Chapter 5, Problem 5.82QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

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

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.

Answer to Problem 5.82QE

Enthalpy change is $178.32\text{kJ}$.

Explanation of Solution

The balanced equation for the given reaction follows as,

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

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of CaCO}}_{\text{3}}\left(s\right)\text{=}-\text{1206}\text{.92}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of CaO}\left(s\right)\text{=}-\text{635}\text{.09}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{635}\text{.09}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times -\text{1206}\text{.92}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}178.32\text{kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

  $\text{2 HI}\left(g\right){\text{ + F}}_2\left(g\right)\underset{}{\to }\text{2 HF}\left(g\right){\text{ + I}}_2\left(s\right)$

Concept introduction:

Refer to (a).

Answer to Problem 5.82QE

Enthalpy change is $-595.16\text{kJ}$.

Explanation of Solution

The balanced equation for the given reaction follows as,

  $\text{2 HI}\left(g\right){\text{ + F}}_2\left(g\right)\underset{}{\to }\text{2 HF}\left(g\right){\text{ + I}}_2\left(s\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o\text{of HI}\left(g\right)\text{=}\text{26}\text{.48}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of F}}_2\left(g\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of HF}\left(g\right)\text{=}-\text{271}\text{.1}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of I}}_2\left(s\right)\text{=}0\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{2 mol}\times -\text{271}\text{.1}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times 0\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{2 mol}\times \text{26}\text{.48}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times 0\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}-595.16\text{kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction should be calculated. The given reaction is,

  ${\text{SF}}_6\left(g\right)\text{ + }3{\text{ H}}_{2}O\left(l\right)\underset{}{\to }\text{6 HF}\left(g\right)\text{ + }S{O}_3\left(g\right)$

Concept introduction:

Refer to (a).

Answer to Problem 5.82QE

Enthalpy change is $44.17\text{kJ}$.

Explanation of Solution

The balanced equation for the given reaction follows as,

  ${\text{SF}}_6\left(g\right)\text{ + }3{\text{ H}}_{2}O\left(l\right)\underset{}{\to }\text{6 HF}\left(g\right)\text{ + }S{O}_3\left(g\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of SF}}_6\left(g\right)\text{=}-\text{1209}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{2}O\left(l\right)\text{=}-\text{285}\text{.83}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of HF}\left(g\right)\text{=}-\text{271}\text{.1}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }S{O}_3\left(g\right)\text{=}-\text{395}\text{.72}\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{6 mol}\times -\text{271}\text{.1}\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol}\times -\text{395}\text{.72}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times -\text{1209}\text{kJ}/\text{mol}\right)\text{+}\left(\text{3 mol}\times -\text{285}\text{.83}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}44.17\text{kJ}\end{array}$