Chapter 14, Problem 14.21QP

Calculate ΔS_surr for each of the reactions in Problem 14.15 and determine if each reaction is spontaneous at 25°C.

Interpretation Introduction

Interpretation:

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{\text{sys}}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{\text{sys}}$ can be calculated by the following equation.

${\text{ ΔS}}_{\text{sys}}{\text{= S}}^{\text{°}}{}_{\text{Products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

Where

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Enthalpy is the amount energy absorbed or released in a process.  The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe ${\text{(ΔS}}_{\text{univ}})$.  If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.21QP

The given reaction is found to be a spontaneous process with   ${\text{ΔS}}_{\text{surr}}{\text{=291JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}\text{(Cu)=33}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{S°}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\text{=188}{\text{.7 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(H}}_{\text{2}}\text{) =131}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(CuO)=43}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(Cu)=0}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\text{=-241}{\text{.8 kJmol}}^{\text{-1}}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}\text{)=0}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(CuO)=-155}{\text{.2 kJmol}}^{\text{-1}}\\ \text{T=298K}\end{array}$

Explanation:

The given data in the problem statement is recorded as shown above.

To calculate ${\text{ΔH}}_{\text{system}}$

Explanation:

The ${\text{ΔH}}_{\text{system}}$ is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{products}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(Cu)+ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\text{- }\left[{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{)+ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(CuO)}\right]\\ \text{= 0+(1)(-241}{\text{.8KJmol}}^{\text{-1}}\text{)-}\left[\text{0+(1)(-155}{\text{.2KJmol}}^{\text{-1}}\text{)}\right]\\ \text{= -86}{\text{.6 kJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

Explanation:

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{\text{(-86}{\text{.6KJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 0}{\text{.291 KJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ ={\text{291JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{sys}}$

Explanation:

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the products and reactants  the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{ = S}}^{\text{°}}{}_{\text{products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= S}}^{\text{°}}\text{(Cu)+S°}\left[{\text{H}}_{\text{2}}\text{O(g)}\right]\text{- }\left[{\text{S}}^{\text{°}}{\text{(H}}_{\text{2}}{\text{)+S}}^{\text{°}}\text{(CuO)}\right]\\ \text{=(1)(33}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(188}{\text{.7 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(131}{\text{.0J K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(1)(43}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{=47}{\text{.5JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{univ}}$

Explanation:

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of ${\text{ΔS}}_{\text{sys}}$ and ${\text{ΔS}}_{\text{surr}}$ in the given equation.

${\text{ΔS}}_{\text{univ}}{\text{=ΔS}}_{\text{sys}}{\text{+ΔS}}_{\text{surr}}$

$\begin{array}{l}\text{= 47}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+291JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{=339 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since , ${\text{ΔS}}_{\text{univ}}\text{> 0}$ the given process is spontaneous

Conclusion

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given reaction are calculated and each reactions was identified to be spontaneous.

Interpretation Introduction

Interpretation:

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{\text{sys}}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{\text{sys}}$ can be calculated by the following equation.

${\text{ ΔS}}_{\text{sys}}{\text{= S}}^{\text{°}}{}_{\text{Products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

Where

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Enthalpy is the amount energy absorbed or released in a process.  The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe ${\text{(ΔS}}_{\text{univ}})$.  If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.21QP

The given reaction is found to be a spontaneous process with    ${\text{ΔS}}_{\text{surr}}\text{=2}\text{.10}\times {\text{10}}^3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Explanation

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(Al}}_2{\text{O}}_3\text{)}=\text{50}{\text{.99 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(Zn)}=\text{41}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(Al) = 28}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}\text{(ZnO)}=\text{43}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{) = -1669}{\text{.8 KJmol}}^{\text{-1}}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(Zn) = 0}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(Al)=0}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(ZnO)=-348}{\text{.0 KJmol}}^{\text{-1}}\\ \text{T=298K}\end{array}$

Explanation:

The values are recored as shown above.

To calculate ${\text{ΔH}}_{\text{system}}$

Explanation:

The ${\text{ΔH}}_{\text{system}}$ is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{products}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{)+3ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(Zn)- }\left[{\text{2ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(Al)+3ΔΗ}}^{\text{°}}{}_{\text{f}}\text{(ZnO)}\right]\\ \text{= (1)(-1669}{\text{.8 KJmol}}^{\text{-1}}\text{)+0-}\left[\text{0+(3)(-348}{\text{.0 KJmol}}^{\text{-1}}\text{)}\right]\\ {\text{= - 626 KJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

Explanation:

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{{\text{(- 626 kJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 2}{\text{.10 kJK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{2}\text{.10}\times {\text{10}}^3{\text{JK}}^{-1}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{sys}}$

Explanation:

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the reactants and products  the given equation.

${\text{(ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{) = S}}^{\text{°}}{}_{\text{products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}\text{= }\left[{\text{S}}^{\text{°}}{\text{(Al}}_2{\text{O}}_3{\text{)+3S}}^{\text{°}}\text{(Zn)}\right]\text{-}\left[{\text{2S}}^{\text{°}}{\text{(Al)+3S}}^{\text{°}}\text{(ZnO)}\right]\\ \text{= (1)(50}{\text{.99 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(3)(41}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(2)(28}{\text{.3 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(3)(43}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ \text{= -12}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{univ}}$

Explanation:

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of ${\text{ΔS}}_{\text{sys}}$ and ${\text{ΔS}}_{\text{surr}}$ in the given equation.

