$($iii$)$ Using the equations $($all of which are found on the Ellingham Diagram$)$ determine the delta $G$ of formation of FeO,$Fe2O3,$ and $Fe3O4$ at $973$ Kelvin.

$4Fe3O4+O2<mn\; id="EditorElement\_1877257">6</mn><mi\; id="EditorElement\_1877259">F</mi><mi\; id="EditorElement\_1877261">e</mi><mn\; id="EditorElement\_1877263">2</mn><mi\; id="EditorElement\_1877265">O</mi><mn\; id="EditorElement\_1877267">3</mn>$

$6$FeO$+O2<mn\; id="EditorElement\_1877289">2</mn><mi\; id="EditorElement\_1877291">F</mi><mi\; id="EditorElement\_1877293">e</mi><mn\; id="EditorElement\_1877295">3</mn><mi\; id="EditorElement\_1877297">O</mi><mn\; id="EditorElement\_1877299">4</mn>$

$2Fe+O2<mn\; id="EditorElement\_1877319">2</mn>$FeO

MTE $4010$ Corrosion

Open textbook and class notes. No access In class time $75$ minutes.

$\frac{H_2}{H_2}\frac{O}{C}{O}_2,{10}^{-8}$obrace$({)}^{{10}^{-8}}$

${10}^{-6}{10}^{\mathrm{.}4}$

${10}^{-4}$

${10}^{-2}$

$-2$

$1$

${10}^{-2}$

$+O_2=2C{O}^O$ M

$2$

${10}^{-4}$

${10}^2-$

$2$

$O_2=2Fe+D_2=2$FeOM

${10}^{-8}$

${10}^{-10}$

$M$

$2C+O_2=2CO$

$O_2=Ti{O}_2$

$\frac{4}{31}$

${10}^{-18}$

$10$

${10}^{12}$

$\frac{H_2}{H_2}O$

$\frac{p_{02}}{a}tm$

${10}^{-100}$

${10}^{-30}$

${10}^{-24}$