\text{Ag}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\to \text{AgBr}\left(s\right)$

It can be observed that reactants include aqueous ions that are hydrated and are in a more disordered state the solid $\text{AgBr}$ lattice in the product side. Thus reaction accompanies a decrease in randomness in the forward direction so the indeed $\text{ΔS}<\text{0}$ for this reaction.