Q98CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the following reaction: 

$\mathbf{3}\mathit{F}{\mathit{e}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

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

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

(c) At $\mathbf{900}{\mathbf{ }}^{\mathbf{0}}\mathbf{C}$, Kc for the reaction is 5.1. If 0.050 mol of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$ and 0.100 mol of Fe(s) are placed in a 1.0-L container at $\mathbf{900}{\mathbf{ }}^{\mathbf{0}}\mathbf{C}$, how many grams of ${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$are present at equilibrium?

Step-by-Step Solution

Answer

(a) The apparent oxidation state of Fe in ${\mathrm{Fe}}_3{\mathrm{O}}_4$  is 2.67.

(b)  ${\mathrm{Fe}}_3{\mathrm{O}}_4$Can be written as $\mathrm{FeO}.{\mathrm{Fe}}_2{\mathrm{O}}_3$

Hence oxidation state of F in FeO = +2 and Fe in  =+3

(c) 2.42 grams of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ are present at equilibrium.

Step 1: (a) Apparent oxidation state of Fe in Fe 3 O 4 and (b) two oxidation states of Fe in Fe 3 O 4 .

(a) The apparent oxidation state of Fe in ${\mathrm{Fe}}_3{\mathrm{O}}_4$ is:

        $$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4=3\times \mathrm{Fe}+4\times \left(-2\right)=0 \\ \mathrm{Fe}=\frac{8}{3} \\ =2.67 \end{gathered}$$

 

(b)  ${\mathrm{Fe}}_3{\mathrm{O}}_4$ Can be written as $\mathrm{FeO}.{\mathrm{Fe}}_2{\mathrm{O}}_3$

Hence oxidation state of F in FeO = +2 and Fe in  =+3

Step 2: (c) How many grams of Fe 3 O 4 are present at equilibrium?

$$\begin{gathered} \mathrm{Concentration} \mathrm{of} \mathrm{water}= \frac{\mathrm{Moles}}{\mathrm{Volume}} \\ \left[\mathrm{H}2\mathrm{O}\right] =\frac{0.050 \mathrm{mol}}{1.0 \mathrm{L}}= 0.050 \mathrm{{\rm M}} \\ \left[\mathrm{Fe}\right] =\frac{0.100 \mathrm{mol} }{1.0 \mathrm{L}}= 0.100 \mathrm{M} \end{gathered}$$

Putting these values in an ICE table:

 

   $\mathrm{KC}=\frac{\left[{\mathrm{Fe}}_3{\mathrm{O}}_4\right][{\mathrm{H}}_2{]}^4}{\left[\mathrm{Fe}{]}^3\right[{\mathrm{H}}_2\mathrm{O}{]}^4}$

Substituting, we get

$$\begin{gathered} 5.1=\frac{\left(\mathrm{x}\right)(4\mathrm{x}{)}^4}{(0.10-3\mathrm{x}{)}^3(0.050-4\mathrm{x}{)}^4)} \\ 5.1=\frac{\left(2556{\mathrm{x}}^5\right)}{(0.10-3\mathrm{x}{)}^3(0.050-4\mathrm{x}{)}^4} \end{gathered}$$

Since x<< 1 

Hence; $0.10-3\mathrm{x} \approx 0.10,\& 0.05-4\mathrm{x} \approx 0.050$

Hence;

$$\begin{gathered} 5.1=\frac{256{\mathrm{x}}^5}{{(0.10-3\mathrm{x})}^3{(0.050-4\mathrm{x})}^4} \\ =\frac{256{\mathrm{x}}^5}{6.25\times {10}^{-9}} \\ 256{\mathrm{x}}^5=3.19\times {10}^{-10} \end{gathered}$$

Thus,

$$\begin{gathered} {\mathrm{x}}^5=1.245\times {10}^{-10} \\ \mathrm{x}=1.045\times {10}^{-2} \mathrm{M} \end{gathered}$$

So, we have $1.045\times {10}^{-2} \mathrm{mol}$  of ${\mathrm{Fe}}_3{\mathrm{O}}_4$  present in 1L of solution.

$$\begin{gathered} \mathrm{Mass} \mathrm{of} {\mathrm{Fe}}_3{\mathrm{O}}_4\mathrm{present} \mathrm{at} \mathrm{equilibrium}=\mathrm{moles}\times \mathrm{molar} \mathrm{mass} \\ =1.045\times {10}^{-12}\times 232 \mathrm{g} {\mathrm{mol}}^{-1} \\ =2.42 \mathrm{g} \end{gathered}$$