# Equilibrium Ratio of Partial Pressures of SO₃ and SO₂ in Air 298 K ## Introduction are asked to find the ratio of the partial pressures of sulfur trioxide (SO₃) to sulfur dioxide (SO₂) in air at 298 K. The partial pressure of oxygen (O₂) in air is given as .21 bar. The key reaction involved is the **contact process**: \[ 2 \, \mathrm{SO_2} (g) + \mathrm{O_2} (g) \rightleftharpoons 2 \, \mathrm{SO_3} (g) \] This reaction reaches an equilibrium that depends on temperature and partial pressures. --- ## Step 1: Write the equilibrium expression The equilibrium constant in terms of partial pressures, \(K_p\), is: \[ K_p = \frac{(P_{\mathrm{SO_3}})^2}{(P_{\mathrm{SO_2}})^2 \times P_{\mathrm{O_2}}} \] Where: - \(P_{\mathrm{SO_3}}\) = partial pressure of SO₃ - \(P_{\mathrm{SO_2}}\) = partial pressure of SO₂ - \(P_{\mathrm{O_2}}\) = partial pressure of O₂ = .21 bar (given) --- ## Step 2: Find \(K_p\) at 298 K The equilibrium constant \(K_p\) can be calculated from thermodynamic data: \[ \Delta G^\circ = -RT \ln K_p \] But a more straightforward approach is to use standard Gibbs free energies of formation or standard enthalpy and entropy changes for the reaction. ### Thermodynamic data at 298 K: | Species | \(\Delta G_f^\circ \) (kJ/mol) | | --------- | ------------------------------ | | SO₂ | -300.4 | | SO₃ | -370.4 | | O₂ | | ### Calculate \(\Delta G^\circ_{\text{reaction}}\): \[ \Delta G^\circ_{\text{reaction}} = \left[2 \times \Delta G_f^\circ (\mathrm{SO_3}) \right] - \left[2 \times \Delta G_f^\circ (\mathrm{SO_2}) + \Delta G_f^\circ (\mathrm{O_2})\right] \] \[ \Delta G^\circ_{\text{reaction}} = 2 \times (-370.4) - \left[2 \times (-300.4) + \right] \] \[ = -740.8 - (-600.8) = -740.8 + 600.8 = -140 \text{ kJ/mol} \] ### Calculate \(K_p\): \[ \Delta G^\circ = -RT \ln K_p \] Where: - \(R = 8.314\, \text{J/(mol·K)}\) - \(T = 298\, \text{K}\) - \(\Delta G^\circ = -140\, \text{kJ/mol} = -140,000\, \text{J/mol}\) \[ -140,000 = - (8.314)(298) \ln K_p \] \[ \ln K_p = \frac{140,000}{8.314 \times 298} \approx \frac{140,000}{2477} \approx 56.5 \] \[ K_p = e^{56.5} \] This is a very large number—indicating the reaction strongly favors SO₃ formation. --- ## Step 3: Calculate the ratio \( \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} \) From the equilibrium expression: \[ K_p = \frac{(P_{\mathrm{SO_3}})^2}{(P_{\mathrm{SO_2}})^2 \times P_{\mathrm{O_2}}} \] Rearranged to find \( \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} \): \[ \left(\frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}}\right)^2 = K_p \times P_{\mathrm{O_2}} \] \[ \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} = \sqrt{K_p \times P_{\mathrm{O_2}}} \] Substitute the known values: \[ \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} = \sqrt{e^{56.5} \times .21} \] Since \(e^{56.5}\) is extremely large, the ratio will be enormous. To express it more practically, take the natural logarithm: \[ \ln \left(\frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}}\right) = \frac{1}{2} \left(\ln K_p + \ln P_{\mathrm{O_2}}\right) \] \[ = \frac{1}{2} \left(56.5 + \ln .21\right) \] Calculate \(\ln .21\): \[ \ln .21 \approx -1.56 \] So: \[ \ln \left(\frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}}\right) = \frac{1}{2} (56.5 - 1.56) = \frac{1}{2} \times 54.94 \approx 27.47 \] Finally: \[ \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} = e^{27.47} \approx 1.2 \times 10^{11} \] --- ## **Final Answer** ### The equilibrium ratio of partial pressures of SO₃ to SO₂ in air at 298 K is approximately: \[ \boxed{ \frac{P_{\mathrm{SO_3}}}{P_{\mathrm{SO_2}}} \approx 1.2 \times 10^{11} } \] This indicates that, under equilibrium conditions at 298 K, the partial pressure of SO₃ vastly exceeds that of SO₂.