Chapter 7, Problem 111AE

Calculate ∆H° for each of the following reactions, which occur in the atmosphere.

a. ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)+{\text{O}}_3\left(g\right)\to {\text{CH}}_3\text{CHO}\left(g\right)+{\text{O}}_2\left(g\right)$

b. ${\text{O}}_3\left(g\right)\text{+NO}\left(g\right)\to {\text{NO}}_2\left(g\right)+{\text{O}}_2\left(g\right)$

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

d. ${\text{2NO(g)+O}}_2\left(g\right)\to 2{\text{NO}}_2\left(g\right)$

Interpretation Introduction

Interpretation: For given reactions, standard enthalpy change has to be calculated.

Concept introduction

Standard Enthalpy change ( ${\text{ΔH}}^{\text{0}}$): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: ${\text{25}}^{\text{0}}\text{C}$ and one atmosphere pressure.

Answer to Problem 111AE

${\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+O}}_{\text{3(g)}}\to {\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}{\text{+O}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -361kJ}$

Explanation of Solution

Given: Standard enthalpy value for given substance in the reactions are,

Substance and state              $\Delta H^{0_{}}$ ${\text{ΔH}}_{\text{f}}^{\text{0}}\text{kJ}$ / mol

${\text{O}}_{\text{3(g)}}$                                               143 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                          -286 ${\text{NO}}_{\text{(g)}}$                                             90 ${\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}$                                   -166 ${\text{C}}_{\text{2}}\text{H}$                                           52 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$                                       -909 ${\text{O}}_{\text{2(g)}}$                                                0 ${\text{NO}}_{\text{2(g)}}$                                             34 ${\text{SO}}_{\text{3(g)}}$                                            -396

    ${\text{ΔH}}^{\text{0}}$= ${\text{H}}^{\text{0}}{}_{\text{product}}{\text{-H}}_{\text{reactant}}^{\text{0}}$

                = $\text{-166}\text{kJ-}\left[\text{143}\text{kJ+52}\text{kJ}\right]$

                = $\text{-361kJ}$

${\text{C}}_{\text{2}}{\text{H}}_{\text{4(g)}}{\text{+O}}_{\text{3(g)}}\to {\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}{\text{+O}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -361kJ}$

The standard enthalpy change for the reaction can be calculated by enthalpy of product versus enthalpy of reactant. The standard enthalpy values for given substances in a reaction are shown (Table.1). By substituting these values in standard enthalpy change equation the standard enthalpy change for the reaction has calculated as -361 k J .

Interpretation Introduction

Interpretation: For given reactions, standard enthalpy change has to be calculated.

Concept introduction

Standard Enthalpy change ( ${\text{ΔH}}^{\text{0}}$): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: ${\text{25}}^{\text{0}}\text{C}$ and one atmosphere pressure.

Answer to Problem 111AE

${\text{O}}_{\text{3(g)}}{\text{+NO}}_{\text{(g)}}\to {\text{NO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -199 kJ}$

Explanation of Solution

Given: Standard enthalpy value for given substance in the reactions are,

Substance and state              $\Delta H^{0_{}}$ ${\text{ΔH}}_{\text{f}}^{\text{0}}\text{kJ}$ / mole

${\text{O}}_{\text{3(g)}}$                                               143 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                          -286 ${\text{NO}}_{\text{(g)}}$                                             90 ${\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}$                                   -166 ${\text{C}}_{\text{2}}\text{H}$                                           52 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$                                       -909 ${\text{O}}_{\text{2(g)}}$                                                0 ${\text{NO}}_{\text{2(g)}}$                                             34 ${\text{SO}}_{\text{3(g)}}$                                            -396

                    ${\text{ΔH}}^{\text{0}}$= ${\text{H}}^{\text{0}}{}_{\text{product}}{\text{-H}}_{\text{reactant}}^{\text{0}}$

                              $\text{=34 kJ-}\left[\text{(90 kJ)+(143 kJ)}\right]$

$\text{= -199 kJ}$

${\text{O}}_{\text{3(g)}}{\text{+NO}}_{\text{(g)}}\to {\text{NO}}_{\text{2(g)}}{\text{+O}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -199 kJ}$

The standard enthalpy change for the reaction can be calculated by enthalpy of product versus enthalpy of reactant. The standard enthalpy values for given substances in a reaction are shown (Table.1). By substituting these values in standard enthalpy change equation the standard enthalpy change for the reaction has calculated as -199kJ.

