91Ó°ÊÓ

Chemical equilibrium is the state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

a)

In this problem we are tasked to evaluate the reaction below.

Oxidation state of $\mathrm{Fe}$in${\text{Fe}}_3{\text{O}}_4$.

We know that the oxidation state of oxygen most of the time is$-2$. Moreover, the compound is neutral, so its overall charge is equal to 0 . The sum of the charges of Fe and $\mathrm{O}$must be equal to 0 . Solve for the charge of $\mathrm{Fe}$.

Overall charge= $3\text{Fe}+4\text{O}$

$\begin{array}{c}0=3\text{Fe}+4(−2)\\ 3\text{Fe}=8\\ \text{Fe}=\frac{8}{3}\\ \text{Fe}=\mathrm{2.67.}\end{array}$

Therefore oxidation state is Fe=2.67.

b)

In this problem we are tasked to evaluate the reaction below.

2 oxidation states of ${\text{Fe}}_3{\text{O}}_4$.

${\text{Fe}}_3{\text{O}}_4$can be written as $\text{FeO}â‹\dots {\text{Fe}}_2{\text{O}}_3$. Do the same procedure from the previous item to obtain the oxidation number of in each subunit Each subunit still have neutral charge. Hence, a 0 overall charge.

Overall charge =

$\begin{array}{c}\left(\text{FeO}\right)=\text{Fe}+\text{O}\\ 0=\text{Fe}+(−2)\\ \text{Fe}=+2.\end{array}$

Overall charge=$\left({\text{Fe}}_2{\text{O}}_3\right)=2\text{Fe}+3\text{O}$

$\begin{array}{c}0=2\text{Fe}+3(−2)\\ 2\text{Fe}=6\\ \text{Fe}=\frac{6}{2}\\ \text{Fe}=+3.\end{array}$

c)

In this problem we are tasked to evaluate the reaction below.

Grams of ${\text{Fe}}_3{\text{O}}_4$present at equilibrium.

First, identify the given values.

$\begin{array}{c}\mathrm{T}={900}^{°}\text{C}\\ =1173.15\text{K}\\ {\mathrm{K}}_{\mathrm{c}}=5.1\\ \left[{\text{H}}_2\text{O}\right]=\frac{0.050\text{mol}}{1.0\text{L}}\\ =0.050{\text{molH}}_2\text{O}/\text{L}\\ {\mathrm{n}}_{\text{Fe}}=0.100\text{molFe}\end{array}$

Write the expression for the equilibrium constant of the reaction in terms of concentration.

$\begin{array}{l}{\mathrm{K}}_{\mathrm{c}}=\frac{\left[\text{products}\right]}{\left[\text{reactants}\right]}\\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\text{H}}_2\right]}^4}{{\left[{\text{H}}_2\text{O}\right]}^4}\end{array}$

Write the reaction table.

Solve for x using the equilibrium formula and the equilibrium expression in terms of concentration.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\text{H}}_2\right]}^4}{{\left[{\text{H}}_2\text{O}\right]}^4}\\ 5.1=\frac{{\left(4\mathrm{x}\right)}^4}{{(0.050−4\mathrm{x})}^4}\\ \sqrt[4]{5.1}=\frac{{\left(4\mathrm{x}\right)}^4}{{(0.050−1\mathrm{x})}^4}\\ 1.50=\frac{4\mathrm{x}}{0.050−4\mathrm{x}}\\ 0.0751−6\mathrm{x}=4\mathrm{x}\\ 10\mathrm{x}=0.0751\\ \mathrm{x}=\frac{0.0751}{10}\\ \mathrm{x}=7.51×{10}^{−3}\end{array}$

Solve for the amount of${\text{H}}_2$ at equilibrium

data-custom-editor="chemistry" $\begin{array}{c}\left[{\text{H}}_2\right]=4x\\ =4(7.51×{10}^{−3})\\ \left[{\text{H}}_2\right]=0.0301\text{molH}\end{array}$

Solve the mass of ${\text{Fe}}_3{\text{O}}_4$from the amount of data-custom-editor="chemistry" ${\mathrm{H}}_2$ produced.

Therefore,

data-custom-editor="chemistry" $\begin{array}{c}{\text{massFe}}_3{\text{O}}_4=\frac{0.0301{\text{molH}}_2}{\text{L}}×\frac{1{\text{molFe}}_3{\text{O}}_4}{4{\text{molH}}_2}×\frac{231.53{\text{gFe}}_3{\text{O}}_4}{1{\text{molFe}}_3{\text{O}}_4}×1.0\text{L}\\ =1.74{\text{gFe}}_3{\text{O}}_4.\end{array}$