Q 87CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Chemosynthetic bacteria reduce ${\mathbf{CO}}_{\mathbf{2}}$  by “splitting” ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{ }\left(g\right)$instead of the ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$used by photosynthetic organisms. Compare the free energy change for splitting ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$with that for splitting ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ . Is there an advantage to using  ${\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ instead of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$?

Step-by-Step Solution

Answer

Due to $∆\mathrm{G}$ is lower, the reaction of ${\mathrm{H}}_2\mathrm{S}$ is more energetically advantageous.

Step 1: Calculation of ∆ G for H 2 O

The photosynthetic reaction of water splitting to remove   is initially described as follows:

 ${\mathrm{nCO}}_2\left(\mathrm{g}\right)+{\mathrm{nH}}_2\mathrm{O}\to {\left({\mathrm{CH}}_2\mathrm{O}\right)}_{\mathrm{n}}+{\mathrm{nO}}_2\left(\mathrm{g}\right)$

If glucose ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ is produced as a result of the photosynthetic process, the equation may be rewritten as

$6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\to {\left({\mathrm{CH}}_2\mathrm{O}\right)}_6+6{\mathrm{O}}_2\left(\mathrm{g}\right)$

Gibb's energy change (free energy change) may thus be written as:
$\begin{array}{rcl}∆{\mathrm{G}}_{{\mathrm{rxn}}_1}& =& \sum _{}^{}∆{\mathrm{G}}_{\mathrm{products}}-\sum _{}^{}∆{\mathrm{G}}_{\mathrm{reactants}}\\ & =& \left[1 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)+6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{O}}_2\right)\right]-\left[\left(6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{CO}}_2\right)+6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{H}}_2\mathrm{O}\right)\right)\right]\\ & =& \left[1 \mathrm{mol}\times -910.56 \mathrm{kJ}/\mathrm{mol}+6 \mathrm{mol}\times 0.00 \mathrm{kJ}/\mathrm{mol}\right]-\left[6 \mathrm{mol}\times -394.40 \mathrm{kJ}/\mathrm{mol}+6 \mathrm{mol}\times -288.60\mathrm{kJ}/\mathrm{mol}\right]\\ & =& 2793.04 \mathrm{kJ}\end{array}$

The process is not spontaneous, but it is catalyzed by numerous enzymes found in the cells.

Calculation of ∆ G for H 2 S

The process may be stated similarly for chemosynthetic bacteria,

${\mathrm{nCO}}_2\left(\mathrm{g}\right)+{\mathrm{nH}}_2\mathrm{S}\to {\left({\mathrm{CH}}_2\mathrm{O}\right)}_{\mathrm{n}}+{\mathrm{nSO}}_2\left(\mathrm{g}\right)$ 

Assume that glucose   is produced as a result of this reaction, then the equation becomes

 $6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{S}\to {\left({\mathrm{CH}}_2\mathrm{O}\right)}_6+6{\mathrm{SO}}_2\left(\mathrm{g}\right)$

Gibb's energy change (free energy change) may thus be written as
$\begin{array}{rcl}∆{\mathrm{G}}_{{\mathrm{rxn}}_2}& =& \sum _{}^{}∆{\mathrm{G}}_{\mathrm{products}}-\sum _{}^{}∆{\mathrm{G}}_{\mathrm{reactants}}\\ & =& \left[1 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)+6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{SO}}_2\right)\right]-\left[\left(6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{CO}}_2\right)+6 \mathrm{mol}\times ∆\mathrm{G}\left({\mathrm{H}}_2\mathrm{S}\right)\right)\right]\\ & =& \left[1 \mathrm{mol}\times -910.56 \mathrm{kJ}/\mathrm{mol}+6 \mathrm{mol}\times -300.20 \mathrm{kJ}/\mathrm{mol}\right]-\left[6 \mathrm{mol}\times -394.40 \mathrm{kJ}/\mathrm{mol}+6 \mathrm{mol}\times -288.60\mathrm{kJ}/\mathrm{mol}\right]\\ & =& 1386.24 \mathrm{kJ}\end{array}$

Because Gibb's energy change is lower, the reaction is more spontaneous, using less energy to create the same type of carbohydrate (as glucose, for example).

Therefore, $∆\mathrm{G}$ is lower, the ${\mathrm{H}}_2\mathrm{S}$ reaction is more energetically advantageous.