Questions: Use standard enthalpies of formation to calculate ΔHr.m° for the following reaction: CaO(s) + CO2(g) → CaCO3(s) Express your answer using four significant figures.

Solution Steps

The given reaction is: \[ \mathrm{CaO}(\mathrm{~s}) + \mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{~s}) \]

We need the standard enthalpies of formation (\(\Delta H_f^\circ\)) for each compound involved in the reaction:

• \(\Delta H_f^\circ (\mathrm{CaO}(\mathrm{~s})) = -635.5 \, \text{kJ/mol}\)
• \(\Delta H_f^\circ (\mathrm{CO}_2(\mathrm{~g})) = -393.5 \, \text{kJ/mol}\)
• \(\Delta H_f^\circ (\mathrm{CaCO}_3(\mathrm{~s})) = -1207.0 \, \text{kJ/mol}\)

The standard enthalpy change of the reaction (\(\Delta H_{\text{r.m}}^\circ\)) can be calculated using the formula: \[ \Delta H_{\text{r.m}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]

Substitute the standard enthalpies of formation into the formula: \[ \Delta H_{\text{r.m}}^\circ = \Delta H_f^\circ (\mathrm{CaCO}_3(\mathrm{~s})) - [\Delta H_f^\circ (\mathrm{CaO}(\mathrm{~s})) + \Delta H_f^\circ (\mathrm{CO}_2(\mathrm{~g}))] \]

\[ \Delta H_{\text{r.m}}^\circ = -1207.0 \, \text{kJ/mol} - [-635.5 \, \text{kJ/mol} + (-393.5 \, \text{kJ/mol})] \]

Calculate the enthalpy change: \[ \Delta H_{\text{r.m}}^\circ = -1207.0 \, \text{kJ/mol} - (-1029.0 \, \text{kJ/mol}) \]

\[ \Delta H_{\text{r.m}}^\circ = -1207.0 \, \text{kJ/mol} + 1029.0 \, \text{kJ/mol} \]

\[ \Delta H_{\text{r.m}}^\circ = -178.0 \, \text{kJ/mol} \]

Final Answer