Chapter 18, Problem 21QP

Calculate $\Delta S_{\text{surr}}$ for each of the reactions in Problem 18.15 and determine if each reaction is spontaneous at $\text{25°C}$.

Interpretation Introduction

Interpretation:

The standard entropy changes of surroundings for the given reactions and spontaneity of the given reactions at ${25}_{}^{\text{o}}\text{C}$ is to be determined.

Concept introduction:

Entropy is the direct measurement of randomness or disorder. Entropy is an extensive property and a state function.

The enthalpy of the system is defined as the addition of the internal energy and the product of the pressure and volume. Enthalpy is a state function and an extensive property.

Change in entropy of the system is defined as the difference between the entropy of initial and final state.

The entropy of a system and the entropy of surroundings constitute entropy of the universe.

Enthalpy change of the reaction is the difference between the enthalpies of the reactants and products.

Answer to Problem 21QP

Solution:

a) $291\text{J}/\text{mol}\cdot \text{K}$, spontaneous

b) $2.10\times {10}^3\text{J}/\text{mol}\cdot \text{K}$, spontaneous

c) $2.99\times {10}^3\text{J}/\text{mol}\cdot \text{K}$, spontaneous

Explanation of Solution

a) ${\text{H}}_2\left(g\right)+\text{CuO}\left(s\right)\to \text{Cu}\left(s\right)+{\text{H}}_2\text{O}\left(g\right)$

The entropy change of the universe for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Here, $\Delta H_{system}$ is the enthalpy change of the system and $T$ is the temperature.

The enthalpy change of the system, $\Delta H_{system}$

is calculated by using the following expression:

$\Delta H_{system}=\Delta H_{rxn}^o$

The standard enthalpy change of the reaction, $\Delta H_{rxn}^o$ is calculated by using the following expression:

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[\text{Cu}\right]+\Delta H_f^o\left[{\text{H}}_2\text{O}\right]\right)-\left(\Delta H_f^o\left[{\text{H}}_2\right]+\Delta H_f^o\left[\text{CuO}\right]\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of the product, and $m$ is the stoichiometric coefficient of the reactant.

From appendix 2, the standard enthalpy change of the formation for the substance is as follows:

$\Delta H_f^o\left[\text{Cu}\left(s\right)\right]=\text{}0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=-241.8\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{H}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[\text{CuO}\left(s\right)\right]=-155.2\text{kJ}/\text{mol}$

Substitute the standard enthalpy change of the formation value of the substance in the above expression,

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(1\right)\times \left(-\text{241}\text{.8}\text{kJ}/\text{mol}\right)+\left(1\right)\left(0\right)\right]-\left[\left(1\right)\left(0\right)+\left(1\right)\left(-155.2\text{kJ}/\text{mol}\right)\right]\\ =-86.6\text{k}\text{J}/\text{mol}\end{array}$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Substitute the value of $T$ and $\Delta H_{rxn}^o$ in the above expression,

$\begin{array}{l}\Delta S_{surrounding}=\frac{-\left(-86.6\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =0.291\text{kJ}/\text{mol}\cdot \text{K}\\ =291\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $291\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

${\text{H}}_2\left(g\right)+\text{CuO}\left(s\right)\to \text{Cu}\left(s\right){\text{+H}}_2\text{O}\left(g\right)$

The standard entropy change for this reaction is calculated by using the following expression:

$\Delta S_{rxn}^o={\displaystyle \sum n}{S}^o\left(\text{product}\right)-{\displaystyle \sum m{S}^o}\left(\text{reactant}\right)$

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

$\Delta S_{rxn}^o=\left(S^0\left[\text{Cu}\right]+S^o\left[{\text{H}}_2\text{O}\right]\right)-\left(S^o\left[{\text{H}}_2\right]+S^o\left[\text{CuO}\right]\right)$

From appendix 2, the standard entropy value of the substance is as follows:

$S^o\left[\text{Cu}\left(s\right)\right]=33.3\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=188.7\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{H}}_2\left(g\right)\right]=131.0\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[\text{CuO}\left(s\right)\right]=43.5\text{J}/\text{K}\cdot \text{mol}$

Substitute the standard entropy value of the substance in the above expression,

$\begin{array}{c}\Delta S_{rxn}^o=\left[\left(1\right)\times \left(\text{33}\text{.3}\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\times \left(\text{188}\text{.7}\text{J}/\text{K}\cdot \text{mol}\right)\right]-\\ \left[\left(1\right)\times \left(\text{131}\text{.0}\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\times \left(43.5\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =47.5\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for this reaction is $47.5\text{J}/\text{K}\cdot \text{mol}$.

