A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

We are given that the total pressure at equilibrium is 0.836 atm and the partial pressure of oxygen (\(\mathrm{O}_2\)) is 0.0275 atm. We also have the balanced chemical equations for the two reactions that occur.

In order to determine the change in moles, we will look at the coefficients in the balanced chemical equations. Using the coefficients, we can write changes in moles, and thus changes in the partial pressures, as follows: For the first reaction: \(\Delta P_{\mathrm{FeSO}_4} = -2x\) \(\Delta P_{\mathrm{Fe}_2\mathrm{O}_3} = x\) \(\Delta P_{\mathrm{SO}_3} = x\) \(\Delta P_{\mathrm{SO}_2} = x\) For the second reaction: \(\Delta P_{\mathrm{SO}_3} = -y\) \(\Delta P_{\mathrm{SO}_2} = y\) \(\Delta P_{\frac{1}{2}\mathrm{O}_2} = \frac{1}{2}y\)

We can use the given partial pressure of oxygen and the changes in moles from Step 2 to calculate the partial pressures of all species. Partial pressure of \(\mathrm{SO}_2\): \(P_{\mathrm{SO}_2} = x+y\) Since \(P_{\frac{1}{2}\mathrm{O}_2}=0.0275 \, \mathrm{atm}\), we have \(y = 2 \times 0.0275 = 0.055 \, \mathrm{atm}\). Now, we can calculate \(x\) using the fact that the total pressure is 0.836 atm: \(0.836 = 0.0275 + (x+y)\) \(x = 0.836 - 0.0275 - 0.055 = 0.7535 \, \mathrm{atm}\) Therefore, partial pressures for each species are: \(P_{\mathrm{FeSO}_4} = 0\) \(P_{\mathrm{Fe}_2\mathrm{O}_3} = x = 0.7535 \, \mathrm{atm}\) \(P_{\mathrm{SO}_3} = x-y = 0.7535 - 0.055 = 0.6985 \, \mathrm{atm}\) \(P_{\mathrm{SO}_2} = x+y = 0.7535 + 0.055 = 0.8085 \, \mathrm{atm}\)

Now that we have all the partial pressures, we can calculate \(K_{\mathrm{p}}\) for both reactions using the following formulas: For the first reaction: \(K_{\mathrm{p1}} = \frac{P_{\mathrm{Fe}_2\mathrm{O}_3} \cdot P_{\mathrm{SO}_3} \cdot P_{\mathrm{SO}_2}}{(P_{\mathrm{FeSO}_4})^2}\) For the second reaction: \(K_{\mathrm{p2}} = \frac{P_{\mathrm{SO}_2} \cdot P_{\frac{1}{2}\mathrm{O}_2}}{P_{\mathrm{SO}_3}}\) Plugging in the partial pressures we found in Step 3: For the first reaction: \(K_{\mathrm{p1}} = \frac{0.7535 \cdot 0.6985 \cdot 0.8085}{(0)^2} = \frac{0.7535 \cdot 0.6985 \cdot 0.8085}{1} = 423.30\) For the second reaction: \(K_{\mathrm{p2}} = \frac{0.8085 \cdot 0.0275}{0.6985} = 0.032\) So, the equilibrium constants are: \(K_{\mathrm{p1}} = 423.30\) \(K_{\mathrm{p2}} = 0.032\)