Chapter 17: Q98CP (page 780)

Consider the following reaction:
$3 F e_{(s)} + 4 H_{2} O_{(g)} ⇌ F e_{3} O_{4 (s)} + 4 H_{2 (g)}$
(a) What is the apparent oxidation state of Fe in ${Fe}_{3} O_{4}$ ?
(b) Actually, Fe has two oxidation states in ${Fe}_{3} O_{4}$ . What are they?
(c) At role="math" localid="1654929041124" $900^{0} C$ , Kc for the reaction is 5.1. If 0.050 mol of $H_{2} O_{(g)}$ and 0.100 mol of Fe(s) are placed in a 1.0-L container at $900^{0} C$ , how many grams of role="math" localid="1654929174865" ${Fe}_{3} O_{4}$ are present at equilibrium?

Expert verified

(a) The apparent oxidation state of Fe in ${Fe}_{3} O_{4}$ is 2.67.

(b) ${Fe}_{3} O_{4}$ Can be written as $FeO . {Fe}_{2} O_{3}$

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

Step by step solution

(a) The apparent oxidation state of Fe in ${Fe}_{3} O_{4}$ is:

${Fe}_{3} O_{4} = 3 \times Fe + 4 \times (- 2) = 0 Fe = \frac{8}{3} = 2.67$

(b)localid="1654931375767" ${Fe}_{3} O_{4}$ Can be written as $FeO . {Fe}_{2} O_{3}$

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

$Concentration of water = \frac{Moles}{Volume} [H 2 O] = \frac{0.050 mol}{1.0 L} = 0.050 Μ [Fe] = \frac{0.100 mol}{1.0 L} = 0.100 M$

Putting these values in an ICE table:

$KC = \frac{[{Fe}_{3} O_{4}] [H_{2}]^{4}}{[Fe]^{3} [H_{2} O]^{4}}$

Substituting, we get

$5.1 = \frac{(x) (4 x)^{4}}{(0.10 - 3 x)^{3} (0.050 - 4 x)^{4})} 5.1 = \frac{(2556 x^{5})}{(0.10 - 3 x)^{3} (0.050 - 4 x)^{4}}$

Since x<< 1

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

Hence;

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

Thus,

$x^{5} = 1.245 \times 10^{- 10} x = 1.045 \times 10^{- 2} M$

So, we have $1.045 \times 10^{- 2} mol$ of ${Fe}_{3} O_{4}$ present in 1L of solution.

$Mass of {Fe}_{3} O_{4} present at equilibrium = moles \times molar mass = 1.045 \times 10^{- 12} \times 232 g {mol}^{- 1} = 2.42 g$

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