At 298 \(\mathrm{K}\) , the standard enthalpies of formation for $\mathrm{C}_{2} \mathrm{H}_{2}(g)$ and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are 227 \(\mathrm{kJ} / \mathrm{mol}\) and \(49 \mathrm{kJ} / \mathrm{mol},\) respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

The formula used to calculate the reaction enthalpy, ΔH°, is as follows: \[ ΔH°_{reaction} = Σ nΔH°_{products} - Σ nΔH°_{reactants} \] where \(Σ\) represents the sum, \(n\) is the stoichiometric coefficient of each species, and \(ΔH°\) is the standard enthalpy of formation of each substance involved in the reaction.

We are given the standard enthalpy of formation for C2H2(g) and C6H6(l) at 298 K: - ΔH° (C2H2) = 227 kJ/mol - ΔH° (C6H6) = 49 kJ/mol

Now we can use the formula to calculate the reaction enthalpy ΔH° for the given reaction. \[ ΔH°_{reaction} = 3 × ΔH°_{C_2H_2} - ΔH°_{C_6H_6} \] \[ ΔH°_{reaction} = 3 × 227 - 49 = 681 - 49 = 632\text{ kJ/mol} \] The standard enthalpy change for the reaction C6H6(l) → 3C2H2(g) is ΔH° = 632 kJ/mol. **Part b: Compare the energy liberated per gram when combusted in air**

For C2H2(g) and C6H6(l), the combustion reactions are: - C2H2(g) + 2.5O2(g) → 2CO2(g) + H2O(g) - C6H6(l) + 7.5O2(g) → 6CO2(g) + 3H2O(g)

To calculate the energy per gram for each combustion reaction, we need to compute the energy per mole (ΔH°) divided by the molar mass of the compound. The molar mass of C2H2 is approximately 26 g/mol, whereas the molar mass of C6H6 is approximately 78 g/mol. Energy per gram for C2H2 combustion: \[ \text{Energy per gram} = \frac{ΔH°}{\text{molar mass}} = \frac{-227\text{ kJ/mol}}{26\text{ g/mol}} = -8.73\text{ kJ/g} \] Energy per gram for C6H6 combustion: \[ \text{Energy per gram} = \frac{ΔH°}{\text{molar mass}} = \frac{-49\text{ kJ/mol}}{78\text{ g/mol}} = -0.63\text{ kJ/g} \] Based on the calculations, acetylene (C2H2) would liberate more energy per gram (-8.73 kJ/g) when combusted in air, compared to benzene (C6H6), which only liberates -0.63 kJ/g.