Chapter 5, Problem 5.1P

(a)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given reaction,

${\text{HCl}}_{\text{(g)}}\text{+}{\text{NH}}_{\text{3(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{HCl}}_{\text{(g)}}\text{+}{\text{NH}}_{\text{3(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{-}\text{314}\text{.43}\right)\text{-}\left(\left(\text{-}\text{92}\text{.31}\right)\text{+}\left(\text{294}\text{.1}\right)\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{-}\text{516}\text{.22}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{HCl}}_{\text{(g)}}\text{+}{\text{NH}}_{\text{3(g)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\text{94}\text{.6}\text{-}\left(\text{186}\text{.91}\text{+}\text{238}\text{.97}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{-}\text{331}\text{.28}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{=}\text{-}\text{516}\text{.22}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\left(\text{298}\text{K}\text{×}\text{-}\text{331}\text{.28}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{=}\text{-}\text{516}\text{.22}\text{kJ}{\text{mol}}^{\text{-1}}\text{+}\text{98}\text{.72}\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{417}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the standard Gibbs energy of the given reaction is $\text{-}\text{417}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given reaction,

$\text{2}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}{\text{Si}}_{\left(\text{s}\right)}\stackrel{}{\to }\text{3}{\text{SiO}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}\text{4}{\text{Al}}_{\left(\text{s}\right)}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}\text{2}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}{\text{Si}}_{\left(\text{s}\right)}\stackrel{}{\to }\text{3}{\text{SiO}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}\text{4}{\text{Al}}_{\left(\text{s}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\left(\text{3}\text{×}\text{-910}\text{.94}\right)\text{+}\left(\text{4}\text{×}\text{0}\right)\right)\text{-}\left(\left(\text{2}\text{×}\text{-}\text{1675}\text{.7}\right)\text{+}\left(\text{3}\text{×}\text{0}\right)\right)\right]{\text{kJmol}}^{\text{-1}}\\ \\ \text{=}\left[\text{-}\text{2732}\text{.82}\text{+}\text{3351}\text{.4}\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{618}\text{.58}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}\text{2}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}\text{3}{\text{Si}}_{\left(\text{s}\right)}\stackrel{}{\to }\text{3}{\text{SiO}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}\text{4}{\text{Al}}_{\left(\text{s}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\left(\text{3}\text{×}\text{41}\text{.84}\right)\text{+}\left(\text{4}\text{×}\text{28}\text{.33}\right)\right)\text{-}\left(\left(\text{2}\text{×}\text{50}\text{.92}\right)\text{+}\left(\text{3}\text{×}\text{18}\text{.83}\right)\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ \\ \text{=}\left[\text{238}\text{.84}\text{-}\text{158}\text{.33}\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{80}\text{.51}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{=}\text{-}\text{618}\text{.58}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\left(\text{298}\text{K}\text{×}\text{80}\text{.51}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{=}\text{-}\text{618}\text{.58}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\text{23}\text{.99}\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{642}\text{.57}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the standard Gibbs energy of the given reaction is $\text{-}\text{642}\text{.57}\text{kJ}{\text{mol}}^{\text{-1}}$.

(c)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given reaction,

${\text{Fe}}_{\text{(s)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{FeS}}_{\left(\text{s}\right)}\text{+}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{Fe}}_{\text{(s)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{FeS}}_{\left(\text{s}\right)}\text{+}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{-}\text{100}\text{.0}\text{+}\text{0}\right)\text{-}\left(\left(\text{0}\right)\text{+}\left(\text{-}\text{20}\text{.63}\right)\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{-}\text{70}\text{.37}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{Fe}}_{\text{(s)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{FeS}}_{\left(\text{s}\right)}\text{+}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{60}\text{.29}\text{+}\text{130}\text{.684}\right)\text{-}\left(\text{27}\text{.28}\text{+}\text{205}\text{.79}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{-}\text{42}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{=}\text{-}\text{70}\text{.37}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\left(\text{298}\text{K}\text{×}\text{-}\text{42}\text{.1}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{=}\text{-}\text{70}\text{.37}\text{kJ}{\text{mol}}^{\text{-1}}\text{+}\text{12}\text{.55}\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{57}\text{.82}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the standard Gibbs energy of the given reaction is $\text{-}\text{57}\text{.82}\text{kJ}{\text{mol}}^{\text{-1}}$.

(d)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given reaction,

${\text{FeS}}_{\text{2}}{}_{\text{(s)}}\text{+}\text{2}{\text{H}}_{\text{2(g)}}\stackrel{}{\to }{\text{Fe}}_{\left(\text{s}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{FeS}}_{\text{2}}{}_{\text{(s)}}\text{+}\text{2}{\text{H}}_{\text{2(g)}}\stackrel{}{\to }{\text{Fe}}_{\left(\text{s}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{0}\text{+}\text{2}\text{×}\text{-}\text{20}\text{.63}\right)\text{-}\left(\text{-}\text{178}\text{.2}\text{+}\left(\text{2}\text{×}\text{0}\right)\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{136}\text{.94}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}{\text{FeS}}_{\text{2}}{}_{\text{(s)}}\text{+}\text{2}{\text{H}}_{\text{2(g)}}\stackrel{}{\to }{\text{Fe}}_{\left(\text{s}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{27}\text{.28}\text{+}\text{2}\text{×}\text{205}\text{.79}\right)\text{-}\left(\text{52}\text{.93}\text{+}\text{2}\text{×}\text{130}\text{.684}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{314}\text{.298}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{=}\text{136}\text{.94}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\left(\text{298}\text{K}\text{×}\text{314}\text{.298}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{=}\text{136}\text{.94}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\text{93}\text{.66}\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{43}\text{.28}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the standard Gibbs energy of the given reaction is $\text{43}\text{.28}\text{kJ}{\text{mol}}^{\text{-1}}$.

(e)

Interpretation Introduction

Interpretation:

For the given reaction, the standard Gibbs energy has to be calculated by calculating the standard reaction enthalpy and entropy.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given reaction,

$\text{2}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{l}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}\text{2}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{l}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{-}\text{813}\text{.99}\text{+}\left(\text{2}\text{×}\text{0}\right)\right)\text{-}\left(\left(\text{2}\text{×}\text{-}\text{187}\text{.78}\right)\text{+}\text{-}\text{20}\text{.63}\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{-}\text{417}\text{.8}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the given reaction,

$\begin{array}{l}\text{2}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\text{+}{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{l}\right)}\text{+}\text{2}{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\left(\text{20}\text{.1}\text{+}\left(\text{2}\text{×}\text{130}\text{.684}\right)\right)\text{-}\left(\left(\text{2}\text{×}\text{109}\text{.6}\right)\text{+}\text{37}\text{.99}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{24}\text{.28}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the reaction Gibbs energy for the given reaction,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{=}\text{-}\text{417}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\left(\text{298}\text{K}\text{×}\text{24}\text{.28}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{=}\text{-}\text{417}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{-}\text{7}\text{.24}\text{kJ}{\text{mol}}^{\text{-1}}\\ \text{=}\text{-}\text{425}\text{.04}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the standard Gibbs energy of the given reaction is $\text{-}\text{425}\text{.04}\text{kJ}{\text{mol}}^{\text{-1}}$.