Multiple Equilibria - Simultaneous Reactions

Setting up the pressure ICE table: initial pressure of SO_3 = 3 atm, SO_2 = 0, O_2 = 0. Let x atm be the decrease in pressure of SO_3. Then: \Delta P(SO_3) = -2x, \Delta P(SO_2) = +2x, \Delta P(O_2) = +x. At equilibrium: P(SO_3) = 3-2x, P(SO_2) = 2x, P(O_2) = x. Total pressure = (3-2x) + 2x + x = 3 + x = 3.75, so x = 0.75 atm. Equilibrium pressures: P(SO_3) = 3 - 2(0.75) = 1.5 atm, P(SO_2) = 1.5 atm, P(O_2) = 0.75 atm. K_p = P(SO_2)^2 \times P(O_2) / P(SO_3)^2 = (1.5)^2 \times (0.75) / (1.5)^2 = 0.75 atm. However, let me recalculate carefully: K_p = (1.5)^2 \times (0.75) / (1.5)^2 = 0.75. But option A is 0.338; let me recheck: if x = 0.75, P(SO_3)=1.5, P(SO_2)=1.5, P(O_2)=0.75. K_p = (1.5^2)(0.75)/(1.5^2) = 0.75 atm. Given options, option B (0.844) and option A (0.338) don't match, but 0.75 is closest to none listed. Re-examining with initial moles approach and partial pressures: the calculated K_p = 0.75 atm by the ICE method is the correct value. Option 0.338 would use incorrect stoichiometric assignments. Option 0.844 uses different x value. The standard JEE approach via pressure ICE table yields K_p = 0.75 atm, illustrating that careful stoichiometric tracking is essential. This is an NCERT-style advanced equilibrium calculation testing simultaneous application of stoichiometry and K_p definition.