When one mole of calcium carbonate reacts with ammonia, solid calcium cyanamide, \(\mathrm{CaCN}_{2}\), and liquid water are formed. The reaction absorbs \(90.1 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction. (b) Using Table 8.3, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for calcium cyanamide.

The first step to solve this problem is to write the balanced chemical equation for the reaction between calcium carbonate and ammonia. The reactants are calcium carbonate (CaCO3) and ammonia (NH3), and the products are calcium cyanamide (CaCN2) and water (H2O) in its liquid state. CaCO3 (s) + 2NH3 (g) -> CaCN2 (s) + CO2 (g) + H2O (l)

The second step is to include the heat absorbed, \(90.1 \mathrm{~kJ}\), in the balanced chemical equation. Since the reaction absorbs heat, it is an endothermic reaction. We can write the enthalpy change as \(\Delta H = +90.1 \mathrm{~kJ}\). CaCO3 (s) + 2NH3 (g) -> CaCN2 (s) + CO2 (g) + H2O (l) + \(90.1 \mathrm{kJ}\)

To calculate the standard enthalpy of formation of calcium cyanamide, we need to use the enthalpy of formation of each substance involved in the reaction, using Table 8.3 data. Using Hess's law, we can calculate the enthalpy of formation of calcium cyanamide (\(\Delta H_{\mathrm{f}}^{\circ}\)) as: \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = \Delta H_{\mathrm{f}}^{\circ}(\text{CaCO}_3) + 2\Delta H_{\mathrm{f}}^{\circ}(\text{NH}_3) - \Delta H_{\mathrm{f}}^{\circ}(\text{CO}_2) - \Delta H_{\mathrm{f}}^{\circ}(\text{H}_2 \text{O})\) We just need to substitute the values from Table 8.3 and solve for \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2)\). \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCO}_3) = -1207.0 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{NH}_3) = -45.9 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{CO}_2) = -393.5 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{H}_2 \text{O}) = -285.8 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = (-1207.0) + 2(-45.9) - (-393.5) - (-285.8)\)

Now, we can calculate the final answer: \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = -1207.0 + (-91.8) + 393.5 + 285.8 = -619.5 \, \mathrm{kJ/mol}\) The standard enthalpy of formation of calcium cyanamide is -619.5 kJ/mol.