Chapter 18, Problem 22QP

Using data from Appendix 2, calculate $\Delta S_{\text{rxn}}^{º}\text{ and }\Delta S_{\text{surr}}$ for each of the reactions in Problem 18.10 and determine if each reaction is spontaneous at $\text{25ºC}$.

Interpretation Introduction

Interpretation:

The standard entropy change, the entropy change of the surroundingsand the spontaneity of the given reactions are to be determined.

Concept introduction:

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

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

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 the product, and $m$ is the stoichiometric coefficient of the reactant.

Answer to Problem 22QP

Solution:

The standard entropy change of the reaction is $\text{188}\text{.9}\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surroundings is $-284\text{J}/\text{mol}\cdot \text{K}$.

Reaction is not spontaneous.

The standard entropy change of the reaction is $-\text{118}\text{.8}\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surroundings is $148\text{J}/\text{mol}\cdot \text{K}$

Reaction is spontaneous

The standard entropy change of the reaction is $-11.4\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surroundings is $1.23\times {10}^3\text{J}/\text{mol}\cdot \text{K}$

Reaction is spontaneous.

The standard entropy change of the reaction is $\text{115}\text{.2}\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surroundings is $-3.16\times {10}^3\text{J}/\text{mol}\cdot \text{K}$

Reaction is not spontaneous.

Explanation of Solution

a) $2{\text{KClO}}_4\left(s\right)\to 2{\text{KClO}}_3\left(s\right){\text{+O}}_2\left(g\right)$

The entropy change of the universe for this reaction is calculated by using the expression as follows:

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

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

The enthalpy change of the system $\Delta H_{sys}$ is calculated by using the expression as follows:

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

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

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

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 the product, and $m$ is the stoichiometric coefficient of the reactant.

$\Delta H_{rxn}^o=\left(2\times \Delta H_f^o\left[{\text{KClO}}_3\right]+\Delta H_f^o\left[{\text{O}}_2\right]\right)-\left(2\times \Delta H_f^o\left[{\text{KClO}}_4\right]\right)$

From appendix 2, the standard enthalpy changesfor the formation of the substances are as follows:

$\Delta H_f^o\left[{\text{KClO}}_3\left(s\right)\right]=\text{}-391.20\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{KClO}}_4\left(s\right)\right]=-433.46\text{kJ}/\text{mol}$

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(2\right)\times \left(-391.\text{20}\text{kJ}/\text{mol}\right)\right]-\left[2\times \left(-433.46\text{kJ}/\text{mol}\right)\right]\\ =84.52\text{kJ}/\text{mol}\\ \Delta H_{sys}=84.52\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=\frac{\left(-84.52\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =-0.284\text{kJ}/\text{mol}\cdot \text{K}\\ =-284\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surroundings is $-284\text{J}/\text{mol}\cdot \text{K}$

Calculate thestandard entropy change of the given reaction.

$2{\text{KClO}}_4\left(s\right)\to 2{\text{KClO}}_3\left(s\right){\text{+O}}_2\left(g\right)$

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

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

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 the product,

and $m$ is the stoichiometric coefficient of the reactant.

$\Delta S_{rxn}^o=\left(2\times S^0\left[{\text{KClO}}_3\right]+S^o\left[{\text{O}}_2\right]\right)-\left(2\times S^o\left[{\text{KClO}}_4\right]\right)$

From appendix 2, the standard entropy values of the substances are as follows:

$S^o\left[{\text{KClO}}_4\left(s\right)\right]=151.0\text{J}/\text{mol}\cdot \text{K}$

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

$S^o\left[{\text{KClO}}_3\left(s\right)\right]=142.97\text{J}/\text{K}\cdot \text{mol}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(2\right)\times \left(142.97\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{205}\text{.0}\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(2\right)\times \left(151.0\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =188.9\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $\text{188}\text{.9}\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

Substitute the values of $\Delta S_{sys}$ and $\Delta S_{surr}$ in the above expression,

$\begin{array}{l}\Delta S_{univ}=188.9\text{J}/\text{mol}\cdot \text{K}+\left(-284\text{J}/\text{mol}.\text{K}\right)\\ =-\text{95}\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the universe for this reaction is $-\text{95}\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of the reaction, the value of $\Delta S_{univ}$ should be positive.

