Chapter 6, Problem 80E

Use the values of $\Delta H_{\text{f}}^{°}$ in Appendix 4 to calculate ∆Hº for the following reactions. (See Exercise 77 .)

a.

b. ${\text{SiCl}}_4\left(l\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)\to {\text{SiO}}_2\left(s\right)+4\text{HCl}\left(aq\right)$

c. $\text{MgO}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)$

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 80E

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}{\text{+3O}}_{\text{2}\left(\text{g}\right)}\to {\text{2CO}}_{\text{2}\left(\text{g}\right)}{\text{+ 3H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)\text{.}}$ ${\text{ΔH}}^{\text{0}}\text{=-1235 kJ}$

Explanation of Solution

Given data          

Standard state for given compound in the reaction are,

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

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}$                           -287 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CO}}_2{}_{\text{(g)}}$                                 -393.5 ${\text{SiCl}}_4{}_{\text{(l)}}$                                -687 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                 -286 ${\text{HCl}}_{\text{(aq)}}$                                -167 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}$                                 -242 $\text{SiO}$                                 -911 ${\text{MgO}}_{\text{(s)}}$                                -602 $\begin{array}{l}{\text{Mg(OH)}}_{\text{2}}{}_{\left(\text{s}\right)}\hfill \end{array}$                         -925

To calculate standard enthalpy change.

          ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}{\text{+3O}}_{\text{2}\left(\text{g}\right)}\to {\text{2CO}}_{\text{2}\left(\text{g}\right)}{\text{+ 3H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)\text{.}}$ ${\text{ΔH}}^{\text{0}}\text{=-1235 kJ}$

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

                    =$\left[\text{2(-393}\text{.5kJ)+3(-242}\text{kJ)}\right]\text{-}\left[\text{1(-278kJ)}\right]$

                    =$\text{-1235 kJ/mol}$

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 80E

${\text{SiCl}}_{\text{4}\left(\text{l}\right)}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{SiO}}_{\text{2}\left(\text{s}\right)}{\text{+ 4HCl}}_{\left(\text{aq}\right)}$          ${\text{ΔH}}^{\text{0}}\text{=-320 kJ}$

Explanation of Solution

Given data          

Standard state for given compound in the reaction are,

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

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}$                           -287 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CO}}_2{}_{\text{(g)}}$                                 -393.5 ${\text{SiCl}}_4{}_{\text{(l)}}$                                -687 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                 -286 ${\text{HCl}}_{\text{(aq)}}$                                -167 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}$                                 -242 $\text{SiO}$                                 -911 ${\text{MgO}}_{\text{(s)}}$                                -602 $\begin{array}{l}{\text{Mg(OH)}}_{\text{2}}{}_{\left(\text{s}\right)}\hfill \end{array}$                         -925

To calculate standard enthalpy change

${\text{SiCl}}_{\text{4}\left(\text{l}\right)}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{SiO}}_{\text{2}\left(\text{s}\right)}{\text{+ 4HCl}}_{\left(\text{aq}\right)}$               ${\text{ΔH}}^{\text{0}}\text{=-320 kJ}$

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

            =$\left[\text{4(-167 kJ)+1(-911kJ)}\right]\text{-}\left[\text{1(-687kJ)+2(-286 kJ)}\right]$

            =$\text{-320 kJ}$

From above reaction, standard state of the compounds are given. By substituting the values in the standard enthalpy change equation, the standard enthalpy change for the reaction calculated as $\text{-320 kJ}\text{.}$

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 80E

${\text{MgO}}_{\left(\text{s}\right)}{\text{+ H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to \text{ Mg}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}$ ${\text{ΔH}}^{\text{0}}\text{=-37kJ}$

Explanation of Solution

Given data

Standard state for given compound in the reaction are,

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

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\text{(l)}}$                           -287 ${\text{O}}_2{}_{\text{(g)}}$                                        0 ${\text{CO}}_2{}_{\text{(g)}}$                                 -393.5 ${\text{SiCl}}_4{}_{\text{(l)}}$                                -687 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$                                 -286 ${\text{HCl}}_{\text{(aq)}}$                                -167 ${\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}$                                 -242 $\text{SiO}$                                 -911 ${\text{MgO}}_{\text{(s)}}$                                -602 $\begin{array}{l}{\text{Mg(OH)}}_{\text{2}}{}_{\left(\text{s}\right)}\hfill \end{array}$                         -925

To calculate standard enthalpy change.

          ${\text{MgO}}_{\left(\text{s}\right)}{\text{+ H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to \text{ Mg}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}$             ${\text{ΔH}}^{\text{0}}\text{=-37kJ}$

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

          =$\left[\text{1(-925kJ)}\right]\text{-}\left[\text{1(-602kJ)+1(-286kJ)}\right]$

          =$\text{-37kJ}$

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 $\text{-37kJ}\text{.}$