Complete combustion of \(1 \mathrm{~mol}\) of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates $$ \begin{array}{l} 1790 \mathrm{~kJ}: \\ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-1790 \mathrm{~kJ} \end{array} $$ Using this information together with data from Appendix C, calculate the enthalpy of formation of acetone.

We are given the following information: 1. The combustion reaction of 1 mol of acetone (C3H6O): \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \) 2. The enthalpy change of this reaction: \( \Delta H^{\circ} = -1790 \, \mathrm{kJ} \) Step 2: Recall the definition of enthalpy of formation

Enthalpy of formation (\( \Delta H_f \)) is the change in enthalpy when 1 mol of a substance is formed from its elements in their standard states. Step 3: Write down the equation relating enthalpy of combustion and enthalpy of formation

The equation that relates enthalpy of combustion (\( \Delta H_c \)) and enthalpy of formation of reactants and products is: \[ \Delta H_c = \sum n_p \Delta H_{f,p} - \sum n_r \Delta H_{f,r} \] Where: - \( \Delta H_{c} \) is the enthalpy of combustion, - \( \Delta H_{f,p} \) is the enthalpy of formation of products, - \( \Delta H_{f,r} \) is the enthalpy of formation of reactants, - \( n_p \) is the moles of each product, - \( n_r \) is the moles of each reactant. Step 4: Compute the enthalpy of combustion using given data

Using the given data from Appendix C, lookup the enthalpy of formation values for CO2 and H2O. We obtain the following values: - \( \Delta H_{f, \mathrm{CO}_{2}} = -393.5\, \mathrm{kJ}\, \mathrm{mol}^{-1} \) - \( \Delta H_{f, \mathrm{H}_{2} \mathrm{O}} = -285.8\, \mathrm{kJ}\, \mathrm{mol}^{-1} \) Now, substitute the above values and the given enthalpy change into Equation 1: \[ -1790 = \left[ 3 \times \left( -393.5 \right) + 3 \times \left( -285.8 \right) \right] - \left[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} + 4 \times \left( 0 \right) \right] \] Since the enthalpy of formation of O2 in its standard state (O2 gas) is zero, we can simplify the equation as: \[ -1790 = \left[ -1180.5 - 857.4 \right] - \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \] Step 5: Solve for the enthalpy of formation of acetone

Rearrange the equation in Step 4 to find \( \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \): \[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} = -1180.5 - 857.4 - \left(-1790\right) \] Now, calculate the final value: \[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} = -2037.9 + 1790 = -247.9\, \mathrm{kJ\, mol^{-1}} \] The enthalpy of formation of acetone, \( \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \), is approximately \( -247.9\, \mathrm{kJ\, mol^{-1}} \).