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

$\Delta S_{univ}\text{is}\text{negative}$

Therefore, the reaction is not spontaneous.

b) ${\text{H}}_2\text{O}\left(g\right)\to {\text{H}}_2\text{O}\left(l\right)$

Calculate the entropy change of the universe for the given reaction.

${\text{H}}_2\text{O}\left(g\right)\to {\text{H}}_2\text{O}\left(l\right)$

The entropy change of the universe for this reaction is calculated by using the expression as follows:

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

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

The enthalpy change of the system $\Delta H_{sys}$ is calculated by using the expression as follows:

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

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

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

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 the product, and $m$ is the stoichiometric coefficient of the reactant.

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

From appendix 2, the standard enthalpy changesfor the formation of the substances are as follows:

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

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(-285.\text{8}\text{kJ}/\text{mol}\right)\right]-\left[\left(-241.8\text{kJ}/\text{mol}\right)\right]\\ =-44.0\text{kJ}/\text{mol}\\ \Delta H_{sys}=-44.0\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(-44.0\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =0.148\text{kJ}/\text{mol}\cdot \text{K}\\ =148\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surroundings is $148\text{J}/\text{mol}\cdot \text{K}$

Calculate thestandard entropy change of the given reaction.

${\text{H}}_2\text{O}\left(g\right)\to {\text{H}}_2\text{O}\left(l\right)$

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

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

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 the product,

and $m$ is the stoichiometric coefficient of the reactant.

$\Delta S_{rxn}^o=\left(S^o\left[{\text{H}}_2\text{O}\left(l\right)\right]\right)-\left(S^o\left[{\text{H}}_2\text{O}\left(g\right)\right]\right)$

From appendix 2, the standard entropy values of the substance are as follows:

$S^o\left[{\text{H}}_2\text{O}\left(g\right)\right]=241.8\text{J}/\text{mol}\cdot \text{K}$

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

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(69.9\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(188.7\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =-118.8\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $-\text{118}\text{.8}\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

Substitute the value of $\Delta S_{sys}$ and $\Delta S_{surr}$ in the above expression,

$\begin{array}{l}\Delta S_{univ}=-118.8\text{J}/\text{mol}\cdot \text{K}+\left(148\text{J}/\text{mol}.\text{K}\right)\\ =29\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the universe for this reaction is $29\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of the reaction, the value of $\Delta S_{univ}$ should be positive.

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

$\Delta S_{univ}\text{is}\text{positive}$

Therefore, the reaction is spontaneous.

c) $\text{2Na}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{NaOH}\left(aq\right){\text{+H}}_2\left(g\right)$

Calculate the entropy change of the universe for the given reaction.

$\text{2Na}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{NaOH}\left(aq\right){\text{+H}}_2\left(g\right)$

The entropy change of the universe for this reaction is calculated by using the expression as follows:

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

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

The enthalpy change of the system $\Delta H_{sys}$ is calculated by using the expression as follows:

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

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

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

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 the product, and $m$ is the stoichiometric coefficient of the reactant.

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

From appendix 2, the standard enthalpy changesfor the formation of the substances are as follows:

$\Delta H_f^o\left[{\text{Na}}^{+}\left(aq\right)\right]=\text{}-239.66\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{OH}}^{-}\left(aq\right)\right]=-229.94\text{kJ}/\text{mol}$

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

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

$\Delta H_f^o\left[{\text{H}}_2\text{O}\left(l\right)\right]=-229.94\text{kJ}/\text{mol}$

Substitute thevalue of the standard enthalpy change for the formation of the substance in the above expression.

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(2\right)\times \left(-239.\text{66}\text{kJ}/\text{mol}\right)+\left(2\right)\times \left(-229.\text{94}\text{kJ}/\text{mol}\right)+\left(0\right)\right]-\left[\left(0\right)+\left(2\right)\times \left(-285.\text{8}\text{kJ}/\text{mol}\right)\right]\\ =-367.6\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(-367.6\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =1.23\text{kJ}/\text{mol}\cdot \text{K}\\ =1.23\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surroundings is $1.23\times {10}^3\text{J}/\text{mol}\cdot \text{K}$

Calculate thestandard entropy change of the given reaction.

$\text{2Na}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2\text{NaOH}\left(aq\right){\text{+H}}_2\left(g\right)$

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

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

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 the product,

and $m$ is the stoichiometric coefficient of the reactant.

