Use the information in Figure and the information below to demonstrate that the average bond energy of the $\mathrm{C}-\mathrm{H}$ bond is $415.9 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left[{\mathrm{CH}}_4\right]=-74.8 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ ∆{\mathrm{H}}_{\mathrm{at}}^{°}\left[\frac{1}{2}\mathrm{H}\right]=+218 \mathrm{kJ} {\mathrm{mol}}^{-1}. \\ ∆{\mathrm{H}}_{\mathrm{at}}^{°}\left[\mathrm{C}\left(\mathrm{graphite}\right)\right]=+716.7 \mathrm{kJ} {\mathrm{mol}}^{-1}. \end{gathered}$$

Look at this equation.

${\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$.

Which one of the following statements is completely correct?

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:

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