Chapter 18, Problem 40GQ

Calculate ΔS°(system), ΔS°(surroundings), and ΔS°(universe) for each of the following processes at 298 K, and comment on how these systems differ.

1. (a) HNO₃(g) → HNO₃(aq)
2. (b) NaOH(s) → NaOH(aq)

Interpretation Introduction

Interpretation:

The entropy change for the system, surroundings and universe for the given reaction should be calculated and commented how this system differs.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{system}\right){\text{=Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change for the reaction.

Answer to Problem 40GQ

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $-119\text{.}98J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $+242.49J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $+122.51J\text{/}K\cdot mol\text{-}rxn$.

Explanation of Solution

The entropy change for the system, surroundings and universe for the given reaction is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

The standard entropy of ${\text{HNO}}_{\text{3}}\left(\text{g}\right)$ is $\text{266}\text{.38 J/K}\cdot \text{mol}$.

The standard entropy of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $\text{146}\text{.4 J/K}\cdot \text{mol}$.

The standard enthalpy of ${\text{HNO}}_{\text{3}}\left(\text{g}\right)$ is $-135.06\text{ kJ/mol}$.

The standard enthalpy of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $-\text{207}\text{.36 kJ/mol}$.

The balanced chemical equation is:

  ${\text{HNO}}_{\text{3}}\left(\text{g}\right)\to {\text{HNO}}_{\text{3}}\left(\text{aq}\right)$

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

$\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{system}\right){\text{=Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ \text{=}\left[\begin{array}{l}\left({\text{1 mol HNO}}_{\text{3}}\left(\text{aq}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{HNO}}_{\text{3}}\left(\text{aq}\right)\right]\text{-}\\ \left({\text{1 mol HNO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{S}}^{\text{°}}\left[{\text{HNO}}_{\text{3}}\left(\text{g}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{system}\right)\text{=}\left[\begin{array}{l}\left({\text{1 mol HNO}}_{\text{3}}\left(\text{aq}\right)\text{/mol-rxn}\right)\left(\text{146}\text{.4 J/K×mol}\right)\text{-}\\ \left({\text{1 mol HNO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{266}\text{.38 J/K×mol}\right)\end{array}\right]\\ \text{=-119}\text{.98 J/K×mol-rxn}\end{array}$

The ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ can be calculated by the following expression,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left({\text{1 mol HNO}}_{\text{3}}\left(\text{aq}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{HNO}}_{\text{3}}\left(\text{aq}\right)\right]\text{-}\\ \left({\text{1 mol HNO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right){\text{Δ}}_{\text{f}}{\text{H}}^{\text{°}}\left[{\text{HNO}}_{\text{3}}\left(\text{g}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left({\text{1 mol HNO}}_{\text{3}}\left(\text{aq}\right)\text{/mol-rxn}\right)\left(\text{-207}\text{.36 kJ/mol}\right)\text{-}\\ \left({\text{1 mol HNO}}_{\text{3}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(\text{-135}\text{.06 kJ/mol}\right)\end{array}\right]\\ \text{=-72}\text{.3 kJ/mol-rxn}\end{array}$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{= }\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}\\ \text{= -}\left[\frac{\text{-72}\text{.3 kJ/mol-rxn}}{\text{298}\text{.15 K}}\right]\left(\text{1000}\frac{\text{J}}{\text{kJ}}\right)\\ \text{= +242}\text{.49 J/K×mol-rxn}\end{array}$

Now,

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)\\ \text{=}\left(\text{-119}\text{.98 J/K×mol-rxn}\right)\text{+}\left(\text{+242}\text{.49 J/K×mol-rxn}\right)\\ \text{=+122}\text{.51 J/K×mol-rxn}\end{array}$

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $-119\text{.}98J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $+242.49J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ for the formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ is $+122.51J\text{/}K\cdot mol\text{-}rxn$.

The given reaction is exothermic. There is more decrease in entropy in formation of ${\text{HNO}}_{\text{3}}\left(\text{aq}\right)$ from ${\text{HNO}}_{\text{3}}\left(\text{g}\right)$ due to decrease in number of moles of gases.

Interpretation Introduction

Interpretation:

The entropy change for the system, surroundings and universe for the given reaction should be calculated and commented how this system differs.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{system}\right){\text{=Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}$

Here, ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ is the enthalpy change for the reaction.

