Question

A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: $$\begin{array}{c}2{\mathrm{FeSO}}_4(s)⟹{\mathrm{Fe}}_2{\mathrm{O}}_3(s)+{\mathrm{SO}}_3(g)+{\mathrm{SO}}_2(g)\\ {\mathrm{SO}}_3(g)\rightleftharpoons {\mathrm{SO}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)\end{array}$$ 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

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

Step-by-step solution

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

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: $P_{SO3}=\frac{n_{SO3}RT}{V}$ $P_{SO2}=\frac{n_{SO2}RT}{V}$ $P_{O2}=\frac{n_{O2}RT}{V}$

Write the expressions for Kp of each reaction

We can now write the expressions for Kp of each reaction. Reaction 1: Kp1 = $\frac{P_{SO3}\ast P_{SO2}}{(P_{FeSO4}{)}^2}$ Reaction 2: Kp2 = $\frac{P_{SO2}\ast P_{O2}^{1/2}}{P_{SO3}}$ 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.

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)

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 = $\frac{P_{SO3}\ast P_{SO2}}{(P_{FeSO4}{)}^2}$ = $\frac{0.0275\ast 0.78098}{(P_{FeSO4}{)}^2}$ Kp2 = $\frac{P_{SO2}\ast P_{O2}^{1/2}}{P_{SO3}}$ = $\frac{0.78098\ast (0.0275{)}^{1/2}}{0.0275}$ 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.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant ($K_p$) is a vital concept. It quantifies the balance point of a reversible reaction. For reactions occurring in the gas phase, like those involving $S{O}_3$ and $S{O}_2$ in the given exercise, we use partial pressures to calculate $K_p$.

The general formula for the equilibrium constant is given by the ratio of the products' partial pressures to the reactants', each raised to the power of their coefficients from the balanced equation.

• For Reaction 1 in our exercise: $K_{p1}=\frac{P_{{\mathrm{SO}}_3}\times P_{{\mathrm{SO}}_2}}{(P_{{\mathrm{FeSO}}_4}{)}^2}$
  
• For Reaction 2: $K_{p2}=\frac{P_{{\mathrm{SO}}_2}\times P_{{\mathrm{O}}_2}^{1/2}}{P_{{\mathrm{SO}}_3}}$

Solids like $FeS{O}_4$ and $F{e}_2{O}_3$ do not appear in these expressions. Their concentrations aren't considered in gas-phase equilibrium because their activity is defined as 1. Remember, $K_p$ will vary with temperature, so it is specific to the conditions at 920 K in this problem.
This equilibrium constant helps us understand how far a process proceeds before reaching a state of balance.

Partial Pressure

Partial pressure is the pressure that each gas in a mixture contributes to the total pressure. In a closed system, the total pressure is the sum of the partial pressures of all gases present.

In the chemical reactions described, partial pressures of $S{O}_3$, $S{O}_2$, and $O_2$ add up to a total pressure of 0.836 atm. Knowing that the partial pressure of oxygen is 0.0275 atm, we can deduce the partial pressures of the other gases using a simple deduction:

• The pressure contribution of $S{O}_3$ is given as 0.0275 atm, identical to $O_2$ due to their stoichiometric relationship.
  
• Therefore, the remainder of the total pressure is due to $S{O}_2$: $P_{{\mathrm{SO}}_2}=0.836-2\times 0.0275=0.78098$ atm.

Understanding partial pressures is crucial for solving equilibrium problems in gas-phase reactions. They allow us to determine the equilibrium constants for reactions, illustrating how gases in a closed system distribute themselves.

Stoichiometry

Stoichiometry in chemical reactions is all about proportions. It dictates the ratio in which substances react and are produced, based on the balanced chemical equation. This concept is fundamental in calculating partial pressures from mole ratios, as seen in this exercise.

Here, Reaction 2 displays a clear stoichiometric relationship:

• 1 mole of $S{O}_3$ disappears to form 1 mole of $S{O}_2$ and $\frac{1}{2}$ mole of $O_2$. This establishes a direct 1:1:0.5 relationship.

This ratio is instrumental in using the given partial pressures to solve for unknown values. Stoichiometry ensures that $S{O}_2$ and $O_2$ are calculated accurately based on the initial conditions laid out, which helps further in determining the equilibrium state.

Accurate stoichiometric relationships are pivotal for deriving precise expressions and calculations in chemical equilibrium scenarios.

Ideal Gas Law

The Ideal Gas Law, represented as $PV=nRT$, links the pressure ($P$), volume ($V$), temperature ($T$), and moles ($n$) of an ideal gas with the ideal gas constant ($R$).

This law is crucial in this exercise for converting between moles and pressure. By rearranging the equation to solve for pressure: $P=\frac{nRT}{V}$, we understand how individual gases relate to their partial pressures.

• Each gas's partial pressure is dependent on its mole count, alongside temperature and volume.
• Since temperature is constant at 920 K in this problem, $R$ and $V$ are consistent factors across calculations.

Using the Ideal Gas Law enables the calculation of the pressure for each gaseous component in the reactions, which is instrumental in determining equilibrium constants ($K_p$) efficiently. In such exercises, it's key to accurately interpret and apply the ideal gas law to link macroscopic gas properties with molecular quantities.