Chapter 18, Problem 23QP

Using data from Appendix 2, calculate $\Delta S_{\text{rxn}}^{º}\text{ and }\Delta S_{\text{surr}}$ for each of the reactions in Problem 18.11 and determine if each reaction is spontaneous at $\text{25ºC}$.

Interpretation Introduction

Interpretation:

The standard entropy change of the reaction and the entropy change of surroundings, for the given reaction and the spontaneity of the given reaction are to be determined.

Concept introduction:

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.

Answer to Problem 23QP

Solution:

The standard entropy change of the reaction is $-75.6\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surrounding is $185\text{J}/\text{mol}\cdot \text{K}$.

The reaction isspontaneous.

The standard entropy change of the reaction is $\text{215}\text{.8}\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surrounding is $-304\text{J}/\text{mol}\cdot \text{K}$.

The reaction is not spontaneous.

The standard entropy change of the reaction is $\text{98}\text{.2}\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surrounding is $-1.46\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

The reaction is not spontaneous.

The standard entropy change of the reaction is $-282\text{J}/\text{mol}\cdot \text{K}$.

The entropy change of the surrounding is $7.20\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

The reaction is spontaneous.

Explanation of Solution

a) ${\text{PCl}}_3\left(l\right){\text{+Cl}}_2\left(g\right)\to {\text{PCl}}_5\left(g\right)$

Calculate entropy change of the universe for the given reaction.

${\text{PCl}}_3\left(l\right){\text{+Cl}}_2\left(g\right)\to {\text{PCl}}_5\left(g\right)$

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

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

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

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

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

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 enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{PCl}}_5\right]\right)-\left(\Delta H_f^o\left[{\text{PCl}}_3\right]+\Delta H_f^o\left[{\text{Cl}}_2\right]\right)$

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

$\Delta H_f^o\left[{\text{PCl}}_5\left(s\right)\right]=\text{}-374.9\text{kJ}/\text{mol}$

$\Delta H_f^o\left[{\text{PCl}}_3\left(l\right)\right]=-319.7\text{kJ}/\text{mol}$

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(-374.\text{9}\text{kJ}/\text{mol}\right)\right]-\left[\left(-319.7\text{kJ}/\text{mol}\right)+\left(0\right)\right]\\ =-55.2\text{kJ}/\text{mol}\\ \Delta H_{sys}=-55.2\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(-55.2\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =0.185\text{kJ}/\text{mol}\cdot \text{K}\\ =185\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $185\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

${\text{PCl}}_3\left(l\right){\text{+Cl}}_2\left(g\right)\to {\text{PCl}}_5\left(g\right)$

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, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product,

and $m$ is the stoichiometric coefficient of reactant.

$\Delta S_{rxn}^o=\left(S^0\left[{\text{PCl}}_5\right]\right)-\left(S^o\left[{\text{PCl}}_3\right]+S^o\left[{\text{Cl}}_2\right]\right)$

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

$S^o\left[{\text{PCl}}_5\left(g\right)\right]=364.5\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[{\text{Cl}}_2\left(g\right)\right]=223.0\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[{\text{PCl}}_3\left(l\right)\right]=217.1\text{J}/\text{mol}\cdot \text{K}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(364.5\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(223.0\text{J}/\text{mol}\cdot \text{K}\right)+\left(\text{217}\text{.1}\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =-75.6\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $-75.6\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

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

$\begin{array}{l}\Delta S_{univ}=185\text{J}/\text{mol}\cdot \text{K}+\left(-75.6\text{J}/\text{mol}.\text{K}\right)\\ =109\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $109\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of 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 isspontaneous.

b) $\text{2HgO}\left(s\right)\to 2\text{Hg}\left(l\right)+{\text{O}}_2\left(g\right)$

Calculate entropy change of the universe for the given reaction.

$\text{2HgO}\left(s\right)\to 2\text{Hg}\left(l\right)+{\text{O}}_2\left(g\right)$

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

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

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

The enthalpy change of the system $\Delta H_{sys}$ is calculated 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 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 enthalpy change for the formation of the substance, $n$ is the stoichiometric coefficient of product, and $m$ is the stoichiometric coefficient of reactant.

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

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

$\Delta H_f^o\left[\text{HgO}\left(s\right)\right]=\text{}-90.7\text{kJ}/\text{mol}$

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

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

Substitute the standard enthalpy change of the formation value of the substances in the above expression.

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(0\right)+\left[\left(0\right)\right]\right]-\left[\left(-90.7\text{kJ}/\text{mol}\right)\right]\\ =90.7\text{kJ}/\text{mol}\\ \Delta H_{sys}=90.7\text{kJ}/\text{mol}\end{array}$

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

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

Substitute the value of $T$ and $\Delta H_{sys}$ in the above expression.

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(90.7\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =-0.304\text{kJ}/\text{mol}\cdot \text{K}\\ =-304\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $-304\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

$\text{2HgO}\left(s\right)\to 2\text{Hg}\left(l\right)+{\text{O}}_2\left(g\right)$

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, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product,

and $m$ is the stoichiometric coefficient of reactant.

