An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ At equilibrium the total pressure and the density of the gaseous mixture were \(1.80 \mathrm{~atm}\) and \(1.60 \mathrm{~g} / \mathrm{L}\), respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

To begin, we must first convert the initial mass of \(\mathrm{SO}_{3}\) to moles. We can do this using the molar mass of \(\mathrm{SO_3}\), which is \(32 + 3 \times 16 = 80\,\text{g/mol}\). Since the initial mass is \(8.00\,\text{g}\), Initial moles of \(\mathrm{SO_3} =\frac{8.00\,\text{g}}{80\,\text{g/mol}} = 0.10\,\text{mol}\).

We are given the density of the gaseous mixture at equilibrium, which is \(1.60\,\text{g/L}\). We can write the equation for the density as follows: Density = \( \frac{\text{Mass of the mixture}}{\text{Volume of the mixture}}\) Before proceeding, let's consider that: - The total mass of the gaseous mixture is \(8.00\,\text{g}\) (since none of the gases are lost) - The total moles of the mixture at equilibrium = \(n_{\mathrm{SO_2}} + n_{\mathrm{O_2}} + n_{\mathrm{SO_3}}\) - The partial pressures of the gases are in proportion to their moles Now, let's use the density to find the moles of the gaseous mixture at equilibrium: \(\frac{8.00\,\text{g}}{\text{Volume of the mixture}}=1.60\,\text{g/L}\) \(\text{Volume of the mixture} = \frac{8.00\,\text{g}}{1.60\,\text{g/L}} = 5.00\,\text{L}\) Assuming ideal gas behavior, we can use the equation \(PV=nRT\) to find the total moles of the gaseous mixture at equilibrium. Given, total pressure \(P = 1.80\,\text{atm}\), volume \(V = 5.00\,\text{L}\), and temperature \(T = 600^{\circ}\text{C}\), we need to convert the temperature to Kelvin. \(T = 600 + 273.15 = 873.15\,\text{K}\). Also, \(R = 0.0821\,\text{atm}\text{L/mol}\text{K}\) Now, we can solve for \(n_{\text{total}}\): \(n_{\text{total}}=\frac{PV}{RT} = \frac{1.80\,\text{atm} \cdot 5.00\,\text{L}}{0.0821\,\text{atm}\text{L/mol}\text{K} \cdot 873.15\,\text{K}}=0.135\,\text{mol}\)

Let \(x\) represent the moles of \(\mathrm{SO_3}\) that decompose. According to the stoichiometry of the reaction given, we can say \(x\) moles of \(\mathrm{SO_2}\) and \(x/2\) moles of \(\mathrm{O_2}\) are produced at equilibrium. Then, we can write the equilibrium expressions in terms of moles as: \(n_{\mathrm{SO_3}} = 0.10 - x\) \(n_{\mathrm{SO_2}} = x\) \(n_{\mathrm{O_2}} =\frac{x}{2}\) Since we found that the total moles at equilibrium are \(0.135\,\text{mol}\), we can write the equation: \(n_{\mathrm{SO_3}} + n_{\mathrm{SO_2}} + n_{\mathrm{O_2}} = 0.135\,\text{mol}\) Plugging the expressions for the moles, we get: \((0.10 - x) + x + \frac{x}{2} = 0.135\) Solving for \(x\): \(0.10 + \frac{x}{2} = 0.135\) \(x = 0.07\,\text{mol}\) Now we can determine the moles of each species at equilibrium: \(n_{\mathrm{SO_3}} = 0.10 - 0.07 = 0.03\,\text{mol}\) \(n_{\mathrm{SO_2}} = 0.07\,\text{mol}\) \(n_{\mathrm{O_2}} =\frac{0.07}{2} = 0.035\,\text{mol}\)

We will now use the mole fractions and the total pressure to determine the partial pressures of the gas species. \(P_{\mathrm{SO_3}} = \frac{n_{\mathrm{SO_3}}}{n_{\text{total}}} \times P\) \(P_{\mathrm{SO_3}} = \frac{0.03\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.40\,\text{atm}\) Similarly, \(P_{\mathrm{SO_2}} = \frac{0.07\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.93\,\text{atm}\) \(P_{\mathrm{O_2}} = \frac{0.035\,\text{mol}}{0.135\,\text{mol}} \times 1.80\,\text{atm} = 0.47\,\text{atm}\)

Finally, we can use the partial pressures to calculate the equilibrium constant \(K_p\) for the reaction. The expression for \(K_p\) is as follows: \(K_p=\frac{P_{\mathrm{SO_2}} \times \sqrt{P_{\mathrm{O_2}}}}{P_{\mathrm{SO_3}}}\) By plugging in the values, we get: \(K_p = \frac{0.93\,\text{atm} \times \sqrt{0.47\,\text{atm}}}{0.40\,\text{atm}} = 1.80\) So, the equilibrium constant \(K_p\) for this reaction is \(1.80\).