An 8.00 -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 calculate the initial amount (in moles) of \(\mathrm{SO_{3}}\), we will use its molar mass: Number of moles of \(\mathrm{SO_{3}} = \dfrac{\text{Given mass of SO3}}{\text{Molar mass of SO3}}\) The molar mass of \(\mathrm{SO_{3}}\) is approximately \(80.07 \, \mathrm{g/mol}\) (32.07 g/mol for S and 3 × 16.00 g/mol for O). Number of moles of $\mathrm{SO_{3}} = \dfrac{8.00 \, \mathrm{g}}{80.07 \, \mathrm{g/mol}} \approx 0.1 \, \mathrm{mol}\)

Let x be the number of moles of \(\mathrm{SO_{3}}\) that decompose at equilibrium. We can then determine the number of moles of \(\mathrm{SO_{2}}\) and \(\mathrm{O_{2}}\) formed: - Moles of \(\mathrm{SO_{3}}\) at equilibrium = \(0.1 - x\) - Moles of \(\mathrm{SO_{2}}\) at equilibrium = \(x\) - Moles of \(\frac{1}{2}\mathrm{O_{2}}\) at equilibrium = \(\dfrac{1}{2}x\)

We are given the density of the gaseous mixture: \(1.60 \, \mathrm{g/L}\). We can use this information to determine the volume of the container: Total mass of the gaseous mixture = 8.00g Density = \(\dfrac{\text{Mass}}{\text{Volume}}\) Therefore, Volume = $\dfrac{\text{Mass}}{\text{Density}} = \dfrac{8.00 \, \mathrm{g}}{1.60 \, \mathrm{g/L}} = 5.00 \, \mathrm{L}\$

The total pressure is given as \(1.80 \, \mathrm{atm}\). As the container volume is 5.00L, we can use the ideal gas equation to find the total moles of gas at equilibrium: Total moles at equilibrium = \(\frac{\text{Pressure × Volume}}{\text{Gas constant (R) × Temperature}}\) Using the appropriate values for the gas constant (R) and given temperature: \(PV = nRT \Rightarrow n = \frac{PV}{RT}\) For this problem, we will use the value of R = 0.0821 atm L/mol K. Temperature = \(600^{\circ} \mathrm{C} + 273.15 = 873.15 \, \mathrm{K}\) Total moles at equilibrium = \(\dfrac{1.80 \, \mathrm{atm} \times 5.00 \, \mathrm{L}}{0.0821 \, \mathrm{atm \, L/mol \, K} \times 873.15 \, \mathrm{K}} \approx 0.125 \, \mathrm{mol}\) From Step 2, at equilibrium we have: $$0.1 - x + x + \dfrac{1}{2}x = 0.125$$ Solving for x: $$x = 0.025 \, \mathrm{mol}$$ Using these values, we can calculate the partial pressures: $$P_{\mathrm{SO_{3}}} = \dfrac{\text{Moles of } \mathrm{SO_{3}}}{\text{Volume}} = \dfrac{0.1 - 0.025}{5.00} \times 1.80 \, \mathrm{atm} \approx 0.27 \, \mathrm{atm}$$ $$P_{\mathrm{SO_{2}}} = \dfrac{\text{Moles of } \mathrm{SO_{2}}}{\text{Volume}} = \dfrac{0.025}{5.00} \times 1.80 \mathrm{atm} \approx 0.09 \, \mathrm{atm}$$ $$P_{\frac{1}{2}\mathrm{O_{2}}} = \dfrac{\text{Moles of } \frac{1}{2}\mathrm{O_{2}}}{\text{Volume}} = \dfrac{0.5 \times 0.025}{5.00} \times 1.80 \, \mathrm{atm} \approx 0.045 \, \mathrm{atm}$$

Now that we have the partial pressures of all the gases involved in the reaction, we can calculate \(K_{\mathrm{p}}\) using the given reaction: $$K_{\mathrm{p}} = \dfrac{P_{\mathrm{SO_{2}}}\times P_{\frac{1}{2}\mathrm{O_{2}}^{\frac{1}{2}}}{P_{\mathrm{SO_{3}}}}$$ Plugging in the values obtained in Step 4: $$K_{\mathrm{p}} = \dfrac{0.09 \, \mathrm{atm} \times \sqrt{0.045 \, \mathrm{atm}}}{0.27 \, \mathrm{atm}} \approx 0.548$$ Therefore, the equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.548.