Question #44682

A highly reactive intermediate \ce{X} (molar mass = $24 \text{ g mol}^{-1}$) is generated photochemically in a solvent \ce{M} at $298 \text{ K}$. It reacts with dissolved \ce{O2} to form an excited adduct \ce{XO2^}, which then undergoes two competitive processes: \begin{align} \ce{XO2^* (solv)} &\xrightarrow{k_1} \ce{XO (solv)} + \ce{O (solv)} \quad \text{(Unimolecular dissociation)} \ \ce{XO2^* (solv)} + \ce{M (l)} &\xrightarrow{k_2} \ce{XO2 (solv)} + \ce{M^* (l)} \quad \text{(Collisional deactivation)} \end{align*} The oxygen atoms (\ce{O}) generated in the first step immediately recombine to form \ce{O2} and do not react with \ce{X}. Given the rate constants: $k_1 = 1.5 \times 10^9 \text{ s}^{-1}$ $k_2 = 5.0 \times 10^7 \text{ M}^{-1}\text{s}^{-1}$ The concentration of the solvent is $[\ce{M}] = 10 \text{ M}$. A student incorrectly assumes that the average mass of oxygen combined with \ce{X} in the bulk mixture must be the arithmetic mean of the masses dictated by the Law of Multiple Proportions for \ce{XO} and \ce{XO2} (i.e., an equimolar mixture). What is the \textbf{actual} total mass of oxygen atoms incorporated into the stable products (\ce{XO} and \ce{XO2}) when $12.0 \text{ g}$ of \ce{X} is completely consumed in this solvent?