Chapter 24, Problem 23PS

Interpretation Introduction

Interpretation:

For the given reaction, the value of ${\Delta }_{\text{r}}{H}^{°}$ for the given oxidation reaction of one mole of glucose is $-2803\text{ kJ/mol-rxn}$ has to be verified.

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

Concept introduction:

The change in the enthalpy of a reaction when reactant is converted into product under standard conditions is called standard enthalpy of reaction.

The expression for standard enthalpy of reaction is,

${\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)$ (1)

Here, ${\Delta }_f{H}^{\text{°}}$ is the standard enthalpy of formation and $n$ is the number of moles of reactant and product in the balanced chemical reaction.

Explanation of Solution

The value of ${\Delta }_{\text{r}}{H}^{°}$ for the given oxidation reaction of one mole of glucose is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpy of formation.

The standard enthalpy of formation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)$ is $-\text{1273}\text{.3 kJ/mol}$.

The standard enthalpy of formation of ${\text{O}}_{\text{2}}\left(\text{g}\right)$ is $\text{0 kJ/mol}$.

The standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ is $-285.8\text{ kJ/mol}$.

The standard enthalpy of formation of ${\text{CO}}_{\text{2}}\left(\text{g}\right)$ is $-\text{393}\text{.5 kJ/mol}$.

The given balanced chemical equation is:

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

The ${\Delta }_{\text{r}}{H}^{°}$ can be calculated by the following expression,

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)\\ =\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{6 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{CO}}_{\text{2}}\left(\text{g}\right)\right]+\\ \left({\text{6 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right]\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\right]+\\ \left({\text{6 mol O}}_2\left(\text{g}\right)\text{/mol-rxn}\right){\Delta }_f{H}^{\text{°}}\left[{\text{O}}_2\left(\text{g}\right)\right]\end{array}\right]\end{array}\right]\end{array}$

Substitute the value of ${\Delta }_f{H}^{\text{°}}$.

$\begin{array}{c}{\text{Δ}}_{\text{r}}{H}^{\text{°}}=\left[\begin{array}{l}\left[\begin{array}{l}\left({\text{6 mol CO}}_{\text{2}}\left(\text{g}\right)\text{/mol-rxn}\right)\left(-\text{393}\text{.5 kJ/mol}\right)+\\ \left({\text{6 mol H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{/mol-rxn}\right)\left(-285.8\text{ kJ/mol}\right)\end{array}\right]-\\ \left[\begin{array}{l}\left({\text{1 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\text{/mol-rxn}\right)\left(-\text{1273}\text{.3 kJ/mol}\right)+\\ \left({\text{6 mol O}}_2\left(\text{g}\right)\text{/mol-rxn}\right)\left(0\right)\end{array}\right]\end{array}\right]\\ =-2803kJ/mol-rxn\end{array}$

The value of ${\Delta }_{\text{r}}{H}^{°}$ for the given oxidation reaction of one mole of glucose is $-2803kJ/mol-rxn$.