Chapter 6, Problem 121AE

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

a. C₂H₄(g) + O₃(g) → CH₃CHO(g) + O₂(g)

b. O₃(g) + NO(g) → NO₂(g) + O₂(g)

c. SO₃(g) + H₂O(l) → H₂SO₄(aq)

d. 2NO(g) + O₂(g) → 2SO₂(g)

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 1 atmosphere pressure.

Answer to Problem 121AE

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

Explanation of Solution

Explanation     

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

Standard enthalpy values for given substances in the reaction are shown above.

To calculate standard enthalpy change for given reaction.

The standard enthalpy change for given equation is -361 kJ .

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

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

                = $\text{-361kJ}$

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

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 1 atmosphere pressure.

Answer to Problem 121AE

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

Explanation of Solution

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

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

Standard enthalpy values for given substances in the reaction are shown above.

To calculate standard enthalpy change for given equation.

The standard enthalpy change for given equation is -199kJ.

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

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

$\text{= -199 kJ}$

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

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 1 atmosphere pressure.

Answer to Problem 121AE

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

Explanation of Solution

Explanation:           

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

Standard enthalpy values for given substances in the reaction are shown above.

To calculate standard enthalpy change for given equation.

The standard enthalpy change for given equation is -227kJ .

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

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

                      $\text{= -227kJ}$

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

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 1 atmosphere pressure.

Answer to Problem 121AE

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

Explanation of Solution

Explanation:           

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

Standard enthalpy values for given substances in the reaction are shown above.

To calculate standard enthalpy change for given equation.

The standard enthalpy change for given equation is -112kJ.

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

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 .