Chapter 5, Problem 5.80QE

(a)

Interpretation Introduction

Interpretation:

Enthalpy change for the photosynthesis of glucose 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 5.80QE

Enthalpy change is $2808\text{kJ}$ and the photosynthesis of glucose is an endothermic reaction.

Explanation of Solution

The balanced equation for the photosynthesis of glucose is given as:

  $6\text{}{\text{ CO}}_{\text{2}}\left(g\right){\text{+ 6 H}}_{2}O\left(l\right)\underset{}{\to }{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right){\text{ + 6 O}}_2\left(g\right)$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of CO}}_{\text{2}}\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{2}O\left(l\right)\text{=}-\text{285}\text{.83}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)\text{=}-\text{1268}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}{\text{O}}_2\left(g\right)\text{=}0\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{1 mol}\times -\text{1268}\text{kJ}/\text{mol}\right)\text{+}\left(\text{6 mol×}0\text{kJ}/\text{mol}\right)\right]-\\ \left[\left(\text{6 mol}\times -\text{393}\text{.51}\text{kJ}/\text{mol}\right)\text{+}\left(\text{6 mol×}-\text{285}\text{.83}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}2808\text{kJ}\end{array}$

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

(b)

Interpretation Introduction

Interpretation:

Enthalpy change for the reduction of iron (III) oxide with carbon should be calculated and has to be labelled as endothermic or exothermic.

Concept introduction:

Refer to (a).

Answer to Problem 5.80QE

Enthalpy change is $467.9\text{kJ}$ and the reduction of iron (III) oxide with carbon is an endothermic reaction.

Explanation of Solution

The balanced equation for the reduction of iron (III) oxide with carbon is given as:

  ${\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\text{ + 3 C}\left(s\right)\underset{}{\to }\text{4 Fe}\left(s\right)\text{+}2{\text{CO}}_2\left(g\right)$.

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\text{=}-\text{824}\text{.2 }\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of C}\left(s\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of CO}}_2\left(g\right)\text{=}-\text{393}\text{.51}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of}\text{Fe}\left(s\right)\text{=}0\text{kJ}/\text{mol}\end{array}$

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,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{4 mol}\times 0\text{kJ}/\text{mol}\right)\text{+}\left(\text{3 mol×}-\text{393}\text{.51}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{3 mol}\times 0\text{kJ}/\text{mol}\right)\text{+}\left(2\text{ mol×}-\text{824}\text{.2}\text{kJ}/\text{mol}\right)\right]\\ \\ \text{=}467.9\text{kJ}\end{array}$

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