${\text{ΔS}}_{\text{univ}}{\text{=ΔS}}_{\text{sys}}{\text{+ ΔS}}_{\text{surr}}$

$\begin{array}{l}=\text{-12}{\text{.5 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+2}\text{.10}\times {\text{10}}^3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{= 2}\text{.09}\times {\text{10}}^3{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Since, ${\text{ΔS}}_{\text{univ}}\text{> 0}$ the given process is spontaneous

Conclusion

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given reaction are calculated and each reactions was identified to be spontaneous.

Interpretation Introduction

Interpretation:

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ${\text{ΔS}}_{\text{sys}}$ is the difference in entropy of the products and reactants.  ${\text{ΔS}}_{\text{sys}}$ can be calculated by the following equation.

${\text{ ΔS}}_{\text{sys}}{\text{= S}}^{\text{°}}{}_{\text{Products}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

Where

${\text{S}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{S}}^{\text{°}}{}_{\text{Products}}$ is the standard entropy of the products

Enthalpy is the amount energy absorbed or released in a process.  The enthalpy change in a system $\text{(Δ}{{\rm H}}_{\text{sys}}\text{)}$ can be calculated by the following equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reactants}}$

Where,

${\text{ΔH}}^{\text{°}}{}_{\text{reactants}}$ is the standard entropy of the reactants

${\text{ΔH}}^{\text{°}}{}_{\text{produdcts}}$ is the standard entropy of the products

Using the value for the change in enthalpy in a system and the temperature, we can calculate ${\text{ΔS}}_{\text{surr}}$

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe ${\text{(ΔS}}_{\text{univ}})$.  If the ${\text{ΔS}}_{\text{univ}}$ is positive, then the reaction will be spontaneous.  If the ${\text{ΔS}}_{\text{univ}}$ is negative then the reaction will be non-spontaneous.

Answer to Problem 14.21QP

(c)  The given reaction is found to be a spontaneous process with   ${\text{ΔS}}_{\text{surr}}\text{=2}{\text{.99×10}}^{\text{3}}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

To record the given data

Given,

$\begin{array}{l}{\text{S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)}=\text{213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \text{S}°{\text{(H}}_2\text{O(}l\text{)) = 69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(CH}}_4\text{)}=\text{186}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{S}}^{\text{°}}{\text{(O}}_2\text{)}=\text{205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

$\begin{array}{l}{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(CO}}_{\text{2}}\text{)=-393}{\text{.5 KJmol}}^{\text{-1}}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{=-285}{\text{.8 KJmol}}^{\text{-1}}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(CH}}_{\text{4}}\text{)=-74}{\text{.85KJmol}}^{\text{-1}}\\ {\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)=0}\\ \text{T=298K}\end{array}$

Explanation:

The given values are recorded as shown above.

To calculate ${\text{ΔH}}_{\text{system}}$

Explanation:

The ${\text{ΔH}}_{\text{system}}$ is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

${\text{ΔH}}_{\text{system}}{\text{ = ΔH}}^{\text{°}}{}_{\text{produdcts}}{\text{- ΔH}}^{\text{°}}{}_{\text{reacttants}}$

$\begin{array}{l}{\text{= ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+2ΔΗ}}^{\text{°}}{}_{\text{f}}\left[{\text{H}}_{\text{2}}\text{O(l)}\right]\text{- }\left[{\text{ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(CH}}_{\text{4}}{\text{)+2ΔΗ}}^{\text{°}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \text{= (1)(-393}{\text{.5 KJmol}}^{\text{-1}}\text{)+(2)(-285}{\text{.8 KJmol}}^{\text{-1}}\text{)-}\left[\text{(1)(-74}{\text{.85 KJmol}}^{\text{-1}}\text{)+0}\right]\\ \text{= -890}{\text{.3 KJmol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{surr}}$

Explanation:

${\text{ΔS}}_{\text{surr}}$ is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

${\text{ΔS}}_{\text{surr}}\text{=-}\frac{{\text{ΔΗ}}_{\text{sys}}}{\text{T}}$

$\begin{array}{l}\text{= -}\frac{\text{(-890}{\text{.3 KJmol}}^{\text{-1}}\text{)}}{\text{298K}}\\ \text{= 2}{\text{.99 KJ K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =\text{2}{\text{.99×10}}^{\text{3}}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{sys}}$

Explanation:

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  Which is calculated by plugging in the values of standard entropy of the reactants and products  the given equation.

${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}{\text{ = S}}^{\text{°}}{}_{\text{produdcts}}{\text{- S}}^{\text{°}}{}_{\text{reactants}}$

$\begin{array}{l}{\text{= S}}^{\text{°}}{\text{(CO}}_{\text{2}}\text{)+2S}°{\text{(H}}_2\text{O(}l\text{)) - }\left[{\text{S}}^{\text{°}}{\text{(CH}}_4{\text{)+2S}}^{\text{°}}{\text{(O}}_2\text{)}\right]\\ =\text{(1)(213}{\text{.6 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(2)(69}{\text{.9 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)-}\left[\text{(1)(186}{\text{.2 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+(2)(205}{\text{.0 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}\right]\\ =-242.8{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate ${\text{ΔS}}_{\text{univ}}$

Explanation:

${\text{ΔS}}_{\text{univ}}$ is calculated by plugging in the values of ${\text{ΔS}}_{\text{sys}}$ and ${\text{ΔS}}_{\text{surr}}$ in the given equation.

${\text{ΔS}}_{\text{univ}}{\text{=ΔS}}_{\text{sys}}{\text{+ΔS}}_{\text{surr}}$

Since, ${\text{ΔS}}_{\text{univ}}\text{>0}$ the given process is spontaneous

Conclusion

The entropy change of surroundings ${\text{(ΔS}}_{\text{surr}}\text{)}$ in the given reaction are calculated and each reactions was identified to be spontaneous.