Problem 129 - Chemistry

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 \(\mathrm{XY}\) (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 \(\mathrm{atm} .\) As \(\mathrm{XY}\) begins to dissociate, the piston moves until 35.0 mole percent of the original \(\mathrm{XY}\) 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

To calculate the density of the gas mixture after the dissociation and the equilibrium constant (K) for the reaction at 25°C, follow these steps: 1. Find the initial moles of XY: 2.00 g / 165 g/mol. 2. Calculate the moles of dissociated XY, X, and Y produced: initial moles of XY * 0.35. 3. Find the final moles of XY, X, and Y and their total moles in the container. 4. Determine the volume of the gas mixture using the Ideal Gas Law (PV = nRT). 5. Calculate the density of the gas mixture: mass / volume. 6. Determine the equilibrium constant (K) by finding the concentrations of XY, X, and Y at equilibrium and using the expression K = [X][Y] / [XY].

Step 1: Calculate the initial moles of XY

Firstly, we need to calculate the initial amount of XY gas present in the container. We can do this using the molar mass formula: Initial moles of XY = mass of XY / molar mass of XY Initial moles of XY = \( \frac{2.00 g}{165 \mathrm{g/mol}} \)

Step 2: Calculate the moles of dissociated XY and moles of X and Y produced

Using the dissociation percent given (35.0 %), we can calculate the moles of dissociated XY and the moles of each product formed. Moles of dissociated XY = Initial moles of XY * 0.35 Now, as the stoichiometry of the reaction is 1:1:1, moles of X produced and moles of Y produced are equal to: Moles of X = Moles of dissociated XY Moles of Y = Moles of dissociated XY

Step 3: Calculate the final moles of XY, X, and Y

We will now calculate the final quantity of each gas present in the container. Final moles of XY = Initial moles of XY – Moles of dissociated XY Final moles of X = Moles of X produced Final moles of Y = Moles of Y produced Total moles in the container = Final moles of XY + Final moles of X + Final moles of Y

Step 4: Calculate the volume of the gas mixture

Using the Ideal Gas Law, we can find the volume of the gas mixture after dissociation, given that the pressure and temperature are constant (P = 0.967 atm and T = 25°C): PV = nRT Where: P = pressure (atm) V = volume (L) n = moles of gas R = ideal gas constant (0.08206 L atm/ K mol) T = temperature (in Kelvin) We know the pressure, temperature and the total moles after dissociation, so we can calculate the volume: V = nRT / P Remember to convert the temperature into Kelvin: Temperature (K) = 25°C + 273.15 = 298.15 K

Step 5: Calculate the density of the gas mixture

Density is defined as mass per unit volume. We can calculate the density of the gas mixture as follows: Density = Mass / Volume Mass of the gas mixture = (Initial moles of XY + Moles of X + Moles of Y) * molar mass of XY Density = \( \frac{Mass}{Volume} \)

Step 6: Determine the equilibrium constant (K) for the reaction

The equilibrium constant expression for the given reaction can be written as: K = \( \frac{[X][Y]}{[XY]} \) Where [X], [Y], and [XY] are the molar concentrations of the respective gases at equilibrium. The concentration can be calculated as: Concentration = \( \frac{Moles}{Volume} \) After finding the concentrations of XY, X, and Y at equilibrium, substitute those values into the K expression and solve for K.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle which helps us relate the pressure, volume, temperature, and number of moles of a gas. It's expressed as:
\( PV = nRT \),
where:

• \( P \) represents pressure measured in atm (atmospheres),
• \( V \) represents volume in liters,
• \( n \) stands for the number of moles,
• \( R \) is the ideal gas constant \( (0.08206 \text{ L atm/ K mol}) \),
• \( T \) is temperature in Kelvin.

This law assumes the gas behaves ideally, meaning particles move randomly without significant volume or interactions. In our exercise, after the reaction achieves equilibrium, we use the ideal gas law to find the volume. You need to remember to convert temperatures from Celsius to Kelvin by adding 273.15, since gas laws require absolute temperatures.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole \((\text{g/mol})\). It's crucial for converting between the mass of a substance and the amount in moles, which becomes particularly useful in chemical calculations. In the exercise, the molar mass of \( \text{XY} \) is given as 165 \( \text{g/mol} \). To find the initial moles of \( \text{XY} \), we use the formula:
\[ \text{Initial moles of XY} = \frac{\text{mass of XY}}{\text{molar mass of XY}} \]This simple relationship allows us to understand how much of \( \text{XY} \) is present initially. Molar mass acts as a bridge between the tangible mass and moles, which are more usable in stoichiometric calculations.

Dissociation Reaction

A dissociation reaction involves the breaking down of a compound into its simpler molecules or atoms. In our example, the gaseous compound \( \text{XY} \) dissociates into \( \text{X} \) and \( \text{Y} \), following the equation:
\[ \text{XY}(g) \rightleftharpoons \text{X}(g) + \text{Y}(g) \]
This reaction achieves a dynamic equilibrium where the rates of forward and reverse reactions are equal. In this exercise, 35% of \( \text{XY} \) dissociates. Calculating moles from the given percentage helps in determining how much \( \text{X} \) and \( \text{Y} \) are produced. Such reactions are significant in understanding the changes in concentration and volumes within a closed system over time.

Equilibrium Constant

The equilibrium constant \( K \) provides insight into the extent of a reaction at equilibrium. It’s a ratio of the concentrations of products to reactants raised to their stoichiometric coefficients in a balanced chemical equation:
\[ K = \frac{[\text{X}][\text{Y}]}{[\text{XY}]} \]
It remains constant for a given reaction at a set temperature. Concentrations are typically expressed in moles per liter (mol/L), calculated by dividing the number of moles by the volume of the container.
In solving our exercise, once we find the concentrations of \( \text{X} \), \( \text{Y} \), and \( \text{XY} \) after dissociation completes, we substitute these into the equilibrium expression to find \( K \).

• A large \( K \) means more products than reactants, indicating a product-favored equilibrium direction.
• A small \( K \) implies the reaction favors reactants.

Understanding \( K \) helps in predicting how a reaction mixture behaves at equilibrium under given conditions.