Calculate the entropy change of the universe.

The entropy change of the universe is calculated by using the following expression:

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surrounding}$

Substitute the value of $\Delta S_{system}$ and $\Delta S_{surrounding}$ in the above expression,

$\begin{array}{l}\Delta S_{universe}=47.5\text{J}/\text{mol}\cdot \text{K}+291\text{J}/\text{mol}.\text{K}\\ \text{ =339}\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $\text{339}\text{J}/\text{K}\cdot \text{mol}$.

For a spontaneous reaction, the value of $\Delta S_{universe}$ should be positive.

The entropy change of the universe for this reaction is positive.

Therefore, the reaction is spontaneous.

b) $\text{2Al}\left(s\right)+3\text{ZnO}\left(s\right)\to {\text{Al}}_2{\text{O}}_3\left(s\right)+\text{Zn}\left(g\right)$

The entropy change of the universe for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Here, $\Delta H_{system}$ is the enthalpy change of the system and $T$ is the temperature.

The enthalpy change of the system, $\Delta H_{system}$, is calculated using the following expression:

$\Delta H_{system}=\Delta H_{rxn}^o$

The standard enthalpy change of the reaction $\Delta H_{rxn}^o$

is calculated using the following expression:

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{Al}}_2{\text{O}}_3\right]+3\times \Delta H_f^o\left[\text{Zn}\right]\right)-\left(2\times \Delta H_f^o\left[\text{Al}\right]+3\times \Delta H_f^o\left[\text{ZnO}\right]\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

From appendix 2, the standard enthalpy change of the formation of the substance is as follows:

$\Delta H_f^o\left[\text{Zn}\left(s\right)\right]=\text{}0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{Al}}_2{\text{O}}_3\left(s\right)\right]=-1669.8\text{kJ}/\text{mol}$

$\Delta H_f^o\left[\text{Al}\left(s\right)\right]=0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[\text{ZnO}\left(g\right)\right]=-348.0\text{kJ}/\text{mol}$

Substitute the standard enthalpy change of the formation value of the substance in the above expression,

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(1\right)\times \left(-1\text{669}\text{.8}\text{kJ}/\text{mol}\right)+\left(0\right)\right]-\left[\left(0\right)+3\times \left(-348.0\text{kJ}/\text{mol}\right)\right]\\ =-626\text{k}\text{J}/\text{mol}\end{array}$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Substitute the value of $T$

and $\Delta H_{rxn}^o$ in the above expression,

$\begin{array}{l}\Delta S_{surrounding}=\frac{\left(-626\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =2.10\text{kJ}/\text{mol}\cdot \text{K}\\ =2.10\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $2.10\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

$\text{2Al}\left(s\right)+3\text{ZnO}\left(s\right)\to {\text{Al}}_2{\text{O}}_3\left(s\right)+3\text{Zn}\left(g\right)$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{rxn}^o=\left(S^0\left[{\text{Al}}_2{\text{O}}_3\right]+3\times S^o\left[\text{Zn}\right]\right)-\left(2\times S^o\left[\text{Al}\right]+3\times S^o\left[\text{ZnO}\right]\right)$

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

From appendix 2, the standard entropy value of the substance is as follows:

$S^o\left[{\text{Al}}_2{\text{O}}_3\left(s\right)\right]=50.99\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[\text{Zn}\left(s\right)\right]=41.6\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[\text{Al}\left(s\right)\right]=28.3\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[\text{ZnO}\left(s\right)\right]=43.9\text{J}/\text{K}\cdot \text{mol}$

Substitute the standard entropy value of the substance in the above expression,

$\begin{array}{c}\Delta S_{rxn}^o=\left[\left(1\right)\times \left(\text{50}\text{.99}\text{J}/\text{K}\cdot \text{mol}\right)+\left(3\right)\times \left(41.6\text{J}/\text{K}\cdot \text{mol}\right)\right]-\\ \left[\left(2\right)\times \left(\text{28}\text{.3}\text{J}/\text{K}\cdot \text{mol}\right)+\left(3\right)\times \left(\text{43}\text{.9}\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =-12.5\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for this reaction is $-12.5\text{J}/\text{K}\cdot \text{mol}$.

