Question 7b270

Apr 21, 2016

Unsurprisingly, the answer given to you is correct.

Explanation:

The balanced chemical equation that describes this redox reaction looks like this

$4 {\text{Fe"_ ((s)) + color(red)(3)"O"_ (2(g)) rightleftharpoons 2"Fe"_ 2"O}}_{3 \left(s\right)}$

You know that the equilibrium mixture contains, in a $\text{2.0-L}$ container

• $\text{1.0 mole Fe}$
• $1.0 \cdot {10}^{- 3} {\text{moles O}}_{2}$
• ${\text{2.0 moles Fe"_2"O}}_{3}$

As you know, solids do not affect the position of the equilibrium because their concentration is presumed constant. In other words, as long as you have some iron metal, $\text{Fe}$, and enough oxygen gas to allow for the reaction to proceed, the position of the equilibrium will be the same at that particular temperature.

This means that the equilibrium constant for this reaction depends exclusively on the concentration of oxygen gas.

By definition, the equilibrium constant will be equal to the ratio between the equilibrium concentrations of the products and the equilibrium concentrations of the reactants, all raised to the power of their corresponding stoichiometric coefficients.

K_c = color(blue)(cancel(color(black)(["Fe"_2"O"_3]^2)))/(color(blue)(cancel(color(black)(["Fe"]^4))) * ["O"_2]^color(red)(3))

which is of course equivalent to

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{c} = \frac{1}{{\left[{\text{O}}_{2}\right]}^{\textcolor{red}{3}}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

All you have to do to find the value of the equilibrium constant is find the concentration of oxygen gas present at equilibrium.

In this case, you'd have

["O"]_2 = (1.0 * 10^(-3)"moles")/"2.0 L" = 5.0 * 10^(-4)"mol L"^(-1)

The equilibrium constant will thus be

${K}_{c} = \frac{1}{5.0 \cdot {10}^{- 4} {\text{mol L}}^{- 1}} ^ \textcolor{red}{3}$

K_c = 1/(125 * 10^(-12)"mol"^(3)"L"^(-3)) = color(green)(|bar(ul(color(white)(a/a)8.0 * 10^9"mol"^(-3)"L"^(3)color(white)(a/a)|)))#

The answer is rounded to two sig figs.