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1. Define standard enthalpy of formation. Explain why the enthalpy changes for the reaction given below are not enthalpies of formation of $CaCO_3$ and HBr:

1. $\text{CaO(s)}+\text{CO}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{CaCO}_3(\text{s});$ $\Delta_\text{r}\text{H}^\circ=-178.3\text{kJ mol}^{-1}$
2. $\text{H}_2(\text{g})+\text{Br}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{HBr(g)};$ $\Delta_\text{r}\text{H}^\circ=-72.8\text{kJ mol}^{-1}$

2. Use the bond enthalpies listed below to determine the enthalpy of reaction:

$\ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H(g)}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ }\ \text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$

Bond enthalpy $(\Delta\text{H}^\circ)/\text{kJ mol}^{-1}$ of C=O = 741; C-H = 414; H-O = 464; O=O = 498.