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Chapter 13: Problem 42

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},\( and 2.0 \)\mathrm{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 \(K\) for the given reaction can be calculated as follows: \[K \approx \frac{1}{(0.50)^{4}(5.0 \times 10^{-4})^{3}} = 1.28 \times 10^{11}\]

Step by step solution

01

Write the expression for K

For the given reaction: \(4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) The equilibrium constant K can be expressed as: \[K = \frac{[\mathrm{Fe}_{2} \mathrm{O}_{3}]^{2}}{[\mathrm{Fe}]^{4}[\mathrm{O}_{2}]^{3}}\] Remember that the concentrations of solids do not affect the equilibrium, so we will only include the concentration terms for the gaseous species.
02

Calculate the molar concentrations of each species

We are given the amount of each species in moles, and the volume of the container is 2.0 L. Molar concentrations can be calculated by dividing the moles of the substance by the volume (in liters) of the container: \[[\mathrm{Fe}] = \frac{1.0\,\text{mol}}{2.0\,\text{L}} = 0.50\,\text{M}\] \[[\mathrm{O}_{2}] = \frac{1.0 \times 10^{-3}\,\text{mol}}{2.0\,\text{L}} = 5.0 \times 10^{-4}\,\text{M}\] \[[\mathrm{Fe}_{2}\mathrm{O}_{3}] = \frac{2.0\,\text{mol}}{2.0\,\text{L}} = 1.0\,\text{M}\]
03

Plug the concentrations into the K expression

Now that we have the molar concentrations of each species, we can substitute these values into the expression for K: \[K = \frac{(1.0)^{2}}{(0.50)^{4}(5.0 \times 10^{-4})^{3}}\]
04

Calculate K

Now, we can compute the value of K: \[K = \frac{1}{(0.50)^{4}(5.0 \times 10^{-4})^{3}} = \frac{1}{(0.0625)(1.25 \times 10^{-10})} = \frac{1}{7.8125 \times 10^{-12}} \approx 1.28 \times 10^{11}\] So, the equilibrium constant K for this reaction is: \[K \approx 1.28 \times 10^{11}\]

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Most popular questions from this chapter

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .^{\circ} \mathrm{C}\) $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ If a 20.0 -g sample of \(\mathrm{CaCO}_{3}\) is put into a 10.0 -L container and heated to \(800 .^{\circ} \mathrm{C},\) what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

At a particular temperature, 8.0 moles of \(\mathrm{NO}_{2}\) is placed into a 1.0 -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is 2.0 \(\mathrm{M}\) . Calculate \(K\) for this reaction.

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$ \(K=400\) . at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and $\mathrm{NH}_{4} \mathrm{HS}\( are placed in a \)5.00-\mathrm{L}$ vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?
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