Chapter 6, Problem 79E

Use the values of $\Delta H_{\text{f}}^{°}$ in Appendix 4 to calculate ∆Hº for the following reactions.

a.

b. ${\text{Ca}}_3\left({\text{PO}}_4\right)\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_4\left(l\right)\to 3{\text{CaSO}}_4\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(l\right)$

c. ${\text{NH}}_3\left(g\right)+\text{HCl}\left(g\right)\to {\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

Interpretation Introduction

Interpretation: Standard enthalpy change has calculated for given reaction.

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 79E

${\text{2NH}}_{\text{3}\left(\text{g}\right)}{\text{+3O}}_{\text{2}\left(\text{g}\right)}{\text{+ 2CH}}_{\text{4}\left(\text{g}\right)}\to {\text{2HCN}}_{\left(\text{g}\right)}{\text{+ 6H}}_{\text{2}}^{}{\text{O}}_{\left(\text{g}\right)\text{.}}$ ${\text{ΔH}}^{\text{0}}\text{=-940kJ}$

Explanation of Solution

Given data

Standard state for given compound in the reaction are,

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

${\text{NH}}_{\text{3(g)}}$                                   -46 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CH}}_{\text{4}}{}_{\text{(g)}}$                                   -75 ${\text{HCN}}_{\text{(g)}}$                                 -135.1 ${\text{H}}_{\text{2}}\text{O(g)}$                                -242 ${\text{Ca(PO}}_{\text{4}}{\text{)}}_{\text{2(s)}}$                           -4126${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}$                               -814 ${\text{CaSO}}_{\text{4(s)}}$                              -1433 ${\text{H}}_{\text{3}}\text{PO}$                              -1267 ${\text{HCl}}_{\text{(g)}}$                                   -92 $\begin{array}{l}{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}\hfill \end{array}$                               -314

To calculate standard enthalpy change.

 The balanced equation is,

${\text{2NH}}_{\text{3}\left(\text{g}\right)}{\text{+3O}}_{\text{2}\left(\text{g}\right)}{\text{+ 2CH}}_{\text{4}\left(\text{g}\right)}\to {\text{2HCN}}_{\left(\text{g}\right)}{\text{+ 6H}}_{\text{2}}^{}{\text{O}}_{\left(\text{g}\right)\text{.}}$            ${\text{ΔH}}^{\text{0}}\text{=-940kJ}$

In generally,

${\text{ΔH}}^{\text{0}}$= ${\displaystyle \sum {\text{n}}_{\text{p}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{product}}-{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{reactant}}$

Here,

${\text{n}}_{\text{p}}$ - Number of moles of product                   

${\text{n}}_{\text{r}}$- Number of moles of reactant

                    =$\left[\text{2(135}\text{.1kJ)+6(-242}\text{kJ)}\right]\text{-}\left[\text{2(-46)+2(-75)}\right]$

                    =$\text{-940}\text{kJ/mol}$

Interpretation Introduction

Interpretation: Standard enthalpy change has calculated for given reaction.

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 79E

${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}\left(\text{s}\right)}{\text{+3H}}_{\text{2}}{\text{SO}}_{\text{4}\left(\text{l}\right)}\to {\text{3CaSO}}_{\text{4}\left(\text{s}\right)}{\text{+2H}}_{\text{3}}^{}{\text{PO}}_{\text{4}\left(\text{l}\right)}$ ${\text{ΔH}}^{\text{0}}\text{=-265kJ}$

Explanation of Solution

Given data

Standard state for given compound in the reaction are,

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

${\text{NH}}_{\text{3(g)}}$                                   -46 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CH}}_{\text{4}}{}_{\text{(g)}}$                                   -75 ${\text{HCN}}_{\text{(g)}}$                                 -135.1 ${\text{H}}_{\text{2}}\text{O(g)}$                                -242 ${\text{Ca(PO}}_{\text{4}}{\text{)}}_{\text{2(s)}}$                           -4126${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}$                               -814 ${\text{CaSO}}_{\text{4(s)}}$                              -1433 ${\text{H}}_{\text{3}}\text{PO}$                              -1267 ${\text{HCl}}_{\text{(g)}}$                                   -92 $\begin{array}{l}{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}\hfill \end{array}$                               -314

The standard state of ammonia gas, oxygen, methane, hydrogen cyanide and water vapour are given. By substituting the values in the standard enthalpy change equation the standard enthalpy change for the reaction calculated as $\text{-940}\text{kJ/mol}\text{.}$

To calculate standard enthalpy change.

The balanced equation is,

         ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}\left(\text{s}\right)}{\text{+3H}}_{\text{2}}{\text{SO}}_{\text{4}\left(\text{l}\right)}\to {\text{3CaSO}}_{\text{4}\left(\text{s}\right)}{\text{+2H}}_{\text{3}}^{}{\text{PO}}_{\text{4}\left(\text{l}\right)}$ ${\text{ΔH}}^{\text{0}}\text{=-265kJ}$

${\text{ΔH}}^{\text{0}}$= ${\displaystyle \sum {\text{n}}_{\text{p}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{product}}-{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{reactant}}$

            =$\left[\text{3(-1433 kJ)+2(-1267kJ)}\right]\text{-}\left[\text{1(-4126kJ)+3(-814 kJ)}\right]$

            =$\text{-265kJ}$

The standard state of some compounds which present in the reaction are given. By substituting the values in the standard enthalpy change equation, the standard enthalpy change for the reaction calculated as $\text{-265kJ}\text{.}$

Interpretation Introduction

Interpretation: Standard enthalpy change has calculated for given reaction.

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 79E

${\text{NH}}_{\text{3}\left(\text{g}\right)}{\text{+HCl}}_{\left(\text{g}\right)}\to {\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}$                                          ${\text{ΔH}}^{\text{0}}\text{=-176kJ}$

Explanation of Solution

Given data

Standard state for given compound in the reaction are,

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

${\text{NH}}_{\text{3(g)}}$                                   -46 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CH}}_{\text{4}}{}_{\text{(g)}}$                                   -75 ${\text{HCN}}_{\text{(g)}}$                                 -135.1 ${\text{H}}_{\text{2}}\text{O(g)}$                                -242 ${\text{Ca(PO}}_{\text{4}}{\text{)}}_{\text{2(s)}}$                           -4126${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(l)}}$                               -814 ${\text{CaSO}}_{\text{4(s)}}$                              -1433 ${\text{H}}_{\text{3}}\text{PO}$                              -1267 ${\text{HCl}}_{\text{(g)}}$                                   -92 $\begin{array}{l}{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}\hfill \end{array}$                               -314

To calculate standard enthalpy change.

The balanced equation is,

                  ${\text{NH}}_{\text{3}\left(\text{g}\right)}{\text{+HCl}}_{\left(\text{g}\right)}\to {\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{s}\right)}$           ${\text{ΔH}}^{\text{0}}\text{=-176kJ}$

${\text{ΔH}}^{\text{0}}$= ${\displaystyle \sum {\text{n}}_{\text{p}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{product}}-{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ ΔH}}_{\text{f}}^{\text{0}}}{\text{,}}_{\text{reactant}}$

          =$\left[\text{1(-314kJ)}\right]\text{-}\left[\text{1(-46kJ)+1(-92kJ)}\right]$

          =$\text{-176kJ}$

The standard state of some compounds which present in the reaction are given. By substituting the values in the standard enthalpy change equation, the standard enthalpy change for the reaction calculated as $\text{-176kJ}\text{.}$