Define standard enthalpy change of combustion.

When red phosphorus burns in oxygen the enthalpy change is $-2967 \mathrm{kJ} {\mathrm{mol}}^{-1}$. For white phosphorus the enthalpy change is $-2984 \mathrm{kJ} {\mathrm{mol}}^{-1}$. For both forms of phosphorus the reaction taking place is: ${\mathrm{P}}_4\left(\mathrm{s}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{P}}_4{\mathrm{O}}_{10}\left(\mathrm{s}\right)$.

Use this information to calculate the enthalpy change for the transformation: ${\mathrm{P}}_4\left(\mathrm{white}\right)\to {\mathrm{P}}_4\left(\mathrm{red}\right)$.

Calculate the standard enthalpy change of combustion of methane using the following bond energies:

$$\begin{gathered} \mathrm{E}\left(\mathrm{C}-\mathrm{H}\right)=+412 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ \mathrm{E}\left(\mathrm{O}=\mathrm{O}\right)=+496 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ \mathrm{E}\left(\mathrm{C}=\mathrm{O}\right)=+805 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ \mathrm{E}\left(\mathrm{O}-\mathrm{H}\right)=+463 \mathrm{kJ} {\mathrm{mol}}^{-1}. \end{gathered}$$

Use the average bond energies that follow to calculate a value for the enthalpy change for the reaction:

$$\begin{gathered} {\mathrm{H}}_2+{\mathrm{I}}_2\to 2\mathrm{HI}. \\ \mathrm{E}\left(\mathrm{H}-\mathrm{H}\right)=+436 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ \mathrm{E}\left(\mathrm{I}-\mathrm{I}\right)=+151 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ \mathrm{E}\left(\mathrm{H}-\mathrm{I}\right)=+299 \mathrm{kJ} {\mathrm{mol}}^{-1}. \end{gathered}$$