Chapter 10, Problem 54E

(a)

Interpretation Introduction

Interpretation:

The value of $\text{ΔG}°$ , $\text{ΔH}°$ and $\text{ΔS}°$ should be calculated for the following reaction at $25°C$.

  ${\text{CH}}_{\text{4}}{\text{(g)+2O}}_{\text{2}}\to {\text{CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(g)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction, whereas if the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Answer to Problem 54E

  $\Delta G{°}_{298}=-801\text{ kJ/mol}$

  $\Delta S{°}_{298}=-4\text{ J/Kmol}$

  $\Delta H{°}_{298}=-802.5\text{ kJ/mol}$

Explanation of Solution

The given reaction is:

  ${\text{CH}}_{\text{4}}{\text{(g)+2O}}_{\text{2}}\to {\text{CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(g)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{H}}_{\text{2}}\text{O}(g)$ is $-229\text{ kJ/mol}$

The value of standard free energy for ${\text{CO}}_2(g)$ is $\text{-394 kJ/mol}$

The value of standard free energy for ${\text{CH}}_4\text{(g)}$ is $-51\text{ kJ/mol}$

The value of standard free energy for ${\text{O}}_2(g)$ is $\text{0 kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(1\times G{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+2\times G{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))-(1\times G{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+2\times G{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g)}))$

  $\Delta G{°}_{298}=[(-394+2(-229))-(-51-2(0))]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-801\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O}(g)$ is $189\text{ J/Kmol}$

The value of standard entropy for ${\text{CO}}_2(g)$ is $214\text{ J/Kmol}$

The value of standard entropy for ${\text{CH}}_4\text{(g)}$ is $\text{186 kJ/mol}$

The value of standard free entropy for ${\text{O}}_2(g)$ is $205\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(1\times S{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+2\times S{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))-(1\times S{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+2\times S{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g)}))$

  $\Delta S{°}_{298}=[(2\times 189+214)-(186\text{ +(2}\times \text{205)})]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-4\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{H}}_{\text{2}}\text{O}(g)$ is $-242\text{ kJ/mol}$

The value of standard enthalpy for ${\text{CO}}_2(g)$ is $-393.5\text{ kJ/mol}$

The value of standard enthalpy for ${\text{CH}}_4\text{(g)}$ is $-75\text{ kJ/mol}$

The value of standard enthalpy for ${\text{O}}_2(g)$ is $0\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(1\times H{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+2\times H{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))-(1\times H{\text{°}}_{\text{298}}{\text{(CH}}_4\text{(g)})+2\times H{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g)}))$

  $\Delta H{°}_{298}=[(-393.5+2\times (-242)\text{)}-((-75)+2(0))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=-802.5\text{ kJ/mol}$

(b)

Interpretation Introduction

Interpretation:

The value of $\text{ΔG}°$ , $\text{ΔH}°$ and $\text{ΔS}°$ should be calculated for the following reaction at $25°C$.

  $\text{6CO}\text{(g)}{\text{+6H}}_{\text{2}}\text{O(l)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s)+6O}}_{\text{2}}\text{(g)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Answer to Problem 54E

  $\Delta G{°}_{298}=+2875\text{ kJ/mol}$

  $\Delta S{°}_{298}=-262\text{ J/Kmol}$

  $\Delta H{°}_{298}=2802\text{ kJ/mol}$

Explanation of Solution

The given reaction is:

  $\text{6CO}\text{(g)}{\text{+6H}}_{\text{2}}\text{O(l)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s)+6O}}_{\text{2}}\text{(g)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{O}}_2(g)$ is $\text{0 kJ/mol}$

The value of standard free energy for ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}$ is $\text{-911 kJ/mol}$

The value of standard free energy for ${\text{CO}}_2(g)$ is $\text{-394 kJ/mol}$

The value of standard free energy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $-237\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(1\times G{\text{°}}_{\text{298}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)})+6\times G{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g))})-(6\times G{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+6\times G{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))$

  $\Delta G{°}_{298}=[(-911+6(0))-(6\times (-394)+6\times (-237))]\text{ kJ/mol}$

  $\Delta G{°}_{298}=+2875\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{O}}_2(g)$ is $\text{205 J/Kmol}$

The value of standard entropy for ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}$ is $\text{212 J/Kmol}$

The value of standard entropy for ${\text{CO}}_2(g)$ is $\text{214 J/Kmol}$

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $70\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(1\times S{\text{°}}_{\text{298}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)})+6\times S{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g))})-(6\times S{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+6\times S{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))$

