Question

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

$\mathbf{2}{\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathbf{Fe}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathit{\Delta }\mathit{G}\mathbf{=}\mathbf{+}\mathbf{840}\mathbf{}\mathbf{kJ}$

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

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

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

Short answer

If the value is negative, it can conclude that the process will run spontaneously

Step-by-step solution

Given data

The value of standard entropy and temperature given below:

$\Delta G_1=+840\mathrm{kJ}$at T=1200⁰C

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.

Simplify the reactions

First, let's set up the given equations:

Step 4: Add equations (i) and (ii)

Now it can assemble the final reaction:

$\begin{array}{l}2F{e}_2{O}_3\left(s\right)\to 4Fe\left(s\right)+3O_2\left(g\right)\Delta G=840kJ\\ 3C\left(s\right)+3O_2\left(g\right)\to 3C{O}_2\left(g\right)\Delta G=-1200kJ\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ 2F{e}_2{O}_3\left(s\right)+3C\left(s\right)+3O_2\left(g\right)\to 4Fe\left(s\right)+3O_2\left(g\right)+3C{O}_2\left(g\right)\end{array}$

After canceling the oxygen out:

$2F{e}_2{O}_3\left(s\right)+3C\left(s\right)\to 4Fe\left(s\right)+3C{O}_2\left(g\right)$

The Gibbs energy for this reaction is:

$\begin{array}{c}\Delta G=840+(-1200)\\ =-360kJ\end{array}$

Since the value is negative, we can conclude that the process will run spontaneously.