Q20.109CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the process of respiration, glucose is oxidized completely. In fermentation, ${\mathbf{O}}_{\mathbf{2}}$ is absent and glucose is broken down to ethanol and ${\mathbf{CO}}_{\mathbf{2}}$. Ethanol is oxidized to ${\mathbf{CO}}_{\mathbf{2}}$ and ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. 

1. Balance the following equations for these processes:

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

Fermentation: ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{6}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Ethanol oxidation: ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

1. Calculate ${\mathbf{\Delta G}}_{\mathbf{rxn}}^{\mathbf{o}}$ for respiration of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{g}$ glucose.
2. Calculate ${\mathbf{\Delta G}}_{\mathbf{rxn}}^{\mathbf{o}}$ for fermentation of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{g}$ glucose. 
3. Calculate ${\mathbf{\Delta G}}_{\mathbf{rxn}}^{\mathbf{o}}$ for oxidation of the ethanol from part (c).

Step-by-Step Solution

Answer

a) The reactions that are balanced are:

$\begin{array}{rcl}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)& \to & 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)& \to & 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)\\ {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)& \to & 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\end{array}$

b) For 1 g of glucose respiration, the Gibbs energy is $-15.981\mathrm{kJ}$.

c) For 1 g of glucose fermentation, the Gibbs energy is $-1.26468\mathrm{kJ}$.

d) From 1 g of glucose fermentation, Gibb's energy for ethanol oxidation is $-12.0827\mathrm{kJ}$.

Step 1: Definition of Oxidation

The loss of electrons by a molecule, atom, or ion during a reaction is known as oxidation. When the oxidation state of a molecule, atom, or ion is enhanced, it is called oxidation.

Step 2: Balancing the equations for Respiration, fermentation, and ethanol oxidation

1. The reactions that are balanced are: 

• Respiration: ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
• Fermentation: ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)\to 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)$
• Ethanol oxidation: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 3: Calculating ΔG rxn o for respiration (b)

• To compute the ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}$per 1 mol of glucose, first, calculate the ${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}$ per 1 mol of glucose.

• Calculate the reaction's Gibbs change under standard conditions (298 K),

$\begin{array}{rcl}{\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}& =& \sum _{}^{}{\mathrm{\Delta G}}_{\mathrm{products} }^{\mathrm{o}}-\sum _{}^{}{\mathrm{\Delta G}}_{\mathrm{reactants} }^{\mathrm{o}}\\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}& =& \left[{6\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{CO}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+6\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)}^{\mathrm{o}}{+6\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{O}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right]\\ & & \begin{array}{l} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=[6\mathrm{mol}\times (-394.40\mathrm{kJ}/\mathrm{mol})+6\mathrm{mol}\times (-237.192\mathrm{kJ}/\mathrm{mol}\left)\right]-[1\mathrm{mol}\times (-910.56\mathrm{kJ}/\mathrm{mol}+6\mathrm{mol}\times (0.00\mathrm{kJ}/\mathrm{mol}\left)\right]\\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}} = -2878.992\mathrm{kJ}/\mathrm{mol} \end{array}\\ & & \end{array}$

• Now, convert 1 g of glucose to grams of material.

$\begin{array}{rcl}\mathrm{n}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)& =& \frac{\mathrm{m}}{{\mathrm{M}}_{\mathrm{R}}}=\frac{1\mathrm{g}}{180.156\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)& =& 5.55075\times {10}^{-3}\mathrm{mol}\\ & & \end{array}$

• Finally, 1 g of glucose equals

$$\begin{gathered} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}= 5.55075\times {10}^{-3}\mathrm{mol}\times \left(-2878.992\mathrm{kJ}/\mathrm{mol}\right) \\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-15.981\mathrm{kJ} \end{gathered}$$

For 1 g of glucose respiration, the Gibbs energy is $-15.981\mathrm{kJ}$.

Step 4: Calculating ΔG rxn o for fermentation

(c) Calculate the Gibbs change of the reaction under standard conditions, as in b) (),

${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=\left[{2\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{CO}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+2\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{C}}_2{\mathrm{H}}_{\mathrm{o}}\mathrm{H}\left(\mathrm{l}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)}^{\mathrm{o}}\right]$

• Using Appendix B as a guide,

$$\begin{gathered} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=[2\mathrm{mol}\times (-394.40\mathrm{kJ}/\mathrm{mol})+2\mathrm{mol}\times (-174.80\mathrm{kJ}/\mathrm{mol}\left)\right]-[1\mathrm{mol}\times (-910.56\mathrm{kJ}/\mathrm{mol}] \\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-227.84\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

• Now that you know that 1 g of glucose equals $5.55075\times {10}^{-3} \mathrm{mol}$,

$$\begin{gathered} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=5.55075\times {10}^{-3}\mathrm{mol}\times \left(-227.84\mathrm{kJ}/\mathrm{mol}\right) \\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1.26468\mathrm{kJ} \end{gathered}$$

For 1 g of glucose fermentation, Gibb's energy is $-1.26468\mathrm{kJ}$.

Step 5: Calculating ΔG rxn o for oxidation of ethanol

(d) Finally, when it comes to ethanol oxidation,

${\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}{=2\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{CO}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+3\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}^{\mathrm{o}}-\left[{1\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)}^{\mathrm{o}}{+2\mathrm{mol}\times \mathrm{\Delta G}}_{{\mathrm{O}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right]$

• Using Appendix B as a guide,

$$\begin{gathered} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=[2\mathrm{mol}\times (-394.40\mathrm{kJ}/\mathrm{mol})+2\mathrm{mol}\times (-237.192\mathrm{kJ}/\mathrm{mol}\left)\right]-[1\mathrm{mol}\times (-174.80\mathrm{kJ}/\mathrm{mol}+2\mathrm{mol}\times (0.00\mathrm{kJ}/\mathrm{mol}] \\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-1088.384\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

• Now that we know that the ethanol obtained from glucose is further oxidized, the ethanol gained from the reaction above is:

$\mathrm{n}\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)=2\times \mathrm{n}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)=1.11015\times {10}^{-2}\mathrm{mol}$

As a result, Gibb's energy

$$\begin{gathered} {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=1.11015\times {10}^{-2}\mathrm{mol}\times \left(-1088.384\mathrm{kJ}/\mathrm{mol}\right) \\ {\mathrm{\Delta G}}_{\mathrm{rxn}}^{\mathrm{o}}=-12.0827\mathrm{kJ} \end{gathered}$$

Gibb's energy for ethanol oxidation is $-12.0827\mathrm{kJ}$, which is obtained from 1 g of glucose fermentation.