Define standard enthalpy change of formation.

Explain the reaction pathway diagram (energy profile diagram) for an endothermic reaction.

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:

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

Given the enthalpy changes $∆{\mathrm{H}}_1$ and $∆{\mathrm{H}}_2$ below, construct Hess's cycle that will enable you to find the enthalpy change,$∆{\mathrm{H}}_{\mathrm{f}}$, for the reaction:

$$\begin{gathered} {\mathrm{MgCO}}_3\left(\mathrm{s}\right)\to \mathrm{MgO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right); ∆{\mathrm{H}}_{\mathrm{r}}. \\ {\mathrm{MgCO}}_3\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{MgCl}}_2 \left(\mathrm{aq}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); ∆{\mathrm{H}}_1. \\ \mathrm{MgO}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{MgCl}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); ∆{\mathrm{H}}_2. \end{gathered}$$