Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$

We have the following reactions and their enthalpy changes: Reaction 1: $$ \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{6}(s) \\ \Delta H_1 = -1640.1\ \mathrm{kJ} $$ Reaction 2: $$ \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) \\ \Delta H_2 = -2940.1\ \mathrm{kJ} $$ We want to calculate the enthalpy change for this reaction: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$

First, let's reverse Reaction 1 so that we have \(\mathrm{P}_{4}\mathrm{O}_{6}(s)\) on the reactant side: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) \longrightarrow \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) \\ \Delta H_1^\prime = 1640.1\ \mathrm{kJ} \qquad (\mathrm{since\ the\ reaction\ has\ been\ reversed}) $$ Now, let's add Reaction 1' and Reaction 2 together: Reaction 1': $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) \longrightarrow \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) $$ Reaction 2: $$ \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$ As we add the two reactions, \(\mathrm{P}_{4}(s)\) on the right of Reaction 1' cancels with \(\mathrm{P}_{4}(s)\) on the left of Reaction 2. Similarly, \(3\mathrm{O}_{2}(g)\) on the right of Reaction 1' combines with \(5\mathrm{O}_{2}(g)\) on the left of Reaction 2, resulting in \(2\mathrm{O}_{2}(g)\) on the left side of the overall reaction. Resulting overall reaction: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$

As we add Reaction 1' and Reaction 2 to obtain the resulting overall reaction, we should also sum their enthalpy changes to obtain the enthalpy change for the target reaction. \(\Delta H_{\mathrm{overall}} = \Delta H_1^\prime + \Delta H_2 = 1640.1\ \mathrm{kJ} - 2940.1\ \mathrm{kJ} = -1300\ \mathrm{kJ}\) So, the enthalpy change for the target reaction is: $$ \Delta H_{\mathrm{overall}} = -1300\ \mathrm{kJ} $$