  $\Delta S{°}_{298}=[(212+6\times 205))-((6\times 214)\text{+(6}\times \text{70)})]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-262\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{O}}_2(g)$ is $\text{0 kJ/mol}$

The value of standard enthalpy for ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}$ is $\text{-1275 kJ/mol}$

The value of standard enthalpy for ${\text{CO}}_2(g)$ is $-\text{393}\text{.5 kJ/mol}$

The value of standard enthalpy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $-286\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(1\times H{\text{°}}_{\text{298}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)})+6\times H{\text{°}}_{\text{298}}{\text{(O}}_2\text{(g))})-(6\times H{\text{°}}_{\text{298}}{\text{(CO}}_2\text{(g)})+6\times H{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)}))$

  $\Delta H{°}_{298}=[(-1275+6\times (0)\text{)}-(6\times (-393.5)+6\times (-286))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=2802\text{ kJ/mol}$

(c)

Interpretation Introduction

Interpretation:

The value of $\text{ΔG}°$ , $\text{ΔH}°$ and $\text{ΔS}°$ should be calculated for the following reaction at $25°C$.

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6H}}_{\text{2}}\text{O(l)}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(s)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Answer to Problem 54E

  $\Delta G{°}_{298}=-356\text{ kJ/mol}$

  $\Delta S{°}_{298}=-209\text{ J/Kmol}$

  $\Delta H{°}_{298}=-416\text{ kJ/mol}$

Explanation of Solution

The given reaction is:

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6H}}_{\text{2}}\text{O(l)}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(s)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{P}}_4{\text{O}}_{10}(s)$ is $\text{-2698 kJ/mol}$

The value of standard free energy for ${\text{H}}_3{\text{PO}}_{\text{4}}(s)$ is $\text{-1119 kJ/mol}$

The value of standard free energy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $-237\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(4\times G{\text{°}}_{\text{298}}{\text{(H}}_3{\text{PO}}_{\text{4}}(s)))-(6\times G{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)})+1\times G{\text{°}}_{\text{298}}{\text{(P}}_4{\text{O}}_{10}(s)\text{)})$

  $\Delta G{°}_{298}=[(4\times (-1119)-((-2698+6\times (-237))]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-356\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{P}}_4{\text{O}}_{10}(s)$ is $\text{229 J/Kmol}$

The value of standard entropy for ${\text{H}}_3{\text{PO}}_{\text{4}}(s)$ is $\text{110 J/Kmol}$

The value of standard entropy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $70\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(4\times S{\text{°}}_{\text{298}}{\text{(H}}_3{\text{PO}}_{\text{4}}(s)))-(6\times S{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)})+1\times S{\text{°}}_{\text{298}}{\text{(P}}_4{\text{O}}_{10}(s)\text{)})$

  $\Delta S{°}_{298}=[(4\times 110))-((229)\text{+(6}\times \text{70)})]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-209\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{P}}_4{\text{O}}_{10}(s)$ is $-\text{2984 kJ/mol}$

The value of standard enthalpy for ${\text{H}}_3{\text{PO}}_{\text{4}}(s)$ is $\text{-1279 kJ/mol}$

The value of standard enthalpy for ${\text{H}}_{\text{2}}\text{O}(l)$ is $-286\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(4\times H{\text{°}}_{\text{298}}{\text{(H}}_3{\text{PO}}_{\text{4}}(s)))-(6\times H{\text{°}}_{\text{298}}{\text{(H}}_2\text{O(g)})+1\times H{\text{°}}_{\text{298}}{\text{(P}}_4{\text{O}}_{10}(s)\text{)})$

  $\Delta H{°}_{298}=[(4\times (-1279\text{))}-(-2984+6\times (-286))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=-416\text{ kJ/mol}$

(d)

Interpretation Introduction

Interpretation:

The value of $\text{ΔG}°$ , $\text{ΔH}°$ and $\text{ΔS}°$ should be calculated for the following reaction at $25°C$.

  ${\text{HCl(g)+NH}}_3\text{(g)}\to {\text{NH}}_4\text{Cl(s)}$

Concept Introduction:

At standard states, the free energy charge which is responsible for the production of 1 mole of a molecule from its elements is known as standard free energy of formation and it is represented by ${\text{ΔG}}_{\text{f}}^{\text{o}}$.

