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

01

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 \(\mathrm{XY}\). $$\text{moles of}\,\mathrm{XY} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.00\,\mathrm{g}}{165\,\mathrm{g/mol}} = 0.0121\,\text{mol}$$

02

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: $$0.0121\,\text{mol} \times 0.350 = 0.00424\,\text{mol}$$

03

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: $$\chi_{XY} = \frac{0.00786\,\text{mol}}{0.01634\,\text{mol}} = 0.481$$ $$\chi_{X} = \frac{0.00424\,\text{mol}}{0.01634\,\text{mol}} = 0.259$$ $$\chi_{Y} = \frac{0.00424\,\text{mol}}{0.01634\,\text{mol}} = 0.259$$

04

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 \(298\,\mathrm{K}\): $$P = \frac{n\bar{M}}{V}\,R\,T \Rightarrow \rho = \frac{P\bar{M}}{R\,T} = \frac{0.967\,\mathrm{atm}\times 165\,\mathrm{g/mol}}{0.0821\,(\mathrm{L\,atm/mol\,K})\times 298\,\mathrm{K}} = 3.15\,\mathrm{g/L}$$

05

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 \(0.139\).

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