Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

The balanced chemical equation for the reaction between sulfur dioxide and oxygen to form sulfur trioxide is: 2 SO2(g) + O2(g) -> 2 SO3(g) This equation shows that 2 moles of SO2 react with 1 mole of O2 to form 2 moles of SO3.

Initially, we have equal moles of SO2 and O2. Let's assume we have x moles of each gas. Before the reaction, there are: - x moles of SO2 - x moles of O2 According to the balanced chemical equation in Step 1, for every 2 moles of SO2, there is 1 mole of O2 consumed. Therefore, in our case, x moles of SO2 will completely react with x/2 moles of O2. This will leave us with x - x/2 = x/2 moles of O2 unreacted. After the reaction goes to completion, there are: - 0 moles of SO2 (all SO2 has reacted) - x/2 moles of O2 (unreacted) - x moles of SO3 (formed) The total moles of gas after the reaction are x + x/2 = 3x/2 moles.

Since the reaction is under constant temperature and pressure, we can use the ideal gas law (PV = nRT) to relate the volumes and moles of the gas mixture. Let V_initial and V_final represent the initial and final volumes of the gas mixture, respectively. Then: PV_initial = n_initial * RT PV_final = n_final * RT Since P, R, and T are constant for both cases, we can write the ratio of the final volume V_final to the initial volume V_initial: V_final / V_initial = (n_final * RT) / (n_initial * RT) The RT terms cancel out: V_final / V_initial = n_final / n_initial From Step 2, n_initial = 2x, and n_final = 3x/2: V_final / V_initial = (3x/2) / (2x) The x terms cancel out: V_final / V_initial = 3/4 So, the ratio of the final volume of the gas mixture to the initial volume of the gas mixture is 3/4.