Question

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

Short answer

In summary, given the reactions and the partial pressures of the gases involved, we calculated the equilibrium constants: \(K_{p1} = 59.42\) and \(K_{p2} = 0.002037\).

Step-by-step solution

Write the expressions for \(K_{p1}\) and \(K_{p2}\)

For each of the given reactions, write the expressions for the equilibrium constants: \(K_{p1} = \frac{P(\mathrm{SO}_3) \cdot P(\mathrm{SO}_2)}{(P(\mathrm{FeSO}_4))^2}\), \(K_{p2} = \frac{P(\mathrm{SO}_2) \cdot P(\frac{1}{2}\mathrm{O}_2)}{P(\mathrm{SO}_3)}\).

Find the partial pressures of other gases

Given the total pressure 0.836 atm and knowing the partial pressure of oxygen gas to be 0.0275 atm, we can find the combined partial pressures of \(\mathrm{SO}_3 (g)\) and \(\mathrm{SO}_2 (g)\) \(P(\mathrm{SO}_3) + P(\mathrm{SO}_2) = 0.836 - 0.0275 = 0.8085\) atm

Express the unknown partial pressures in terms of the known value

Let \(x\) be the partial pressure of \(\mathrm{SO}_3 (g)\). Then, the partial pressure of \(\mathrm{SO}_2 (g)\) is \(0.8085 - x\). Now we can write the expression for \(K_{p1}\) in terms of x: \(K_{p1} = \frac{x (0.8085 - x)}{(0.0275)^2}\).

Use the second reaction to find the relationship between the partial pressures of \(\mathrm{SO}_3\) and \(\mathrm{SO}_2\)

In the second reaction, we can express the amount of \(\mathrm{SO}_3\) produced in terms of the amount of \(\mathrm{SO}_2\). The mole ratio of the gases in the second reaction is 1:1:\(\frac{1}{2}\). Since twice the amount of \(\mathrm{SO}_2\) is produced for every mole of O2, we get: \(x = 0.8085 - (2 \cdot 0.0275) = 0.8085 - 0.055 = 0.7535\) atm

Calculate the partial pressure of \(\mathrm{SO}_2 (g)\)

Now, we can calculate the partial pressure of \(\mathrm{SO}_2 (g)\): \(P(\mathrm{SO}_2) = 0.8085 - 0.7535 = 0.055\) atm

Calculate the equilibrium constants \(K_{p1}\) and \(K_{p2}\)

Now that we have the partial pressures of all gases, we can calculate \(K_{p1}\) and \(K_{p2}\): \(K_{p1} = \frac{(0.7535)(0.055)}{(0.0275)^2} = 59.42\) \(K_{p2} = \frac{(0.055)(0.0275)}{0.7535} = 0.002037\) So, the equilibrium constants for the given reactions are: \(K_{p1} = 59.42\) \(K_{p2} = 0.002037\)

Key concepts

chemical reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. In the context of our exercise, we have two important reactions involving iron(II) sulfate:

• The first reaction is: \(2 \mathrm{FeSO}_4 (s) \rightleftharpoons \mathrm{Fe}_2\mathrm{O}_3(s) + \mathrm{SO}_3(g) + \mathrm{SO}_2(g)\).
• The second reaction is: \(\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g)\).

In these reactions, solid iron(II) sulfate decomposes into solid iron(III) oxide, while gaseous sulfur trioxide and sulfur dioxide are formed. The second reaction further shows the breakdown of sulfur trioxide into sulfur dioxide and oxygen gas.
These processes are examples of equilibrium reactions, meaning they reach a state where the forward and reverse reactions occur at the same rate, resulting in constant concentrations (or partial pressures in the case of gases) of reactants and products.

partial pressure

In gas-phase reactions like those involving sulfur trioxide, sulfur dioxide, and oxygen, partial pressure is a critical concept. Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. Each gas in the reaction has its own partial pressure, contributing to the total pressure of the system.
Given in the exercise, the total pressure in the container is 0.836 atm, and the partial pressure of oxygen is 0.0275 atm. By knowing the partial pressure of one component, we can calculate the partial pressures of other gases. For instance, the sum of the partial pressures of SO\(_3\) and SO\(_2\) is found by subtracting the partial pressure of oxygen from the total pressure:

• \(P(\mathrm{SO}_3) + P(\mathrm{SO}_2) = 0.836 - 0.0275 = 0.8085 \ \mathrm{atm}\).
• This information is essential for determining the equilibrium constants (\(K_p\)) for the reactions involved.

iron(II) sulfate

Iron(II) sulfate, also known as ferrous sulfate, is a solid compound typically represented by the formula \(\mathrm{FeSO}_4\). It is often used in various applications, including as a precursor in the preparation of iron-based reactions.
In the exercise, iron(II) sulfate undergoes thermal decomposition at high temperatures (920 K) in an evacuated container. This decomposition is the starting point of the chemical reactions explored in the exercise. The breakdown of iron(II) sulfate results in the production of three different substances:

• Solid iron(III) oxide \(\mathrm{Fe}_2\mathrm{O}_3\),
• Gaseous sulfur dioxide \(\mathrm{SO}_2\), and
• Gaseous sulfur trioxide \(\mathrm{SO}_3\).

These reactions are important in understanding how equilibrium is established and how different components interact under certain conditions.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that explains the behavior of a system at equilibrium when it is subjected to a change in conditions. It states that if a system is disturbed by changes in temperature, pressure, or concentration of components, the system will adjust itself to partially counteract the effect of the disturbance and a new equilibrium will be established.
In the context of the reactions we are investigating, this principle helps us understand how changes in pressure or temperature might affect the equilibrium positions of the reactions involved. For instance:

• An increase in the overall pressure in the system would typically favor the formation of fewer gas molecules, potentially shifting the equilibrium to the side with fewer moles of gas.
• Likewise, a temperature increase could favor the endothermic reaction, absorbing the additional heat.

Understanding Le Chatelier's Principle is key to predicting and explaining the effects of external factors on the equilibrium of reactions and helps in designing processes where controlling the yield of products is crucial.