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Chapter 13: Problem 120

A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: $$\begin{array}{c}{2 \mathrm{FeSO}_{4}(s) \Longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g)} \\\ {\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)}\end{array}$$ 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.

Short Answer

Expert verified
The equilibrium constant Kp for reaction 1 is approximately 0.8239, and for reaction 2, it is approximately 5.322.

Step by step solution

01

Calculate the initial moles of each species

We are given the partial pressure of oxygen is 0.0275 atm. We use this as a base to calculate moles for other species according to the stoichiometry of the reactions. For Reaction 1: 2 FeSO4 (s) -> Fe2O3 (s) + SO3 (g) + SO2 (g) For Reaction 2: SO3 (g) <=> SO2 (g) + 1/2 O2 (g) From reaction 2, we understand that mole ratio between SO3 and SO2 is 1:1. The partial pressure of oxygen given is 0.0275 atm. Thus, there will be 0.0275 atm of SO3 and 0.0275 atm of SO2. Now we can find the total pressure by adding the partial pressures of all gases. Total pressure = 0.836 atm
02

Write the expressions for the partial pressures of each species using moles and the ideal gas law

The ideal gas law is \(PV = nRT\). Rearranging for pressure, we have \(P = \frac{nRT}{V}\). We can now write the expressions for the partial pressures of each gas: \(P_{SO3} = \frac{n_{SO3}RT}{V}\) \(P_{SO2} = \frac{n_{SO2}RT}{V}\) \(P_{O2} = \frac{n_{O2}RT}{V}\)
03

Write the expressions for Kp of each reaction

We can now write the expressions for Kp of each reaction. Reaction 1: Kp1 = \(\frac{P_{SO3} * P_{SO2}}{(P_{FeSO4})^2}\) Reaction 2: Kp2 = \(\frac{P_{SO2} * P_{O2}^{1/2}}{P_{SO3}}\) Notice that the iron(II) sulfate and iron(III) oxide terms are not in the Kp expressions because they are solids, and their concentrations do not affect the equilibrium constant.
04

Solve the two equilibrium equations for the partial pressures

Since we already calculated the total pressure and partial pressure of oxygen, we can solve the two equilibrium equations for the partial pressures: Total pressure = 0.836 atm Partial pressure of Oxygen (P_O2) = 0.0275 atm Partial pressure of SO3 (P_SO3) = 0.0275 atm Partial pressure of SO2 (P_SO2) = 0.78098 atm (From total pressure, P_SO2 + 0.0275 + 0.0275 = 0.836)
05

Calculate Kp values using the expressions found in step 3

Now, we can plug the partial pressures back into the Kp expressions for each reaction. Kp1 = \(\frac{P_{SO3} * P_{SO2}}{(P_{FeSO4})^2}\) = \(\frac{0.0275 * 0.78098}{(P_{FeSO4})^2}\) Kp2 = \(\frac{P_{SO2} * P_{O2}^{1/2}}{P_{SO3}}\) = \(\frac{0.78098 * (0.0275)^{1/2}}{0.0275}\) Solving for Kp1 and Kp2: Kp1 ≈ 0.8239 Kp2 ≈ 5.322 The equilibrium constant Kp for reaction 1 is approximately 0.8239, and for reaction 2, it is approximately 5.322.

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Most popular questions from this chapter

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

The reaction to prepare methanol from carbon monoxide and hydrogen $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ is exothermic. If you wanted to use this reaction to produce methanol commercially, would high or low temperatures favor a maximum yield? Explain.

Hydrogen for use in ammonia production is produced by the reaction $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \frac{\text { Nicatalyst }}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants $\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\( are\)\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}$ Is the reaction exothermic or endothermic?

The partial pressures of an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)\mathrm{NO}_{2}(g)\( are \)P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\( atm and \)P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}$ at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.
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