$\Delta S_{rxn}^o=\left(2\times S^o\left[\text{Hg}\left(l\right)\right]+S^o\left[{\text{O}}_2\left(g\right)\right]\right)-\left(2\times S^o\left[\text{HgO}\left(s\right)\right]\right)$

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

$S^o\left[\text{Hg}\left(l\right)\right]=77.4\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[\text{HgO}\left(g\right)\right]=72.0\text{J}/\text{mol}\cdot \text{K}$

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

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[2\times \left(77.4\text{J}/\text{mol}\cdot \text{K}\right)+\left(205.0\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[2\times \left(72.0\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =215.8\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $\text{215}\text{.8}\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

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

$\begin{array}{l}\Delta S_{univ}=215.8\text{J}/\text{mol}\cdot \text{K}+\left(-3\text{04}\text{J}/\text{mol}.\text{K}\right)\\ =-88\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $-88\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of 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.

c) ${\text{H}}_2\left(g\right)\to 2\text{H}\left(g\right)$

Calculate entropy change of the universe for the given reaction.

${\text{H}}_2\left(g\right)\to 2\text{H}\left(g\right)$

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

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

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

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

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

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

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

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

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

From table 8.6, bond enthalpy of the elements are as follows:

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

$\Delta H_f^o\left[{\text{H}}_2\left(g\right)\right]=436.4\text{kJ}/\text{mol}$

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[\left(\text{436}\text{.4}\text{kJ}/\text{mol}\right)\right]-\left[\left(0\right)\right]\\ \Delta H_{system}=\text{436}\text{.4}\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(436.4\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =-1.46\text{kJ}/\text{mol}\cdot \text{K}\\ =-1.46\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $-1.36\times {10}^3\text{J}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

${\text{H}}_2\left(g\right)\to 2\text{H}\left(g\right)$

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, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product,

and $m$ is the stoichiometric coefficient of reactant.

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

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

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

$S^o\left[\text{H}\left(g\right)\right]=114.6\text{J}/\text{mol}\cdot \text{K}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(2\right)\times \left(114.6\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(131.0\text{J}/\text{mol}\cdot \text{K}\right)\right]\\ =98.2\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $\text{98}\text{.2}\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

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

$\begin{array}{l}\Delta S_{univ}=98.2\text{J}/\text{mol}\cdot \text{K}+\left(-1.46\times {10}^3\text{J}/\text{mol}.\text{K}\right)\\ =-1.36\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $-1.36\times {10}^3\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of 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.

d) $\text{U}\left(s\right)+3{\text{F}}_2\left(g\right)\to {\text{UF}}_6\left(s\right)$

Calculate entropy change of the universe for the given reaction.

$\text{U}\left(s\right)+3{\text{F}}_2\left(g\right)\to {\text{UF}}_6\left(s\right)$

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

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

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 using the following expression:

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

The standard enthalpy changes of the reaction $\Delta H_{rxn}^o$ is calculated by using bond enthalpy of the element expression 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 product, and $m$ is the stoichiometric coefficient of reactant.

$\Delta H_{rxn}^o=\left(\Delta H_f^o\left[{\text{UF}}_6\right]\right)-\left(\Delta H_f^o\left[\text{U}\right]+3\times \Delta H_f^o\left[{\text{F}}_2\right]\right)$

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

$\Delta H_f^o\left[{\text{UF}}_6\left(s\right)\right]=-2147\text{kJ}/\text{mol}$

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

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

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

$\begin{array}{l}\Delta H_{rxn}^o=\left[-\left(1\right)\times 2147\text{kJ}/\text{mol}\right]-\left[3\times \left(0\right)+\left(0\right)\right]\\ =-2147\text{kJ}/\text{mol}\\ \Delta H_{sys}=-2147\text{kJ}/\text{mol}\end{array}$

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

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

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

$\begin{array}{l}\Delta S_{surr}=-\frac{\left(-2147\text{kJ}/\text{mol}\right)}{298\text{K}}\\ =7.20\text{kJ}/\text{mol}\cdot \text{K}\\ =7.20\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the entropy change of the surrounding is $7.20\text{kJ}/\text{mol}\cdot \text{K}$.

Calculate the standard entropy change of the given reaction.

$\text{U}\left(s\right)+3{\text{F}}_2\left(g\right)\to {\text{UF}}_6\left(s\right)$

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, $\Delta S^o$ is the standard entropy change of the substance, $n$ is the stoichiometric coefficient of product,

and $m$ is the stoichiometric coefficient of reactant.

$\Delta S_{rxn}^o=\left(S^0\left[{\text{UF}}_6\right]\right)-\left(S^o\left[\text{U}\right]+3\times S^o\left[{\text{F}}_2\right]\right)$

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

$S^o\left[{\text{UF}}_6\left(s\right)\right]=378\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[\text{U}\left(s\right)\right]=50.21\text{J}/\text{mol}\cdot \text{K}$

$S^o\left[{\text{F}}_2\left(g\right)\right]=0\text{J}/\text{mol}\cdot \text{K}$

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

$\begin{array}{l}\Delta S_{rxn}^o=\left[\left(378\text{J}/\text{mol}\cdot \text{K}\right)\right]-\left[\left(1\right)\times \left(50.21\text{J}/\text{mol}\cdot \text{K}\right)+\left[\left(3\right)\times \left(203.34\text{J}/\text{mol}\cdot \text{K}\right)\right]\right]\\ =-282\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, the standard entropy change for this reaction is $-282\text{J}/\text{mol}\cdot \text{K}$.

Calculate the entropy change of the universe.

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

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

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

$\begin{array}{c}\Delta S_{univ}=7.20\times {10}^3\text{J}/\text{mol}\cdot \text{K}+\left(-282\text{J}/\text{mol}.\text{K}\right)\\ =6.92\times {10}^3\text{J}/\text{mol}\cdot \text{K}\end{array}$

Therefore, entropy change of the universe for this reaction is $6.92\times {10}^3\text{J}/\text{mol}\cdot \text{K}$ .

For the spontaneity of 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.