Question

At 1200⁰C, the reduction of iron oxide to elemental iron and oxygen is not spontaneous:

$\mathbf{2}\mathit{F}{\mathit{e}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathit{F}\mathit{e}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathit{\Delta }\mathit{G}\mathbf{=}\mathbf{+}\mathbf{840}\mathbf{}\mathit{k}\mathit{J}$

Show how this process can be made to proceed if all the oxygen generated reacts with carbon:

$\mathit{C}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathit{\Delta }\mathit{G}\mathbf{=}\mathbf{-}\mathbf{400}\mathbf{}\mathit{k}\mathit{J}$

This observation is the basis for the smelting of iron ore with coke to extract metallic iron.

Short answer

The fraction is45.65%.

Step-by-step solution

Given data

The reaction of formation of ATP:

${\mathrm{ADP}}^{2-}\left(aq\right)+{\mathrm{HPO}}_4^{2-}\left(aq\right)+{\mathrm{H}}^{+}\left(aq\right)\to {\mathrm{ATP}}^{+}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

$\Delta G=+34.5\mathrm{kJ}$

Concept of the reduction

Reduction is the exchange of electrons between species in a chemical reaction involving the gain of electrons or a reduction in an element's oxidation state.

 Step 3: Simplify the reactions

Since 38 molecules of ADP^3- gives 38 molecules of ATP^4-, multiply both sides with 38

Reaction for the metabolism of glucose.

Step 4: Add equations (i) and (ii) 

Combining equations 1 and 2 to obtain the coupling reaction between ATP and glucose

Step 4: Calculate the fraction of glucose

The fraction of glucose oxidation energy is the ratio of energy released in the oxidation reaction 2 to the energy of formation of ATP(1)

Fraction of glucose energy in ATP$=\frac{\mathrm{Energyofglucoseoxidation}}{\mathrm{EnergyofATPformation}}\times 100\%$

Put the value of the given data in the above formula:

Fraction of glucose energy data-custom-editor="chemistry" $\begin{array}{c}=\frac{1311kJ}{2872kJ}\times 100\%\\ =45.65\%\end{array}$