Chapter 18, Problem 20QP

Calculate $\Delta S_{\text{surr}}$ for each of the reactions in Problem 18.14 and determine if each reaction is spontaneous at $\text{25C}\text{.}$

Interpretation Introduction

Interpretation:

The standard entropy changesin thesurroundings of the given reaction and spontaneity of the given reactionis to be determined.

Concept introduction:

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

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

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

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

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

For a spontaneous reaction, $\Delta {\text{G}}^0$ must be negative, $\Delta {\text{S}}^0$ must be positive, and $\Delta {\text{H}}^0$ must be negative.

Answer to Problem 20QP

Solution:

a)

$1.007\times {10}^3\text{J}/\text{K}\cdot \text{mol}$

The reaction is spontaneous.

b)

$-220\text{J}/\text{K}\cdot \text{mol}$

The reaction is not spontaneous.

c)

$9.9\times {10}^3\text{J}/\text{K}\cdot \text{mol}$ .

The reaction is spontaneous.

Explanation of Solution

a) $\text{S}\left(\text{Rhombic}\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_2\left(g\right)$ at $T={25}^o\text{C}$

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

$\Delta S_{surrounding}=\frac{-\Delta H_{system}}{T}$ …… (1)

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:

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

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

The enthalpy of reaction is as follows:

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

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

$\Delta H_f^o\left[{\text{SO}}_2\left(g\right)\right]=-296.4\text{kJ}/\text{mol}$

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

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

Substitute the standard enthalpy change of the formationvalues of the substancesin the above expression,

$\begin{array}{c}\Delta H_{rxn}^o=\left[\left(1\right)\times \left(-\text{296}\text{.4}\text{kJ}/\text{mol}\right)\right]-\left[\left(0\right)+\left(0\right)\right]\\ \Delta H_{rxn}^o=-296.4\text{k}\text{J}/\text{mol}\\ \Delta H_{system}=-296.4\text{kJ}/\text{mol}\end{array}$

Substitute the values of $T$ and $\Delta H_{rxn}^o$ in equation (1),

$\begin{array}{l}\Delta S_{surrounding}=\frac{\left(-296.4\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =9.95\times {10}^2\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the system is $\text{9}\text{.95}\times {\text{10}}^2\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change for the system.

The entropy change for the system is calculated using the following expression:

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

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction.

Calculate the standard entropy change of the given reaction.

$\text{S}\left(\text{Rhombic}\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_2\left(g\right)$

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

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

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

$S^o\left[{\text{SO}}_2\left(g\right)\right]=248.5\text{J}/\text{K}\cdot \text{mol}$

$S^o\left[\text{S}\left(\text{Rhombic}\right)\right]=31.88\text{J}/\text{K}\cdot \text{mol}$

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

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(1\right)\times \left(\text{248}\text{.5}\text{J}/\text{K}\cdot \text{mol}\right)\right]-\left[\left(1\right)\times \left(\text{205}\text{.0}\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\times \left(31.88\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =11.6\text{J}/\text{K}\cdot \text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{universe}=\left(11.6\text{J}/\text{mol}\cdot \text{K}+9.95\text{J}/\text{mol}.\text{K}\right)\\ =\text{1}\text{.007}\times {\text{10}}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $1.007\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

For a spontaneous of 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) ${\text{MgCO}}_3\left(S\right)\to \text{MgO}\left(S\right)+{\text{CO}}_2\left(g\right)$

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

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

Here, $\Delta H_{system}$ is the enthalpy change of the system, $T$ 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 followingexpression:

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

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.

The enthalpy change for the reaction is as follows:

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[\text{MgO}\right]+\Delta H_f^o\left[{\text{CO}}_2\right]\right)-\left(\Delta H_f^o\left[{\text{MgCO}}_3\right]\right)$

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

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

$\Delta H_f^o\left[\text{MgO}\left(s\right)\right]=-601.8\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{MgCO}}_3\left(s\right)\right]=-1112.9\text{kJ}/\text{mol}$

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(1\right)\times \left(-601.\text{8}\text{kJ}/\text{mol}\right)+\left(-393\text{kJ}/\text{mol}\right)\right]-\left[\left(-1112.9\text{kJ}/\text{mol}\right)\right]\\ =117.6\text{k}\text{J}/\text{mol}\end{array}$

$\Delta H_{system}=117.6\text{kJ}/\text{mol}$

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

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

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

$\begin{array}{l}\Delta S_{surrounding}=\frac{\left(-117.6\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =-.395\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the system is $-0.395\text{J}/\text{mol}\cdot \text{K}$

Calculate the entropy change for the system.

