Chapter 17, Problem 76A

(a)

Interpretation Introduction

Interpretation: The change in enthalpy using heat of formation 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 76A

The change in enthalpy for the reaction is ${\text{CH}}_4\text{(g)+}\frac{3}{2}{\text{O}}_2\text{(g)}\to {\text{CO(g)+2H}}_{\text{2}}\text{O(l)}$ is $\text{-607}\text{.25 kJ}$

Explanation of Solution

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

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

  ${\text{ΔH}}^{\text{0}}{}_{\text{rxn}}{\text{= [1×H}}^{\text{0}}{}_{\text{f}}\text{(CO)+2}\times {\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_2{\text{O)]-[1×H}}^{\text{0}}{}_{\text{f}}{\text{(CH}}_4\text{)+}\frac{3}{2}{\text{×H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)]}$

  ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(CO)= -110}\text{.5 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{)= -285}\text{.8 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CH}}_4\text{)= -74}\text{.85 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_{\text{2}}\text{)= 0 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [(1mol×- 110}\text{.5kJ/mol)+(2mol× -285}\text{.8kJ/mol)]-[(1mol×-74}\text{.85kJ/mol)+(}\frac{3}{2}\text{mol×0kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-110}\text{.5kJ+(-571}\text{.6kJ)]-[-74}\text{.85kJ+0kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-682}\text{.1kJ-(-74}\text{.85kJ)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= -607}\text{.25 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation: The change in enthalpy using heat of formation 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 76A

The change in enthalpy for the reaction is ${\text{2CO(g)+O}}_{\text{2}}\text{(g) }\to 2{\text{CO}}_{\text{2}}\text{(g)}$ is $\text{-765 kJ}$

Explanation of Solution

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

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

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

  ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(CO}}_2\text{)= -393}\text{.5 kJ/mol}$ , ${\text{H}}^{\text{0}}{}_{\text{f}}\text{(CO)= -110}\text{.5 kJ/mol}$ and ${\text{H}}^{\text{0}}{}_{\text{f}}{\text{(O}}_2\text{)= 0 kJ/mol}$

Putting in the values, we get

  $\begin{array}{l}{\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [2mol×(-393}\text{.5kJ/mol)]-[2mol×(-110}\text{.5kJ/mol)+1mol×(0kJ/mol)]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-787kJ]-[-221kJ+0kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= [-787kJ]-[-221kJ]}\\ {\text{ΔH}}^{\text{0}}{}_{\text{rxn}}\text{= -765 kJ}\end{array}$