Chapter 4, Problem 4D.5AST

Interpretation Introduction

Interpretation:

The standard enthalpy of combustion of glucose has to be determined.

Concept Introduction:

The standard enthalpy of the reaction is the difference of the sum of standard enthalpy of formation of all the products and the sum of standard enthalpy of formation of all the reactants.  The mathematical equation for the calculation of standard enthalpy of the reaction is shown below.

$\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(reac\tan ts\right)$

Explanation of Solution

The balanced chemical equation for the combustion of glucose is shown below.

    $C_6{H}_{12}{O}_6\left(s\right)+6O_{\text{2}}\left(g\right)\to 6C{O}_{\text{2}}\left(g\right)+6H_2\text{O}\left(l\right)$

The standard enthalpy of formation of products that are $C{O}_{\text{2}}\left(g\right)$ and $H_2\text{O}\left(l\right)$ is $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$ and $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{C{O}_2}\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{C{O}_2}$ is the number of moles of $C{O}_{\text{2}}\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is the standard enthalpy of formation of $C{O}_{\text{2}}\left(g\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_2\text{O}\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_2\text{O}\left(l\right)$.

The value of ${\text{n}}_{C{O}_2}$ is $6mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{2}O}$ is $6mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{C{O}_2}$, $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(6mol\right)\times \left(-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(6mol\right)\times \left(-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-2361.06\text{kJ}+\left(-1714.98\text{kJ}\right)\\ =-2361.06\text{kJ}-1714.98\text{kJ}\\ =-4076.04kJ\end{array}$

The standard enthalpy of formation of reactants that are $C_6{H}_{12}{O}_6\left(s\right)$ and $O_2\left(g\right)$ is $-\text{1268}\text{.0}\text{kJ}\cdot {\text{mol}}^{-1}$ and $0.0\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}={\text{n}}_{C_6{H}_{12}{O}_6}\Delta {\text{H}}_f^{\circ }\left(C_6{H}_{12}{O}_6,s\right)+{\text{n}}_{O_2}\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$        (2)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{C_6{H}_{12}{O}_6}$ is the number of moles of $C_6{H}_{12}{O}_6\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(C_6{H}_{12}{O}_6,s\right)$ is the standard enthalpy of formation of $C_6{H}_{12}{O}_6\left(s\right)$.
• ${\text{n}}_{O_2}$ is the number of moles of $O_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is the standard enthalpy of formation of $O_2\left(g\right)$.

The value of ${\text{n}}_{C_6{H}_{12}{O}_6}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(C_6{H}_{12}{O}_6,s\right)$ is $-\text{1268}\text{.0}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{O_2}$ is $6mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{C_6{H}_{12}{O}_6}$, $\Delta {\text{H}}_f^{\circ }\left(C_6{H}_{12}{O}_6,s\right)$, ${\text{n}}_{O_2}$ and $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}=\left(1mol\right)\times \left(-\text{1268}\text{.0}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(6mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =-\text{1268}\text{.0}\text{kJ}+0.0\text{kJ}\\ =-\text{1268}\text{.0}\text{kJ}\end{array}$

The standard enthalpy of combustion of glucose is calculated by the expression shown below.

  $\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(reac\tan ts\right)$        (3)

Where,

• $\Delta \text{H}°$ is standard enthalpy of reaction.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-4076.04kJ$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is $-\text{1268}\text{.0}\text{kJ}$.

Substitute the value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ in equation (3).

  $\begin{array}{c}\Delta \text{H}°=\left(-4076.04kJ\right)-\left(-\text{1268}\text{.0}\text{kJ}\right)\\ =-4076.04kJ+\text{1268}\text{.0}\text{kJ}\\ =-2808.04kJ\end{array}$

Thus, after rounding to four significant figures, the standard enthalpy of combustion of glucose is $\underset{\_}{-2808kJ\cdot mol}$.