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Chapter 13: Problem 37

Consider the following reaction at a certain temperature: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ An equilibrium mixture contains \(1.0\) mole of \(\mathrm{Fe}, 1.0 \times 10^{-3}\) mole of \(\mathrm{O}_{2}\), and \(2.0\) moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a \(2.0-\mathrm{L}\) container. Calculate the value of \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant for the given reaction is \(K = 8.0 \times 10^9\).

Step by step solution

01

Setup the reaction and K expression

Write down the balanced chemical reaction for the system and set up the expression for the equilibrium constant, K. The given reaction is: \[ 4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \] The K expression for this reaction is: \[ K = \frac{[\mathrm{Fe}_{2}\mathrm{O}_3]^2}{[\mathrm{O}_{2}]^3} \]
02

Calculate the molarity of the components

The given equilibrium concentrations are 1.0 mol of Fe, 1.0 x 10^{-3} mol of O_2, and 2.0 mol of Fe_2O_3, each in a 2.0 L container. We will now convert these quantities into molarities. \[ [\mathrm{Fe}] = \frac{1.0\mathrm{~mol}}{2.0\mathrm{~L}} = 0.5\mathrm{~M} \] \[ [\mathrm{O}_{2}] = \frac{1.0 \times 10^{-3}\mathrm{~mol}}{2.0\mathrm{~L}} = 5.0 \times 10^{-4}\mathrm{~M} \] \[ [\mathrm{Fe}_{2}\mathrm{O}_3] = \frac{2.0\mathrm{~mol}}{2.0\mathrm{~L}} = 1.0\mathrm{~M} \]
03

Use the molarity to calculate K

Substitute the molarities into the K expression and solve for K. Note that the concentration of Fe(s) does not appear in the K expression because it is a solid. The equilibrium expression for this reaction is: \[ K = \frac{[\mathrm{Fe}_{2}\mathrm{O}_3]^2}{[\mathrm{O}_{2}]^3} \] \[ K = \frac{(1.0\mathrm{~M})^2}{(5.0 \times 10^{-4}\mathrm{~M})^3} \] \[ K = \frac{1.0}{5.0^3 \times 10^{-12}} \] \[ K = \frac{1.0}{125 \times 10^{-12}} \] \[ K = \frac{1.0}{1.25 \times 10^{-10}} \] \[ K = 8.0 \times 10^9 \] Therefore, the equilibrium constant for this reaction is \(K = 8.0 \times 10^9\).

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8}\). What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{array}{l} 2 \mathrm{FeSO}_{4}(s) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{array} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

Consider the following reaction: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) $$ Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2}\), and \(\mathrm{CO}_{2}\) are put into \(\underline{\mathrm{a}}\) flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \({ }^{14} \mathrm{C}\), will \({ }^{14} \mathrm{C}\) be found only in CO molecules for an indefinite period of time? Explain.

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ In an experiment, \(1.00\) mole of \(\mathrm{H}_{2}, 1.00 \mathrm{~mole}\) of \(\mathrm{I}_{2}\), and \(1.00\) mole of HI are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.
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