Chapter 5, Problem 5.79QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the fermentation of glucose to ethyl alcohol and carbon dioxide should be calculated and has to be labelled as endothermic or exothermic.

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 5.79QE

Enthalpy change is $-74\text{kJ}$ and the fermentation of glucose to ethyl alcohol and carbon dioxide is an exothermic reaction.

Explanation of Solution

The balanced equation for the fermentation of glucose to ethyl alcohol and carbon dioxide is given as:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)\underset{}{\to }2\text{}{\text{ CO}}_{\text{2}}\left(g\right)\text{+}2{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)$.

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)\text{=}-\text{1268}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)\text{=}-\text{277}\text{.69}\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(2\text{ mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\text{+}\left(2\text{ mol×}-\text{277}\text{.69}\text{kJ}/\text{mol}\right)\right]-\left[-\text{1268}\text{kJ}/\text{mol}\right]\\ \\ \text{=}-74\text{kJ}\end{array}$

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

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the combustion of normal butane should be calculated and has to be labelled as endothermic or exothermic.

Concept introduction:

Refer to (a).

Answer to Problem 5.79QE

Enthalpy change is $-2878.46\text{kJ}$ and the combustion of normal butane is an exothermic reaction.

Explanation of Solution

The balanced equation for the combustion of normal butane is given as:

  ${\text{n-C}}_{\text{4}}{\text{H}}_{\text{10}}\left(g\right)\text{ + }\frac{\text{13}}{2}{O}_2\left(g\right)\underset{}{\to }2{\text{CO}}_2\left(g\right){\text{+ 5 H}}_{2}O\left(l\right)$.

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{ofn-C}}_{\text{4}}{\text{H}}_{\text{10}}\left(g\right)\text{=}-\text{124}\text{.73}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }{O}_2\left(g\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of CO}}_2\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}{\text{H}}_{2}O\left(l\right)\text{=}-\text{285}\text{.83}\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{4 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\text{+}\left(5\text{ mol×}-\text{285}\text{.83}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{1 mol}\times -\text{124}\text{.73}\text{kJ}/\text{mol}\right)\text{+}\left(\frac{13}{2}\text{ mol×}0\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}-2878.46\text{kJ}\end{array}$

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