$\Delta S_{rxn}^o=\left(2\times S_{}^o\left[{\text{Na}}^{+}\left(aq\right)\right]+2\times S_{}^o\left[{\text{OH}}^{-}\left(aq\right)\right]+S_{}^o\left[{\text{H}}_2\right]\right)-\left(S_{}^o\left[\text{Na}\right]+2\times S_{}^o\left[{\text{H}}_2\text{O}\left(l\right)\right]\right)$

From appendix 2, the standard entropy values of the substances are as follows:

$S^o\left[{\text{Na}}^{+}\left(s\right)\right]=60.25\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[O{H}^{-}\left(aq\right)\right]=-10.5\text{J}/\text{mol}\cdot \text{K}$

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

$S^o\left[\text{Na}\left(s\right)\right]=51.05\text{J}/\text{mol}\cdot \text{K}$

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

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(2\right)\times \left(60.25\text{J}/\text{mol}\cdot \text{K}\right)+2\times \left(-1\text{0}\text{.5}\text{J}/\text{mol}\cdot \text{K}\right)+\left(131.0\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(2\right)\times \left(51.05\text{J}/\text{mol}\cdot \text{K}\right)+\left(2\right)\times \left(69.9\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =-11.4\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $-11.4\text{J}/\text{mol}\cdot \text{K}$

Calculate the entropy change of the universe.

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

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

Substitute the values of $\Delta S_{sys}$ and $\Delta S_{surr}$ in the above expression,

$\begin{array}{c}\Delta S_{univ}=1.23\times {10}^3\text{J}/\text{mol}\cdot \text{K}+\left(-11.4\text{J}/\text{mol}.\text{K}\right)\\ =1.22\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the universe for this reaction is $1.22\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

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

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

$\Delta S_{univ}\text{is}\text{positive}$.

Therefore, the reaction is spontaneous.

d) ${\text{N}}_2\left(g\right)\to \text{N}\left(g\right)$

Calculate the entropy change of the universe for the given reaction.

${\text{N}}_2\left(g\right)\to \text{N}\left(g\right)$

The entropy change of the universe for this reaction is calculated by using the expression as follows:

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

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

The enthalpy change of the system $\Delta H_{sys}$ is calculated by using the expression as follows:

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

The standard enthalpy change of the reaction $\Delta H_{rxn}^o$ is calculated by using theexpression for thebond enthalpy of the element, as follows:

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

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.

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{N}}_2\left(g\right)\right]\right)-\left(2\times \Delta H_f^o\left[\text{N}\left(g\right)\right]\right)$

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

$\Delta H_f^o\left[{\text{N}}_2\left(g\right)\right]=941.4\text{kJ}/\text{mol}$

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(1\right)\times 941.4\text{kJ}/\text{mol}\right]-\left[2\times \left(0\right)\right]\\ =941.4\text{kJ}/\text{mol}\\ \Delta H_{sys}=941.4\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(941.4\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =-3.16\text{kJ}/\text{mol}\cdot \text{K}\\ =-3.16\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surroundings is $-3.16\times {10}^3\text{J}/\text{mol}\cdot \text{K}$

Calculate thestandard entropy change of the given reaction.

${\text{N}}_2\left(g\right)\to \text{N}\left(g\right)$

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

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

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 the product,

and $m$ is the stoichiometric coefficient of the reactant.

$\Delta S_{rxn}^o=\left(2\times S^0\left[\text{N}\left(g\right)\right]\right)-\left(S^o\left[{\text{N}}_2\left(g\right)\right]\right)$

From appendix 2, the standard entropy values of the substances are as follows:

$S^o\left[\text{N}\left(g\right)\right]=153.3\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[{\text{N}}_2\left(g\right)\right]=191.5\text{J}/\text{mol}\cdot \text{K}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(2\right)\times \left(153.3\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(1\right)\times \left(191.5\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =115.2\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $\text{115}\text{.2}\text{J}/\text{mol}\cdot \text{K}$

Calculate the entropy change of the universe.

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

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

Substitute the values of $\Delta S_{sys}$ and $\Delta S_{surr}$ in the above expression,

$\begin{array}{l}\Delta S_{univ}=115.1\text{J}/\text{mol}\cdot \text{K}+\left(-3.16\times {10}^3\text{J}/\text{mol}.\text{K}\right)\\ =-3.04\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the universe for this reaction is $-304\times {10}^3\text{J}/\text{mol}\cdot \text{K}$ .

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

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

$\Delta S_{univ}\text{is}\text{negative}$.

Therefore, the reaction is not spontaneous.