The entropy change for the system is calculated using the following expression:

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

Here, $\Delta S_{rxn}^o$ is the standard entropy change for the reaction.

Calculate the standard entropy change of the given reaction.

${\text{MgCO}}_3\left(S\right)\to \text{MgO}\left(S\right)+{\text{CO}}_2\left(g\right)$

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

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

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.

$\Delta S_{rxn}^o=\left(S^0\left[\text{MgO}\right]+S^o\left[{\text{CO}}_2\right]\right)-\left(S^o\left[{\text{MgCO}}_3\right]\right)$

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

$S^o\left[\text{MgO}\left(S\right)\right]=26.78\text{J}/\text{K}\cdot \text{mol}$

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

$S^o\left[{\text{MgCO}}_3\left(g\right)\right]=65.69\text{J}/\text{K}\cdot \text{mol}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(1\right)\times \left(\text{26}\text{.78}\text{J}/\text{K}\cdot \text{mol}\right)+\left(1\right)\times \left(213.6\text{J}/\text{K}\cdot \text{mol}\right)\right]-\left[\left(1\right)\times \left(\text{65}\text{.69}\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =174.7\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for this reaction is $174.7\text{J}/\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

The entropy change of the universe is calculate 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,

$\begin{array}{l}\Delta S_{universe}=174.7\text{J}/\text{mol}\cdot \text{K}+\left(-395\text{J}/\text{mol}\cdot \text{K}\right)\\ =-220\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $-220\text{J}/\text{mol}\cdot \text{K}$ .

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

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

Therefore, the reaction is not spontaneous.

c) ${\text{2C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)$

Calculate entropy change of the universe for the given reaction.

${\text{2C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)$

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 $\Delta H_{system}$ iscalculated 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 followingexpression:

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

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.

$\Delta H_{rxn}^o=\left(\left(4\right)\times \Delta H_f^o\left[{\text{CO}}_2\right]+\left(6\right)\times \Delta H_f^o\left[{\text{H}}_2\text{O}\right]\right)-\left(\left(2\right)\times \Delta H_f^o\left[C_2{H}_6\right]+\left(7\right)\times \Delta H_f^o\left[{\text{O}}_2\right]\right)$

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

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

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

$\Delta H_f^o\left[{\text{C}}_2{\text{H}}_6\left(\right)\right]=-84.7\text{k}\text{J}/\text{mol}$

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(4\right)\times \left(-393.\text{5}\text{kJ}/\text{mol}\right)+\left(6\right)\left(-285.8\text{kJ}/\text{mol}\right)\right]-\left[\left(2\right)\times \left(-84.7\text{kJ}/\text{mol+0}\right)\right]\\ =-3.119\times {10}^3\text{k}\text{J}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surrounding}=\frac{\left(-3119\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =1.05\times {10}^4\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the system is $1.05\times {10}^4\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

${\text{2C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)$

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

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

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

$\Delta S_{rxn}^o=\left(\left(4\right)\times S^0\left[{\text{CO}}_2\right]+\left(6\right)\times S^o\left[{\text{H}}_2\text{O}\right]\right)-\left(\left(2\right)\times S^o\left[C_2{H}_6\right]+\left(7\right)\times S^o\left[{\text{O}}_2\right]\right)$

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

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

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

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

$S^o\left[{\text{C}}_2{\text{H}}_6\left(g\right)\right]=229.5\text{J}/\text{K}\cdot \text{mol}$

Substitute the values of standard entropy value of the substances in the above expression,

$\begin{array}{c}\Delta S_{rxn}^o=\left[\left(4\right)\times \left(\text{213}\text{.6}\text{J}/\text{K}\cdot \text{mol}\right)+\left(6\right)\times \left(\text{69}\text{.9}\text{J}/\text{K}\cdot \text{mol}\right)\right]-\\ \left[\left(2\right)\times \left(\text{229}\text{.5}\text{J}/\text{K}\cdot \text{mol}\right)+\left(7\right)\times \left(205.0\text{J}/\text{K}\cdot \text{mol}\right)\right]\\ =-620.2\text{J}/\text{K}\cdot \text{mol}\end{array}$

Therefore, the standard entropy change for this reaction is $-620.2\text{J}/\text{mol}\cdot \text{K}$.

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

$\begin{array}{l}\Delta S_{universe}=-620.3\text{J}/\text{mol}\cdot \text{K}+1.05\times {10}^4\text{J}/\text{mol}.\text{K}\\ \text{ =9}\text{.9}\times {\text{10}}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $9.9\times {10}^3\text{J}/\text{mol}\cdot \text{K}$ .

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.