Chapter 9, Problem 80E

(a)

Interpretation Introduction

Interpretation: The value of $\Delta {\text{H}}^{\text{o}}$ for the following reaction needs to be determined:

  ${\text{C}}_2{\text{H}}_4({\text{g) + O}}_3(\text{g})\to {\text{CH}}_3{\text{CHO(g) + O}}_2(\text{g)}$

Concept Introduction:The difference between the sum of enthalpies of the products and the sum of enthalpies of formation of reactants is known as reaction enthalpy. The general formula for reaction enthalpy is:

  $\Delta {\text{H}}_{\text{reaction}}=\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}$

Answer to Problem 80E

  $\Delta {\text{H}}^{\text{o}}=-361\text{ kJ}$

Explanation of Solution

The given reaction is:

  ${\text{C}}_2{\text{H}}_4({\text{g) + O}}_3(\text{g})\to {\text{CH}}_3{\text{CHO(g) + O}}_2(\text{g)}$

The value of standard enthalpy of formation for:

  $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g) = +52 kJ}\\ {\text{O}}_{\text{3}}\text{(g) = +143 kJ}\\ {\text{CH}}_{\text{3}}\text{CHO(g) = -166 kJ}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ}\end{array}$

The value of $\Delta {\text{H}}^{\text{o}}$ for the given reaction is calculated as:

  $\begin{array}{l}\Delta {\text{H}}_{\text{reaction}}=\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = [(-166 kJ) + (0 kJ)] - [(+143 kJ) + (+52 kJ)]}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = -361 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta {\text{H}}^{\text{o}}$ for the following reaction needs to be determined:

  ${\text{O}}_3\text{(g) + NO(g) }\to {\text{NO}}_2(g){\text{ + O}}_2(\text{g)}$

Concept Introduction: The difference between the sum of enthalpies of the products and the sum of enthalpies of formation of reactants is known as reaction enthalpy. The general formula for reaction enthalpy is:

  $\Delta {\text{H}}_{\text{reaction}}=\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}$

Answer to Problem 80E

  $\Delta {\text{H}}^{\text{o}}=-199\text{ kJ}$

Explanation of Solution

The given reaction is:

  ${\text{O}}_3\text{(g) + NO(g) }\to {\text{NO}}_2(g){\text{ + O}}_2(\text{g)}$

The value of standard enthalpy of formation for:

  $\begin{array}{l}\text{NO(g) = +90 kJ}\\ {\text{O}}_{\text{3}}\text{(g) = +143 kJ}\\ {\text{NO}}_2(g)\text{ = +34 kJ}\\ {\text{O}}_{\text{2}}\text{(g) = 0 kJ}\end{array}$

The value of $\Delta {\text{H}}^{\text{o}}$ for the given reaction is calculated as:

  $\begin{array}{l}\Delta {\text{H}}_{\text{reaction}}=\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = [(0) + (+34 kJ)] - [(143 kJ) + (90 kJ)]}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = -199 kJ }\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta {\text{H}}^{\text{o}}$ for the following reaction needs to be determined:

  ${\text{SO}}_3({\text{g) + H}}_2\text{O(l)}\to {\text{H}}_2{\text{SO}}_4(\text{aq)}$

Concept Introduction: The difference between the sum of enthalpies of the products and the sum of enthalpies of formation of reactants is known as reaction enthalpy. The general formula for reaction enthalpy is:

  $\Delta {\text{H}}_{\text{reaction}}=\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}$

Answer to Problem 80E

  $\Delta {\text{H}}^{\text{o}}=-132\text{ kJ}$

Explanation of Solution

The given reaction is:

  ${\text{SO}}_3({\text{g) + H}}_2\text{O(l)}\to {\text{H}}_2{\text{SO}}_4(\text{aq)}$

The value of standard enthalpy of formation for:

  $\begin{array}{l}{\text{SO}}_3(\text{g) = -396 kJ}\\ {\text{H}}_2\text{O(l) = -286 kJ}\\ {\text{H}}_2{\text{SO}}_4(\text{aq) = -814 kJ}\end{array}$

The value of $\Delta {\text{H}}^{\text{o}}$ for the given reaction is calculated as:

  $\begin{array}{l}\Delta {\text{H}}_{\text{reaction}}=\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = [(-814 kJ)] - [(-396 kJ)] + [(-286 kJ)]}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = -132 kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of $\Delta {\text{H}}^{\text{o}}$ for the following reaction needs to be determined:

  $2{\text{NO(g) + O}}_2(\text{g)}\to {\text{2NO}}_2(\text{g)}$

Concept Introduction: The difference between the sum of enthalpies of the products and the sum of enthalpies of formation of reactants is known as reaction enthalpy. The general formula for reaction enthalpy is:

  $\Delta {\text{H}}_{\text{reaction}}=\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Sigma \Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}$

Answer to Problem 80E

  $\Delta {\text{H}}^{\text{o}}=-112\text{ kJ}$

Explanation of Solution

The given reaction is:

  $2{\text{NO(g) + O}}_2(\text{g)}\to {\text{2NO}}_2(\text{g)}$

The value of standard enthalpy of formation for:

  $\begin{array}{l}\text{NO}(\text{g) = 90 kJ}\\ {\text{O}}_2\text{(g) = 0 kJ}\\ {\text{NO}}_{\text{2}}(\text{aq) = +34 kJ}\end{array}$

The value of $\Delta {\text{H}}^{\text{o}}$ for the given reaction is calculated as:

  $\begin{array}{l}\Delta {\text{H}}_{\text{reaction}}=\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{products) - }\Delta {\text{H}}_{\text{f}}^{\text{o}}(\text{reactants)}\\ \Delta {\text{H}}_{\text{reaction}}\text{= [2(34 kJ)] - [2(90 kJ) + 0]}\\ \Delta {\text{H}}_{\text{reaction}}\text{ = -}112\text{ kJ }\end{array}$