Problem 83 - Chemistry: Principles and Reactions

Question

At a certain temperature, the reaction $$\mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightleftharpoons \mathrm{XeF}_{4}(g)$$ gives a \(50.0 \%\) yield of \(\mathrm{XeF}_{4}\), starting with \(\mathrm{Xe}\left(P_{\mathrm{X}_{e}}=0.20 \mathrm{~atm}\right)\) and \(\mathrm{F}_{2}\) \(\left(P_{\mathrm{F}_{2}}=0.40 \mathrm{~atm}\right)\). Calculate \(K\) at this temperature. What must the initial pressure of \(\mathrm{F}_{2}\) be to convert \(75.0 \%\) of the xenon to \(\mathrm{XeF}_{4} ?\)

Step-by-Step Solution

Answer

Answer: The initial pressure of \(F_2\) needed to convert \(75 \%\) of the xenon to \(XeF_4\) is \(0.60\, \text{atm}\).

Step 1: Write the equilibrium constant expression

The equilibrium constant \(K_p\) is given by the expression: $$K_p = \frac{P_{XeF_4}}{P_{Xe}\times P_{F_2}^2}$$

Step 2: Define the pressure changes for the reaction

Let the initial pressures be given as \(P_{Xe}=0.20\,\text{atm}\) and \(P_{F2}=0.40\,\text{atm}\). Let \(x\) represent the change in pressure of the reactants and products at equilibrium. Then the changes in pressure are: - Xenon: \(0.20 - x\) - Fluorine: \(0.40 - 2x\) - Xenon hexafluoride: \(0 + x\)

Step 3: Use the reaction yield to find the equilibrium pressures

Since the reaction yield is \(50.0\%\), we know that only half of the initial pressure of xenon is actually converted. Therefore, the equilibrium pressure of \(Xe\), \(XeF_4\), and \(F_2\) are: $$P_{Xe} = 0.20 - 0.10 = 0.10\,\text{atm}$$ $$P_{XeF_4} = 0.10\,\text{atm}$$ $$P_{F_2} = 0.40 - 2(0.10) = 0.20\,\text{atm}$$

Step 4: Calculate the equilibrium constant \(K_p\)

Substitute the equilibrium pressures into the equilibrium constant expression: $$K_p = \frac{0.10}{(0.10)(0.20)^2} = 25$$

Step 5: Set up the new pressure changes for 75% conversion

Now we have to find the initial pressure of \(F_2\), denoted as \(P'_{F_2}\), required to convert \(75.0\%\) of xenon to \(XeF_4\). Let \(x'\) represent the new change in pressure of the reactants and products at equilibrium: - Xenon: \(0.20 - x'\) - Fluorine: \(P'_{F_2} - 2x'\) - Xenon hexafluoride: \(0 + x'\) The new equilibrium pressures for 75.0% conversion are: $$P'_{Xe} = 0.20\,\text{atm} - \frac{3}{4} (0.20\,\text{atm}) = 0.05\,\text{atm}$$ $$P'_{XeF_4} = \frac{3}{4} (0.20\,\text{atm}) = 0.15\,\text{atm}$$ $$P'_{F_2} = P'_{F_2} - 2x'$$

Step 6: Calculate the initial pressure of \(F_2\) for 75% conversion

Substitute the new equilibrium pressures into the equilibrium constant expression and solve for the initial pressure of \(F_2\): $$K_p = \frac{0.15}{(0.05)(P'_{F_2} - 2(0.15))^2} = 25$$ Solving for \(P'_{F_2}\), we get: $$P'_{F_2} = 0.60\,\text{atm}$$ The initial pressure of \(F_2\) needed to convert \(75 \%\) of the xenon to \(XeF_4\) is \(0.60\, \text{atm}\).

Key Concepts

Equilibrium Constant

The equilibrium constant \(K_p\) is a crucial value in chemical reactions that occur in a closed system. It represents the ratio of product concentration to reactant concentration at equilibrium, each raised to the power of their coefficients in the balanced equation. In our given reaction of xenon and fluorine gases forming xenon hexafluoride \(XeF_4\), \(K_p\) is expressed as: \[K_p = \frac{P_{XeF_4}}{P_{Xe} \times P_{F_2}^2} \]This expression indicates that the equilibrium constant depends on the partial pressures of the reactants and products.

• If \(K_p\) is large, the reaction heavily favors products at equilibrium, indicating a more complete forward reaction.
• If \(K_p\) is small, the equilibrium favors reactants, with less conversion to products.

Understanding \(K_p\) allows chemists to predict the position of equilibrium and adjust conditions to optimize the desired reaction outcome.

Pressure Changes

In gas-phase reactions, the change in pressures of reactants and products as the reaction reaches equilibrium is a key concept. For our specific reaction, these pressure changes help us understand how reactant pressures transform into product pressures.Initially, the pressures for xenon (Xe) and fluorine \(F_2\) are given, and we use the variable \(x\) to denote the change in these pressures. For the conversion of 50\% yield:

• Xe changes by \(0.20 - x\)
• F_2 changes by \(0.40 - 2x\)
• XeF_4 emerges with a pressure increase of \(+x\)

For 75\% conversion, a similar method with \(x'\) showcases the needed adjustments to find the initial pressure necessary for such yield. This analysis of pressure changes is essential for predicting how initial conditions and conversions affect equilibrium.

Reaction Yield

Reaction yield is a measure of the efficiency of a chemical process and is defined as the percentage of reactants converted into products. In this exercise, two yields are assessed:- **50\% Yield:** Here, half of the initial xenon pressure converts to \(XeF_4\). Knowing the yield helps calculate the equilibrium pressures of all involved components.- **75\% Yield:** Requires improved conditions, like altering initial pressures, to convert three-fourths of xenon into the product. Here, the challenge is calculating the required initial pressure of \(F_2\) to ensure this increased conversion.Reaction yield is pivotal in industrial applications to maximize productivity and ensure economic viability of chemical processes. By achieving higher yields, less material is wasted, and more product is formed.

Equilibrium Pressure

The concept of equilibrium pressure relates to the specific pressures of reactants and products once a chemical reaction reaches a state of balance. At equilibrium, the rate of the forward reaction matches the rate of the reverse reaction, leading to constant pressure values of all species involved.To find these pressures, we start with the initial conditions and apply known yields:

• For 50\% yield: \(P_{Xe} = 0.10 \text{ atm}\), \(P_{F_2} = 0.20 \text{ atm}\), and \(P_{XeF_4} = 0.10 \text{ atm}\).
• For 75\% yield: \(P_{Xe} = 0.05 \text{ atm}\), \(P_{XeF_4} = 0.15 \text{ atm}\), with adjustments guided by the equilibrium constant to determine initial \(P_{F_2}\).

Knowing equilibrium pressures allows us to analyze and predict the behavior of the chemical system as well as modify parameters to achieve desired outcomes.