Chapter 5, Problem 5.81QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated and has to be labelled as endothermic or exothermic. The given reaction is,

  ${\text{NaHCO}}_{\text{3}}\left(s\right)\underset{}{\to }\text{NaOH}\left(s\right){\text{ + CO}}_2\left(g\right)$

Concept introduction:

Enthalpy$\left(H\right)$: It is the total amount of heat in a particular system.

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

${\text{ΔH}}_{\text{f}}^{\text{ο}}$ is the standard enthalpies of formation.

n is the number of moles.

$\text{ΔH}$ in a reaction is positive then it is an endothermic reaction whereas the value obtained for $\text{ΔH}$ is negative it is an exothermic reaction.

Answer to Problem 5.81QE

Enthalpy change is $131.69\text{kJ}$ and it is an endothermic reaction.

Explanation of Solution

The balanced equation for the given reaction follows as,

  ${\text{NaHCO}}_{\text{3}}\left(s\right)\underset{}{\to }\text{NaOH}\left(s\right){\text{ + CO}}_2\left(g\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of NaHCO}}_{\text{3}}\left(s\right)\text{=}-\text{950}\text{.81}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of NaOH}\left(s\right)\text{=}-\text{425}\text{.61}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{425}\text{.61}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times -\text{950}\text{.81}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}131.69\text{kJ}\end{array}$

The sign of enthalpy change is positive. Hence, it is an endothermic reaction.

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated and has to be labelled as endothermic or exothermic. The given reaction is,

  ${\text{H}}_{\text{2}}\text{O}\left(l\right){\text{ + SO}}_3\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)$

Concept introduction:

Refer to (a).

Answer to Problem 5.81QE

Enthalpy change is $-132.44\text{kJ}$ and it is an exothermic reaction.

Explanation of Solution

The balanced equation for the given reaction follows as,

  ${\text{H}}_{\text{2}}\text{O}\left(l\right){\text{ + SO}}_3\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of H}}_{\text{2}}\text{O}\left(l\right)\text{=}-\text{285}\text{.83}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of SO}}_3\left(g\right)\text{=}-\text{395}\text{.72}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)\text{=}-\text{813}\text{.99}\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{813}\text{.99}\text{kJ}/\text{mol}\right)\right]-\left[\left(\text{1 mol}\times -\text{285}\text{.83}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times -\text{395}\text{.72}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}-132.44\text{kJ}\end{array}$

The sign of enthalpy change is negative. Hence, it is an exothermic reaction.

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated and has to be labelled as endothermic or exothermic. The given reaction is,

  ${\text{H}}_{\text{2}}\text{O}\left(g\right){\text{ + SO}}_3\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)$

Concept introduction:

Refer to (a).

Answer to Problem 5.81QE

Enthalpy change is $-176.45\text{kJ}$ and it is an exothermic reaction.

Explanation of Solution

The balanced equation for the given reaction follows as,

  ${\text{H}}_{\text{2}}\text{O}\left(g\right){\text{ + SO}}_3\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of H}}_{\text{2}}\text{O}\left(g\right)\text{=}-\text{241}\text{.82}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of SO}}_3\left(g\right)\text{=}-\text{395}\text{.72}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(l\right)\text{=}-\text{813}\text{.99}\text{kJ}/\text{mol}\end{array}$

Change in enthalpy can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{813}\text{.99}\text{kJ}/\text{mol}\right)\right]-\left[\left(\text{1 mol}\times -\text{241}\text{.82}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times -\text{395}\text{.72}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}-176.45\text{kJ}\end{array}$

The sign of enthalpy change is negative. Hence, it is an exothermic reaction.