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?

Write the balanced chemical equation for the reaction between sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3): \[2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)\]

The mole ratio between the reactants and products in the balanced chemical equation is: 2 moles of SO2 reacts with 1 mole of O2 to produce 2 moles of SO3.

The Ideal Gas Law is given by the equation: \(PV = nRT\) Since we are dealing with the same temperature (T) and pressure (P), we can use a simplified form of the Ideal Gas Law relating the initial and final volume (V) to the number of moles (n) and the gas constant (R). Thus, we have two equations: Initial Volume (V1) = (n_SO2 + n_O2)RT Final Volume (V2) = n_SO3RT

Divide V2 by V1 to find the ratio of final volume to initial volume: \(\frac{V2}{V1} = \frac{n_{SO3}RT}{(n_{SO2} + n_{O2})RT}\) Now we can cancel out the T and R terms from both numerator and denominator, and substitute the mole ratios from the balanced chemical equation (Step 2): \(\frac{V2}{V1} = \frac{2n_{SO2}}{2n_{SO2} + n_{O2}}\) Since equal moles of sulfur dioxide and oxygen have been mixed, we can say that: \(n_{SO2} = n_{O2} = n\) Substitute this into our volume ratio equation and simplify: \(\frac{V2}{V1} = \frac{2n}{2n + n} = \frac{2n}{3n}\) Now, cancel out the n term from both numerator and denominator: \(\frac{V2}{V1} = \frac{2}{3}\) Thus, the ratio of the final volume of the gas mixture to the initial volume of the gas mixture is 2:3.