If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is more than zero, then non-spontaneous will be the reaction whereas If the value of ${\text{ΔG}}_{298}^{\text{o}}$ is less than zero, then spontaneous will be the reaction.

The mathematical expression for ${\text{ΔG}}_{\text{f}}^{\text{o}}$ at standard state is given by:

  ${\text{ΔG}}_{\text{298}}^{\text{o}}\text{=ΔH°-TΔS°}$ or,

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

Answer to Problem 54E

  $\Delta G{°}_{298}=-91\text{ kJ/mol}$

  $\Delta S{°}_{298}=-284\text{ J/Kmol}$

  $\Delta H{°}_{298}=-176\text{ kJ/mol}$

Explanation of Solution

The given reaction is:

  ${\text{HCl(g)+NH}}_3\text{(g)}\to {\text{NH}}_4\text{Cl(s)}$

The mathematical expression for the standard free energy at room temperature is:

  $\Delta G°=\Delta G{°}_{298}={\displaystyle \sum \text{n}}{\text{G°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{G°}}_{\text{298}}\text{(reactants)}$

The value of standard free energy for ${\text{NH}}_{\text{4}}\text{Cl(s)}$ is $\text{-203 kJ/mol}$

The value of standard free energy for ${\text{NH}}_{\text{3}}\text{(g)}$ is $\text{-17 kJ/mol}$

The value of standard free energy for $\text{HCl}(g)$ is $-95\text{ kJ/mol}$

Put the values,

  $\Delta G{°}_{298}=(1\times G{\text{°}}_{\text{298}}{\text{(NH}}_{\text{4}}\text{Cl(s})))-(1\times G{\text{°}}_{\text{298}}{\text{(NH}}_{\text{3}}\text{(g))}+1\times G{\text{°}}_{\text{298}}\text{(HCl}(g)\text{)})$

  $\Delta G{°}_{298}=[(-203)-(-95+(-17))]\text{ kJ/mol}$

  $\Delta G{°}_{298}=-91\text{ kJ/mol}$

The mathematical expression for the standard entropy at room temperature is:

  $\Delta S°=\Delta S{°}_{298}={\displaystyle \sum \text{n}}{\text{S°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{S°}}_{\text{298}}\text{(reactants)}$

The value of standard entropy for ${\text{NH}}_{\text{4}}\text{Cl(s)}$ is $\text{96 J/Kmol}$

The value of standard entropy for ${\text{NH}}_{\text{3}}\text{(g)}$ is $\text{193 J/Kmol}$

The value of standard entropy for $\text{HCl}(g)$ is $187\text{ J/Kmol}$

Put the values,

  $\Delta S{°}_{298}=(1\times S{\text{°}}_{\text{298}}{\text{(NH}}_{\text{4}}\text{Cl(s})))-(1\times S{\text{°}}_{\text{298}}{\text{(NH}}_{\text{3}}\text{(g))}+1\times S{\text{°}}_{\text{298}}\text{(HCl}(g)\text{)})$

  $\Delta S{°}_{298}=[(96))-(193\text{+}187)]\text{ J/Kmol}$

  $\Delta S{°}_{298}=-284\text{ J/Kmol}$

The mathematical expression for the standard enthalpy at room temperature is:

  $\Delta H°=\Delta H{°}_{298}={\displaystyle \sum \text{n}}{\text{H°}}_{\text{298}}\text{(products})-{\displaystyle \sum \text{p}}{\text{H°}}_{\text{298}}\text{(reactants)}$

The value of standard enthalpy for ${\text{NH}}_{\text{4}}\text{Cl(s)}$ is $\text{-314 kJ/mol}$

The value of standard enthalpy for ${\text{NH}}_{\text{3}}\text{(g)}$ is $\text{-46 kJ/mol}$

The value of standard enthalpy for $\text{HCl}(g)$ is $-92\text{ kJ/mol}$

Put the values,

  $\Delta H{°}_{298}=(1\times H{\text{°}}_{\text{298}}{\text{(NH}}_{\text{4}}\text{Cl(s})))-(1\times H{\text{°}}_{\text{298}}{\text{(NH}}_{\text{3}}\text{(g))}+1\times H{\text{°}}_{\text{298}}\text{(HCl}(g)\text{)})$

  $\Delta H{°}_{298}=[(-314\text{)}-(-92+(-46))]\text{ kJ/mol}$

  $\Delta H{°}_{298}=-176\text{ kJ/mol}$