Problem 10 - Aircraft Propulsion

Question

One mole of oxygen, \(\mathrm{O}_{2(\mathrm{~g})}\), is heated to \(4000 \mathrm{~K}\) at the pressure of \(p_{m}\). A fraction of the oxygen dissociates to oxygen atoms according to $$ x \mathrm{O}_{2} \rightarrow 2 x \mathrm{O} $$ Assuming a state of equilibrium is reached in the mixture, calculate (a) mole fraction of \(\mathrm{O}_{2}\) at equilibrium when \(p_{m}\) is \(1 \mathrm{~atm}\). (b) mole fraction of \(\mathrm{O}_{2}\) at equilibrium when \(p_{m}\) is \(10 \mathrm{~atm}\). Assume the equilibrium constant for the reaction $$ \mathrm{O}_{2} \leftrightarrow 2 \mathrm{O} $$ is \(K_{p}=2.19\) atm at the temperature of \(4000 \mathrm{~K}\). Explain the effect of pressure on dissociation.

Step-by-Step Solution

Answer

The mole fraction of \(\mathrm{O}_{2}\) at 1 atm equilibrium is 0.787, while at 10 atm is 0.16. The increase in pressure decreases the dissociation of \(\mathrm{O}_{2}\), hence increasing its mole fraction.

Step 1: Understand the Equilibrium Relation

The given equilibrium reaction is \(\mathrm{O}_{2} \leftrightarrow 2 \mathrm{O}\). Let \(x\) be the mole fraction of \(\mathrm{O}_{2}\) that dissociates. So, this would leave \(1-x\) moles of \(\mathrm{O}_{2}\) and \(2x\) moles of \(\mathrm{O}\) in the mixture at equilibrium. From the general form of the ideal gas law \(PV=nRT\), we can derive the mole fractions of \(\mathrm{O}_{2}\) and \(\mathrm{O}\) by the total moles in the mixture.

Step 2: Calculate the Mole Fractions at 1 atm

The total moles would be \(1-x+2x = 1+x\). The mole fraction of \(\mathrm{O}_{2}\) is given by its moles divided by total moles or \((1-x)/(1+x)\) and the mole fraction of \(\mathrm{O}\) will be \(2x/(1+x)\). Since we know that the equilibrium constant is \(K_{p}=2.19\) and it is derived by taking the mole fraction of products divided by the mole fraction of reactants, we have \(K_{p}=(2x/(1+x))^2 / (1-x)/(1+x)=2.19\). Solving for \(x\) will give us \(x=0.213\). Therefore, the mole fraction of \(\mathrm{O}_{2}\) at 1 atm is \(1-x = 1-0.213 = 0.787\).

Step 3: Calculate the Mole Fractions at 10 atm

We know that at higher pressures, the equilibrium will shift to decrease pressure, in this case by favoring the side of the reaction with fewer moles of gas. But to quantify this, we need to adjust the equilibrium constant \(K_{p}'=K_{p}*P=2.19*10\). Solving for \(x\) will give us \(x=0.84\). Therefore, the mole fraction of \(\mathrm{O}_{2}\) at 10 atm is \(1-x = 1-0.84 = 0.16\).

Step 4: Explain the Effect of Pressure on Dissociation

Increasing the pressure moves the equilibrium to decrease the pressure. It favours the reactant side of the equation which has fewer gas molecules (oxygen molecule \(\mathrm{O}_{2}\)). This causes an increase in the mole fraction of oxygen molecule, \(\mathrm{O}_{2}\), hence decreasing the dissociation.

Key Concepts

Mole Fraction

The concept of mole fraction is fundamental when studying chemical reactions in a mixture. In simple terms, the mole fraction (\( X_i \)) of a component in a mixture is the ratio of the number of moles of that component to the total number of moles in the mixture. This can be calculated using the formula:
\[ X_i = \frac{n_i}{n_{total}} \]where \(n_i\) is the number of moles of the component and \(n_{total}\) is the total number of moles in the mixture.

In the problem provided, oxygen gas (\( O_2 \)) dissociates into oxygen atoms (\( O \)), influencing the mole fractions of both reactants and products at equilibrium. Calculating these mole fractions involves understanding the quantities involved in the dissociation and re-calculating the quantities at state of equilibrium.

Dissociation

Dissociation refers to the process whereby a compound breaks down into simpler components, which in molecular terms often means transforming multi-atom molecules into individual atoms or ions. In the exercise, oxygen molecules dissociate into oxygen atoms:
\[ \mathrm{O}_2 \leftrightarrow 2 \mathrm{O} \]
At equilibrium, some fraction of the original oxygen (\( O_2 \)) molecules has transformed into individual oxygen atoms.

The extent of dissociation is quantified by the variable \( x \), representing the proportion of molecule \( O_2 \) that dissociates. Hence, for every mole of \( O_2 \) initially present, \( 1-x \) moles of \( O_2 \) remain at equilibrium, while \( 2x \) moles of \( O \) are produced. Understanding dissociation is critical for analyzing how reactions change over time and when various pressures and temperatures are applied.

Equilibrium Constant

The equilibrium constant (\( K_p \)) is a measure that expresses the ratio of the concentrations of products to reactants for a reaction at equilibrium, each raised to the power of their coefficients in the reaction equation. This value is key to determining the position of equilibrium and the extent of reaction progression.
The equilibrium constant for the reversible reaction given in the exercise:
\[ \mathrm{O}_2 \leftrightarrow 2 \mathrm{O} \]
is \( K_p = 2.19 \) atm.

Using the relation:
\[ K_p = \frac{(P_{\mathrm{O}})^2}{P_{\mathrm{O}_2}} \]
where \( P_{\mathrm{O}_2} \) and \( P_{\mathrm{O}} \) represent the partial pressures of \( O_2 \) and \( O \) respectively, we substitute the mole fraction to find the equilibrium state. Given the equilibrium constant, we can also predict how changes in pressure and temperature will shift the equilibrium.

Pressure Effects

Pressure significantly influences chemical reactions, especially those involving gases. According to Le Chatelier's Principle, if a system at equilibrium experiences a change in pressure, the system will adjust to minimize that change.
For gaseous reactions like the dissociation of \( O_2 \) into oxygen atoms:
\[ \mathrm{O}_2 \leftrightarrow 2 \mathrm{O} \]
increasing pressure typically shifts the equilibrium toward reducing the number of gas molecules to lower the system's overall pressure.

This means that at higher pressure, the equilibrium will favor the formation of more \( O_2 \) because it results in fewer moles of gas. For example, at 10 atm, there is a greater tendency for \( O_2 \) molecules to form than at 1 atm. Consequently, higher pressures will not favor dissociation, resulting in a lower mole fraction of \( O \), thereby retaining more \( O_2 \) in the mixture at equilibrium.