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

01

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 = \( \frac{2.00 g}{165 \mathrm{g/mol}} \)

02

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

03

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

04

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

05

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 = \( \frac{Mass}{Volume} \)

06

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.

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