Answer to Problem 40GQ

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $-16.36J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $+144.96J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $+128.6J\text{/}K\cdot mol\text{-}rxn$.

Explanation of Solution

The entropy change for the system, surroundings and universe for the given reaction is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

The standard entropy of $\text{NaOH}\left(\text{s}\right)$ is $\text{64}\text{.46 J/K}\cdot \text{mol}$.

The standard entropy of $\text{NaOH}\left(\text{aq}\right)$ is $\text{48}\text{.1 J/K}\cdot \text{mol}$.

The standard enthalpy of $\text{NaOH}\left(\text{s}\right)$ is $-425.93\text{ kJ/mol}$.

The standard enthalpy of $\text{NaOH}\left(\text{aq}\right)$ is $-\text{469}\text{.15 kJ/mol}$.

The balanced chemical equation is:

  $\text{NaOH}\left(\text{s}\right)\to \text{NaOH}\left(\text{aq}\right)$

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ can be calculated by the following expression,

  $\begin{array}{c}\Delta S^{\circ }\left(\text{system}\right)={\text{Δ}}_{\text{r}}{\text{S}}^{\text{°}}\text{= }\sum {\text{nS}}^{\text{°}}\left(\text{products}\right)\text{-}\sum {\text{nS}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{aq}\right)\text{/mol-rxn}\right)S^{°}\left[\text{NaOH}\left(\text{aq}\right)\right]-\\ \left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right)S^{°}\left[\text{NaOH}\left(\text{s}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values

$\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{system}\right)\text{=}\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{aq}\right)\text{/mol-rxn}\right)\left(\text{48}\text{.1 J/K×mol}\right)\text{-}\\ \left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{64}\text{.46 J/K×mol}\right)\end{array}\right]\\ \text{=-16}\text{.36 J/K×mol-rxn}\end{array}$

The ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}$ can be calculated by the following expression,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{H}}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{aq}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[\text{NaOH}\left(\text{aq}\right)\right]-\\ \left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{°}\left[\text{NaOH}\left(\text{s}\right)\right]\end{array}\right]\end{array}$

Substituting the respective values

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{°}}\text{=}\left[\begin{array}{l}\left(\text{1 mol NaOH}\left(\text{aq}\right)\text{/mol-rxn}\right)\left(\text{-469}\text{.15 kJ/mol}\right)\text{-}\\ \left(\text{1 mol NaOH}\left(\text{s}\right)\text{/mol-rxn}\right)\left(\text{-425}\text{.93 kJ/mol}\right)\end{array}\right]\\ \text{= -43}\text{.22 kJ/mol-rxn}\end{array}$

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ can be calculated by the following expression,

  $\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)\text{=}\frac{{\text{-Δ}}_{\text{r}}{\text{H}}^{\text{o}}}{\text{T}}\\ \text{=-}\left[\frac{\text{-43}\text{.22 kJ/mol-rxn}}{\text{298}\text{.15 K}}\right]\left(\text{1000}\frac{\text{J}}{\text{kJ}}\right)\\ \text{=+144}\text{.96 J/K×mol-rxn}\end{array}$

Now,

$\begin{array}{c}{\text{ΔS}}^{\text{o}}\left(\text{universe}\right){\text{= ΔS}}^{\text{o}}\left(\text{system}\right){\text{+ΔS}}^{\text{o}}\left(\text{surroundings}\right)\\ \text{= }\left(\text{-16}\text{.36 J/K×mol-rxn}\right)\text{+}\left(\text{+144}\text{.96 J/K×mol-rxn}\right)\\ \text{= +128}\text{.6 J/K×mol-rxn}\end{array}$

The ${\text{ ΔS}}^{\text{o}}\left(\text{system}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $-16.36J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{surroundings}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $+144.96J\text{/}K\cdot mol\text{-}rxn$.

The ${\text{ΔS}}^{\text{o}}\left(\text{universe}\right)$ for the formation of $\text{NaOH}\left(\text{aq}\right)$ is $+128.6J\text{/}K\cdot mol\text{-}rxn$.

The given reactions is exothermic. The entropy change for the formation of $\text{NaOH}\left(\text{aq}\right)$ from $\text{NaOH}\left(\text{s}\right)$ should be positive. However, the value is negative due to hydrogen bonding.