Problem 119 - Chemistry An Atoms First Approach

Question

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g):\).$$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$.A 2.00 -g sample of XY (molar mass \(=165 \mathrm{g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The density of the gas mixture after dissociation is \(3.15\,\mathrm{g/L}\), and the equilibrium constant for this reaction at \(25^{\circ}\mathrm{C}\) is \(0.139\).

Step 1: Calculate the initial moles of \(\mathrm{XY}\)

Given that we have a \(2.00\,\mathrm{g}\) sample of \(\mathrm{XY}\) and its molar mass is \(165\,\mathrm{g/mol}\), we can calculate the initial moles of \(\mathrm{XY}\). $$\text{moles of}\,\mathrm{XY} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.00\,\mathrm{g}}{165\,\mathrm{g/mol}} = 0.0121\,\text{mol}$$

Step 2: Calculate the moles of \(\mathrm{XY}\), \(\mathrm{X}\), and \(\mathrm{Y}\) after dissociation

We are also given that \(35.0\%\) of the original \(\mathrm{XY}\) has dissociated. Therefore, we can find the moles of each compound after dissociation. Moles of \(\mathrm{XY}\) remaining: $$0.0121\,\text{mol} \times (1 - 0.350) = 0.00786\,\text{mol}$$ Since \(1\) molecule of \(\mathrm{XY}\) dissociates into \(1\) molecule of \(\mathrm{X}\) and \(1\) molecule of \(\mathrm{Y}\), the moles of \(\mathrm{X}\) and \(\mathrm{Y}\) produced are as follows: Moles of \(\mathrm{X}\) and \(\mathrm{Y}\) produced: $$0.0121\,\text{mol} \times 0.350 = 0.00424\,\text{mol}$$

Step 3: Find the total moles and the mole fraction of each gas after dissociation

To find the mole fraction of each gas, we need to find the total moles after dissociation: Total moles: $$0.00786\,\text{mol} (\mathrm{XY}) + 0.00424\,\text{mol}(\mathrm{X}) + 0.00424\,\text{mol} (\mathrm{Y}) = 0.01634\,\text{mol}$$ Now, we can find the mole fraction of each gas: $$\chi_{XY} = \frac{0.00786\,\text{mol}}{0.01634\,\text{mol}} = 0.481$$ $$\chi_{X} = \frac{0.00424\,\text{mol}}{0.01634\,\text{mol}} = 0.259$$ $$\chi_{Y} = \frac{0.00424\,\text{mol}}{0.01634\,\text{mol}} = 0.259$$

Step 4: Calculate the density of the gas mixture

Now that we have the mole fraction of each gas, we can find the average molar mass of the mixture: $$\bar{M} = 165\,\mathrm{g/mol}(\chi_{XY})+165\,\mathrm{g/mol}\left(\frac{\chi_{X}+\chi_{Y}}{2}\right) = 165\,\mathrm{g/mol}$$ We can now find the density of this gas mixture using the ideal gas law, given that the pressure is constant at \(0.967\,\mathrm{atm}\) and the temperature is \(25^{\circ}\mathrm{C}\) which is equivalent to \(298\,\mathrm{K}\): $$P = \frac{n\bar{M}}{V}\,R\,T \Rightarrow \rho = \frac{P\bar{M}}{R\,T} = \frac{0.967\,\mathrm{atm}\times 165\,\mathrm{g/mol}}{0.0821\,(\mathrm{L\,atm/mol\,K})\times 298\,\mathrm{K}} = 3.15\,\mathrm{g/L}$$

Step 5: Find the value of \(K\)

To find the value of \(K\), we need to use the equilibrium concentrations: $$K = \frac{[\mathrm{X}][\mathrm{Y}]}{[\mathrm{XY}]} = \frac{\chi_{X}\chi_{Y}/V}{\chi_{XY}/V} = \frac{\chi_{X}\chi_{Y}}{\chi_{XY}} = \frac{0.259\times 0.259}{0.481} = 0.139$$ So at \(25^{\circ}\mathrm{C}\), the equilibrium constant, \(K = 0.139\). To summarize, the density of the gas mixture after dissociation is \(3.15\,\mathrm{g/L}\), and the equilibrium constant for this reaction at \(25^{\circ}\mathrm{C}\) is \(0.139\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential tool in chemistry that helps us understand the behavior of gases under various conditions. It's expressed with the equation: \(PV = nRT\). Here, \(P\) stands for pressure, \(V\) is the volume, \(n\) represents the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

In the context of this problem, the Ideal Gas Law is used to determine the gas density after you know the total moles of gas as well as its average molar mass. When gases dissociate or react, the total volume and pressure can change in a system with a movable piston. To find the density, or \(\rho\), the equation can be rewritten by rearranging terms to isolate \(\rho\):

\(\rho = \frac{P\bar{M}}{RT}\)

This formula tells us how gas density relates directly to pressure and molar mass, while having an inverse relationship with temperature.

Equilibrium Constant

The Equilibrium Constant, represented as \(K\), is a key concept in chemical reactions. It provides a measure of the relative amounts of reactants and products at equilibrium.

For a simple dissociation reaction like \(\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g) + \mathrm{Y}(g)\), \(K\) is calculated using the concentrations (or pressures) of the gases at equilibrium. The formula is:

\[ K = \frac{[X][Y]}{[XY]} \]

In this exercise, after calculating the mole fractions of \(\mathrm{X}, \mathrm{Y},\) and \(\mathrm{XY}\), you can solve for \(K\) using:

\[K = \frac{\chi_{X}\chi_{Y}}{\chi_{XY}}\]

This gives a ratio that tells us the position of equilibrium. A low \(K\) value implies more reactants are present, while a high \(K\) value indicates that products are favored in the reaction.

Mole Fraction

The mole fraction, denoted as \(\chi\), is a way to express the concentration of a particular component in a gas mixture. It is defined as the ratio of the number of moles of one component to the total number of moles in the mixture.

Here's how you calculate the mole fraction for each gas in the dissociation of \(\mathrm{XY}\):

• \(\chi_{XY} = \frac{\text{moles of } XY}{\text{total moles}}\)
• \(\chi_{X} = \frac{\text{moles of } X}{\text{total moles}}\)
• \(\chi_{Y} = \frac{\text{moles of } Y}{\text{total moles}}\)

In the given problem, calculating these mole fractions helps to determine how much of the original gas dissociated and how it's distributed between the products \(\mathrm{X}\) and \(\mathrm{Y}\). Understanding mole fractions is critical for further calculations such as determining the equilibrium constant.

Gas Density

Gas Density, represented as \(\rho\), describes the mass of gas per unit volume and is typically measured in grams per liter (g/L). For gases, density is a functional combination of the Ideal Gas Law components.

The density of a gas can change as its composition changes, especially in reactions involving dissociation. Using the equation \(\rho = \frac{P\bar{M}}{RT}\), where \(\bar{M}\) is the average molar mass, we can find the density of a gaseous mixture post-dissociation.

In the exercise, knowing the conditions such as pressure (0.967 atm), temperature (298 K), and calculated molar mass is crucial to finding that the gas density is \(3.15 \, \mathrm{g/L}\). Calculating gas density helps predict the behavior of gas mixtures in real-world applications.