Q17.98CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the following reaction:

 $\mathbf{3}\mathbf{\text{Fe}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(a) What is the apparent oxidation state of $\mathbf{\text{Fe}}$ in ${\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}$  ?

(b) Actually, Fe has two oxidation states in ${\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}$ . What are they?

(c) At  ${\mathbf{900}}^{\mathbf{°}}\mathbf{\text{C}}\mathbf{,}{\mathbf{K}}_{\mathbf{\text{c}}}$ for the reaction is $\mathbf{5}\mathbf{.1}$ . If   $\mathbf{0}\mathbf{.050}\mathbf{ }\mathbf{\text{mol}}$ of ${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)}$   and  $\mathbf{0}\mathbf{.100}\mathbf{ }\mathbf{\text{mol}}$  of $\mathbf{\text{Fe}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}$  are placed in a $\mathbf{1}\mathbf{.0}\mathbf{-}\mathbf{\text{L}}$   container at ${\mathbf{900}}^{\mathbf{°}}\mathbf{\text{C}}$  , how many grams of  ${\mathbf{\text{Fe}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}$  are present at equilibrium?

Note: The synthesis of ammonia is a major process throughout the industrialized world. Problems  $\mathbf{17}\mathbf{.99}$  to $\mathbf{17}\mathbf{.105}$  refer to various aspects of this all-important reaction:

 ${\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{\text{NH}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{\Delta H}}_{\mathbf{\text{rxn}}}^{\mathbf{°}}\mathbf{=}\mathbf{-}\mathbf{91}\mathbf{.8}\mathbf{ }\mathbf{\text{kJ}}$

Step-by-Step Solution

Answer

a) The apparent oxidation state  $\text{Fe}=2.67$ .

b) The oxidations states are $\text{Fe}=+3$  .

c)  $\mathrm{L}=1.74 {\text{g Fe}}_3{\text{O}}_4$  is present at equilibrium.

Step 1: Definition of equilibrium

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

Step 2: Find apparent oxidation state

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.

Step 3: Which are oxidation state

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  $\text{FeO}\cdot {\text{Fe}}_2{\text{O}}_3$. 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= $\left({\text{Fe}}_2{\text{O}}_3\right)=2\text{Fe}+3\text{O}$ 

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

Step 4: Calculate total grams at equilibrium

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.

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

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

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

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" $\begin{array}{c}{\text{ mass Fe}}_3{\text{O}}_4=\frac{0.0301 {\text{molH}}_2}{ \text{L}}\times \frac{1 {\text{molFe}}_3{\text{O}}_4}{4 {\text{molH}}_2}\times \frac{231.53 {\text{gFe}}_3{\text{O}}_4}{1 {\text{molFe}}_3{\text{O}}_4}\times 1.0 \text{L}\\ =1.74 {\text{gFe}}_3{\text{O}}_4.\end{array}$