Chapter 17, Problem 79A

(a)

Interpretation Introduction

Interpretation: The change in enthalpy during given chemical reaction is to be calculated.

Concept Introduction : The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants.

Answer to Problem 79A

The change in enthalpy for the reaction ${\text{2C(s,graphite)+O}}_{\text{2}}\text{(g)}\to 2\text{CO(g)}$ is $\text{-198 kJ}$

Explanation of Solution

To calculate the heat released during the reaction, we need to know heat of formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(products)-}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [2×H}}^{\text{0}}{}_{\text{f}}{\text{(CO)]-[2×H}}^{\text{0}}{}_{\text{f}}{\text{(C}}_{graphite}{\text{)+1×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(CO)= -99 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(C}}_{\text{graphite}}\text{)= 0 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [2mol×(-99 kJ/mol)]-[1mol×(0kJ/mol)+1mol×(0 kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-198kJ]-[0kJ+0kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-198-0kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= -198 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The change in enthalpy during given chemical reaction is to be calculated.

Concept Introduction :

The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants.

Answer to Problem 79A

The change in enthalpy for the reaction ${\text{2H}}_2{\text{O}}_2\text{(l)}\to 2{\text{H}}_{\text{2}}{\text{O(l)+2O}}_{\text{2}}\text{(g)}$ is $\text{-947}\text{.6 kJ}$

Explanation of Solution

To calculate the heat released during the reaction, we need to know heat of formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(products)-}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [2×H}}^{\text{0}}{}_{\text{f}}{\text{(CO)]-[2×H}}^{\text{0}}{}_{\text{f}}{\text{(C}}_{graphite}{\text{)+1×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_2{\text{O}}_2\text{)= -187}\text{.8 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_2\text{O)= -286 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [2mol×(-286 kJ/mol)+2mol×(0 kJ/mol)]-[2mol×(-187}\text{.8kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-572kJ+0kJ]-[375}\text{.6kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-572kJ-375}\text{.6kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= -947}\text{.6 kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation: The change in enthalpy during given chemical reaction is to be calculated.

Concept Introduction : The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants.

Answer to Problem 79A

The change in enthalpy for the reaction ${\text{4NH}}_3{\text{(g)+5O}}_2\text{(g)}\to {\text{5NO(g)+6H}}_{\text{2}}\text{O(g)}$ is $\text{-1183}\text{.55 kJ}$

Explanation of Solution

To calculate the heat released during the reaction, we need to know heat of formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(products)-}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [5×H}}^{\text{0}}{}_{\text{f}}\text{(NO)+6}\times {\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_2{\text{O)]-[4×H}}^{\text{0}}{}_{\text{f}}{\text{(NH}}_3{\text{)+5×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(NO)= 90}\text{.25 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_2{\text{O}}_{(g)}\text{)= -241}\text{.8 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(NH}}_3{}_{(g)}\text{)= 46}\text{.0 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [5mol×(90}\text{.25kJ/mol)+6mol×(-241}\text{.8kJ/mol)]-[4mol×(46}\text{.0kJ/mol)+5mol×(0kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [451}\text{.25kJ+(-1450}\text{.8kJ)]-[184kJ+(0kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-999}\text{.55kJ]-[184kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= -1183}\text{.55 kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation: The change in enthalpy during given chemical reaction is to be calculated.

Concept Introduction:

The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants.

Answer to Problem 79A

The change in enthalpy for the reaction ${\text{CaCO}}_{\text{3}}\text{(s)}\to {\text{CaO(s)+CO}}_{\text{2}}\text{(g) }$ is $\text{178}\text{.3 kJ}$

Explanation of Solution

To calculate the heat released during the reaction, we need to know heat of formation of all reactants and products.

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(products)-}\Delta {\text{H}}^{\text{0}}{}_{\text{f}}\text{(reactants)]}$

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [1×H}}^{\text{0}}{}_{\text{f}}\text{(CaO)+1}\times {\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CO}}_2{\text{)]-[1×H}}^{\text{0}}{}_{\text{f}}{\text{(CaCO}}_3\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(CaO)= -635}\text{.1 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CO}}_2\text{)= -393}\text{.5 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CaCO}}_3\text{)= -1206}\text{.9 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [1mol×(-635}\text{.1kJ/mol)+1mol×(-393}\text{.5kJ/mol)]-[1mol×(-1206}\text{.9kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-635}\text{.1kJ+(-393}\text{.5kJ)]-[1206}\text{.9kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-1028}\text{.6kJ]-[-1206}\text{.9kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= 178}\text{.3 kJ}\end{array}$