Chapter 14.3, Problem 2PPB

In each of the following reactions, there is one species for which the standard entropy is not listed in Appendix 2. In each case, using the values that are in Appendix 2 and the $\Delta S{°}_{\text{rxn}}$that is given, determine the value of the missing standard entropy at 25°C:

(a)     $\text{K}(s)\to \text{K(}l\text{), }\Delta S{°}_{\text{rxn}}=7.9\text{J/K}\cdot \text{mol}$,

(b)     $2{\text{S(rhombic) + Cl}}_{\text{2}}(g)\to {\text{S}}_{\text{2}}{\text{Cl}}_2(g),\Delta S{°}_{\text{rxn}}=44.74\text{J/K }\cdot \text{mol}$,

(c)     ${\text{O}}_{\text{2}}(g)+2{\text{MgF}}_2(s)\to 2\text{MgO}(s)+2{\text{F}}_2(g),\Delta S{°}_{\text{rxn}}=140.76\text{J/K}\cdot \text{mol}$

Interpretation Introduction

Interpretation:

The missing standard entropy values of the reactants and products has to be calculated for the given set of reactions.

Concept introduction:

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

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

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

Standard entropy change in a reaction and entropy change in the system are same.

Answer to Problem 2PPB

Standard entropy of the product, $\text{K(l)}$ $\text{=71}\text{.5}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Explanation

To record the given data

Given,

$\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{reaction}}=7.9{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}\text{(K(s))}=63.6{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the unknown standard entropy of the product $\text{(K(l))}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  The standard entropy of one of the reactant or product can be calculated by plugging in the known standard entropy values of the products and reactants and the entropy change of the reaction in the given equation.

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

$\begin{array}{l}\text{7}{\text{.9JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{ = }\left[{\text{S}}^{\text{°}}\text{(K(l))}\right]\text{- }\left[{\text{S}}^{\text{°}}\text{(K(s))}\right]\\ \text{7}{\text{.9JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{= }\left[{\text{S}}^{\text{°}}\text{(K(l))}\right]\text{-}\left[\text{63}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\end{array}$

Hence,

$\begin{array}{l}\left[{\text{S}}^{\text{°}}\text{(K(l))}\right]=\text{63}{\text{.6JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+7}{\text{.9JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =71.5{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Conclusion

The missing standard entropy values of the reactants and products has been calculated for the given reaction.

Interpretation Introduction

Interpretation:

The missing standard entropy values of the reactants and products has to be calculated for the given set of reactions.

Concept introduction:

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

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

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

Standard entropy change in a reaction and entropy change in the system are same.

Answer to Problem 2PPB

Standard entropy of the product, ${\text{S}}_2{\text{Cl}}_2\text{(g)}$ $\text{=331}\text{.5}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Explanation

To record the given data

Given,

$\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{reaction}}=44.74{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}{\text{(Cl}}_2\text{(g))}=223.0{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}\text{(S(rhombic))=}\text{31}{\text{.88JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the unknown standard entropy of the product ${\text{S}}_2{\text{Cl}}_2\text{(g)}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  The standard entropy of one of the reactant or product can be calculated by plugging in the known standard entropy values of the products and reactants and the entropy change of the reaction in the given equation.

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

$\begin{array}{l}\text{44}{\text{.74JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{ = }\left[{\text{S}}^{\text{°}}{\text{(S}}_2{\text{Cl}}_2\text{(g))}\right]\text{- }\left[{\text{S}}^{\text{°}}{\text{(Cl}}_2{\text{(g))+S}}^{\text{°}}\text{(S(rhombic))}\right]\\ \text{44}{\text{.74JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{= }\left[{\text{S}}^{\text{°}}{\text{(S}}_2{\text{Cl}}_2\text{(g))}\right]\text{-}\left[223.0{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}+\text{31}{\text{.88JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\end{array}$

Hence,

$\begin{array}{l}\left[{\text{S}}^{\text{°}}{\text{(S}}_2{\text{Cl}}_2\text{(g))}\right]=\text{44}{\text{.74JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}+223.0{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}+\text{31}{\text{.88JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =331.5{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Conclusion

The missing standard entropy values of the reactants and products has been calculated for the given reaction.

Interpretation Introduction

Interpretation:

The missing standard entropy values of the reactants and products has to be calculated for the given set of reactions.

Concept introduction:

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants.  ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$  can be calculated by the following equation.

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

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

Standard entropy change in a reaction and entropy change in the system are same.

Answer to Problem 2PPB

Standard entropy of the reactant, ${\text{MgF}}_{\text{2}}\text{(s)}$, $\text{=57}{\text{.24JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Explanation

To record the given data

Given,

$\begin{array}{l}{\text{ΔS}}^{\text{°}}{}_{\text{reaction}}=140.76{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}{\text{(O}}_2\text{(g))}=205.0{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}{\text{(F}}_2\text{(g))}\text{=}203.34{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ {\text{S}}^{\text{°}}\text{(MgO(s))}\text{=}26.78{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the unknown standard entropy of the reactant ${\text{(MgF}}_{\text{2}}\text{(s))}$

The ${\text{ΔS}}^{\text{°}}{}_{\text{reaction}}$ is the difference in entropy of the reactants and products.  The standard entropy of one of the reactant or product can be calculated by plugging in the known standard entropy values of the products and reactants and the entropy change of the reaction in the given equation.

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

$\begin{array}{l}\text{140}{\text{.76JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{ = 2S}}^{\text{°}}{\text{(MgO(s))+2S}}^{\text{°}}{\text{(F}}_{\text{2}}\text{(g))- }\left[{\text{S}}^{\text{°}}{\text{(O}}_{\text{2}}{\text{(g))+2S}}^{\text{°}}{\text{(MgF}}_{\text{2}}\text{(s))}\right]\\ \text{140}{\text{.76JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{= }\left[\text{(2)}\left(\text{26}{\text{.78JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{+(2)}\left(\text{203}{\text{.34JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\right]\text{-}\left[\text{205}{\text{.0JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}{\text{+2S}}^{\text{°}}{\text{(MgF}}_{\text{2}}\text{(s))}\right]\end{array}$

Hence,

$\begin{array}{l}{\text{2S}}^{\text{°}}{\text{(MgF}}_{\text{2}}\text{(s))=}\left[\text{(2)(26}{\text{.78JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)+}\text{(2)(203}{\text{.34JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})\right]-\left[\text{140}{\text{.76JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+205}{\text{.0JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\\ \text{=114}{\text{.48JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$ ${\text{S}}^{\text{°}}{\text{(MgF}}_{\text{2}}\text{(s))=57}{\text{.24JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

Conclusion

The missing standard entropy values of the reactants and products has been calculated for the given reaction.