The entropy change of the universe is calculated using the following expression:

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surrounding}$

Substitute the value of $\Delta S_{system}$ and $\Delta S_{surrounding}$ in the above expression,

$\begin{array}{l}\Delta S_{universe}=\left(-12.5\text{J}/\text{mol}\cdot \text{K}+2.10\times {10}^3\text{J}/\text{mol}.\text{K}\right)\\ =\text{2}\text{.09}\times {\text{10}}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $2.09\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

For a spontaneous reaction, the value of $\Delta S_{universe}$ should be positive.

The entropy change of the universe for this reaction is positive.

Therefore, the reaction is spontaneous.

c) ${\text{CH}}_4\left(g\right)+2{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The entropy change of the universe for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Here, $\Delta H_{system}$ is the enthalpy change of the system and $T$ is the temperature.

The enthalpy change of the system, $\Delta H_{system}$, is calculated using the following expression:

$\Delta H_{system}=\Delta H_{rxn}^o$

The standard enthalpy change of the reaction $\Delta H_{rxn}^o$

is calculated using the following expression:

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{CO}}_2\right]+2\times \Delta H_f^o\left[{\text{H}}_2\text{O}\left(l\right)\right]\right)-\left(\Delta H_f^o\left[{\text{CH}}_4\right]+2\times \Delta H_f^o\left[{\text{O}}_2\right]\right)$

Here, $\Delta H_{rxn}^o$ is the standard enthalpy change for the reaction, $\Delta H_f^o$ is the standard enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

From appendix 2, the standard enthalpy change of the formation for the substance are as follows:

$\Delta H_f^o\left[{\text{CO}}_2\left(g\right)\right]=\text{}-393.5\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{O}}_2\left(g\right)\right]=0\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{CH}}_4\left(g\right)\right]=-74.85\text{kJ}/\text{mol}$

Substitute the standard enthalpy change of the formation value of the substance in the above expression,

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(1\right)\times \left(-393.\text{5}\text{kJ}/\text{mol}\right)+\left(2\right)\left(-285.8\right)\right]-\left[\left(0\right)+\left(-74.85\text{kJ}/\text{mol}\right)\right]\\ =-890.3\text{k}\text{J}/\text{mol}\end{array}$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$

Substitute the value of $T$

and $\Delta H_{rxn}^o$ in the above expression,

$\begin{array}{l}\Delta S_{surrounding}=\frac{\left(-890.3\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =2.99\text{kJ}/\text{mol}\cdot \text{K}\\ =2.99\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the system is $2.99\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

${\text{CH}}_4\left(g\right)+2{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The standard entropy change for this reaction is calculated using the following expression:

$\Delta S_{rxn}^o=\left(S^0\left[{\text{CO}}_2\right]+\left(2\right)\times S^o\left[{\text{H}}_2\text{O}\right]\right)-\left(S^o\left[{\text{CH}}_4\right]+\left(2\right)\times S^o\left[{\text{O}}_2\right]\right)$

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction, $S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

From appendix 2, the standard entropy value of the substance is as follows:

$S^o\left[{\text{CO}}_2\left(g\right)\right]=213.6\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{H}}_2\text{O}\left(l\right)\right]=69.9\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{O}}_2\left(g\right)\right]=205.0\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[{\text{CH}}_4\left(g\right)\right]=186.2\text{J}/\text{K}\cdot \text{mol}$

Substitute the standard entropy value of the substance in the above expression,

$\begin{array}{c}\Delta S_{rxn}^o=\left[\left(1\right)\times \left(\text{213}\text{.6}\text{J}/\text{K}\cdot \text{mol}\right)+\left(2\right)\times \left(\text{69}\text{.9}\text{J}/\text{K}\cdot \text{mol}\right)\right]-\\ \left[\left(1\right)\times \left(\text{186}\text{.2}\text{J}/\text{K}\cdot \text{mol}\right)+\left(2\right)\times \left(205.0\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =-242.8\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for this reaction is $-242.8\text{J}/\text{K}\cdot \text{mol}$.

Calculate the entropy change of the universe.

The entropy change of the universe is calculated using the following expression:

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surrounding}$

Substitute the value of $\Delta S_{system}$ and $\Delta S_{surrounding}$ in the above expression,

$\begin{array}{l}\Delta S_{universe}=-242.8\text{J}/\text{mol}\cdot \text{K}+2.99\times {10}^3\text{J}/\text{mol}.\text{K}\\ \text{ =2}\text{.75}\times {\text{10}}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $2.75\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

For a spontaneous reaction, the value of $\Delta S_{universe}$ should be positive. The entropy change of the universe for this reaction is positive.

Therefore, the reaction is spontaneous.