For the gas-phase oxidation of sulfur dioxide,2 SO₂(g)+ O₂(g)$\leftrightarrows$ 2 SO₃(g),K_p= 4.17 $\times$ 10² at 200.0 $\circ$ C.If a closed vessel is filled with SO₃ at an initial pressure of 0.033 atm at 200 $\circ$ C,what is the equilibrium partial pressure of SO₃?

Question

Quizplus · Accepted Answer

According to Le Chatelier's principle, the equilibrium will shift to the side with fewer moles of gas when the pressure is increased. In this case, the forward reaction (2 SO2 + O2 → 2 SO3) produces 2 moles of gas, while the reverse reaction (2 SO3 → 2 SO2 + O2) produces only 3 moles of gas. Therefore, increasing the pressure will shift the equilibrium to the left (toward the side with fewer moles of gas), resulting in a decrease in the partial pressure of SO3. Since the initial pressure of SO3 is very low compared to the equilibrium partial pressure, we can assume that the shift in equilibrium will be significant enough to cause a decrease in SO3 partial pressure below the initial value of 0.033 atm. Therefore, the correct answer is B) less than 0.033 atm.

For the Gas-Phase Oxidation of Sulfur Dioxide,2 SO2(g)+ O2(g) ⇆\leftrightarrows⇆ 2 SO3(g),Kp = 4

For the Gas-Phase Oxidation of Sulfur Dioxide,2 SO₂(g)+ O₂(g) $\leftrightarrows$ 2 SO₃(g),K_p= 4