Interpretation Introduction

Interpretation: For given reactions, standard enthalpy change has to be calculated.

Concept introduction

Standard Enthalpy change ( ${\text{ΔH}}^{\text{0}}$): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: ${\text{25}}^{\text{0}}\text{C}$ and one atmosphere pressure.

Answer to Problem 111AE

${\text{SO}}_{\text{3(g)}}\text{+H}\text{O}{}_{\text{(l)}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$ ${\text{ΔH}}^{\text{0}}\text{= -227kJ}$

Explanation of Solution

Given: Standard enthalpy value for given substance in the reactions are,

Substance and state              $\Delta H^{0_{}}$ ${\text{ΔH}}_{\text{f}}^{\text{0}}\text{kJ}$ / mole

${\text{O}}_{\text{3(g)}}$                                               143 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                          -286 ${\text{NO}}_{\text{(g)}}$                                             90 ${\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}$                                   -166 ${\text{C}}_{\text{2}}\text{H}$                                           52 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$                                       -909 ${\text{O}}_{\text{2(g)}}$                                                0 ${\text{NO}}_{\text{2(g)}}$                                             34 ${\text{SO}}_{\text{3(g)}}$                                            -396

${\text{ΔH}}^{\text{0}}$= ${\text{H}}^{\text{0}}{}_{\text{product}}{\text{-H}}_{\text{reactant}}^{\text{0}}$

                      $\text{= -909}\text{kJ-}\left[\text{(-396}\text{kJ)+(-286}\text{kJ)}\right]$

                      $\text{= -227kJ}$

                    ${\text{SO}}_{\text{3(g)}}\text{+H}\text{O}{}_{\text{(l)}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$              ${\text{ΔH}}^{\text{0}}\text{= -227kJ}$

The standard enthalpy change for the reaction can be calculated by enthalpy of product versus enthalpy of reactant. The standard enthalpy values for given substances in a reaction are shown (Table.1). By substituting these values in standard enthalpy change equation the standard enthalpy change for the reaction has calculated as -227kJ.

Interpretation Introduction

Interpretation: For given reactions, standard enthalpy change has to be calculated.

Concept introduction

Standard Enthalpy change ( ${\text{ΔH}}^{\text{0}}$): The heat change when molar quantities of reactants as specified by chemical equation to form a product at standard conditions. Standard condition: ${\text{25}}^{\text{0}}\text{C}$ and one atmosphere pressure.

Answer to Problem 111AE

${\text{2NO}}_{\text{(g)}}{\text{+O}}_{\text{2(g)}}\to {\text{2NO}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -112kJ}$

Explanation of Solution

Given: Standard enthalpy value for given substance in the reactions are,

Substance and state              $\Delta H^{0_{}}$ ${\text{ΔH}}_{\text{f}}^{\text{0}}\text{kJ}$ / mole

${\text{O}}_{\text{3(g)}}$                                               143 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                          -286 ${\text{NO}}_{\text{(g)}}$                                             90 ${\text{CH}}_{\text{3}}{\text{CHO}}_{\text{(g)}}$                                   -166 ${\text{C}}_{\text{2}}\text{H}$                                           52 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$                                       -909 ${\text{O}}_{\text{2(g)}}$                                                0 ${\text{NO}}_{\text{2(g)}}$                                             34 ${\text{SO}}_{\text{3(g)}}$                                            -396

${\text{2NO}}_{\text{(g)}}{\text{+O}}_{\text{2(g)}}\to {\text{2NO}}_{\text{2(g)}}$ ${\text{ΔH}}^{\text{0}}\text{= -112 kJ}$

The standard enthalpy change for the reaction can be calculated by enthalpy of product versus enthalpy of reactant. The standard enthalpy values for given substances in a reaction are shown (Table.1). By substituting these values in standard enthalpy change equation the standard enthalpy change for the reaction has calculated as -112kJ.