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?

To find the standard enthalpy change for \(C_6H_6(l) \longrightarrow 3C_2H_2(g)\), we will use the formula: \[\Delta H^\circ = \sum^n_{i=1} \Delta H^\circ_{\text{products}} - \sum^n_{i=1} \Delta H^\circ_{\text{reactants}}\] In our case, the standard enthalpy of formation for \(C_6H_6 (l)\) is 49 kJ/mol, and for \(C_2H_2 (g)\) is 227 kJ/mol. Plugging in the values, we get \[\Delta H^\circ = 3\times (227) - 1 \times(49)\]

Now let's calculate the value of ΔH° from the expression above: \[\Delta H^\circ = (681) - (49)\] \[\Delta H^\circ = 632 \mathrm{kJ/mol}\] So, the standard enthalpy change for the reaction \(C_6H_6(l) \longrightarrow 3C_2H_2(g)\) is 632 kJ/mol.

In this step, we'll first calculate the energy released per gram upon combustion for each compound by finding their specific combustion enthalpies. After that, we'll compare which compound releases more energy per gram. For acetylene: The combustion reaction for acetylene is: \(2 C_2H_2(g) + 5 O_2(g) \longrightarrow 4 CO_2(g) + 2 H_2O(g)\) It has a ΔH° = -1256 kJ/mol (based on the given enthalpy of formation of acetylene). Molar mass of acetylene: \(M_{C_2H_2} = 2 \times (12.01) + 2 \times (1) = 26.04 \mathrm{g/mol}\) Specific combustion enthalpy for acetylene: \(-1256\, \mathrm{kJ/mol} / 26.04\, \mathrm{g/mol} \approx -48.25\, \mathrm{kJ/g}\) For benzene: The combustion reaction for benzene is: \(C_6H_6(l) + \frac{15}{2} O_2(g) \longrightarrow 6 CO_2(g) + 3 H_2O(g)\) It has a ΔH° = -3268 kJ/mol (based on the given enthalpy of formation of benzene). Molar mass of benzene: \(M_{C_6H_6} = 6 \times (12.01) + 6 \times (1) = 78.12 \mathrm{g/mol}\) Specific combustion enthalpy for benzene: \(-3268\, \mathrm{kJ/mol} / 78.12\, \mathrm{g/mol} \approx -41.83\, \mathrm{kJ/g}\)

Comparing both specific combustion enthalpies, we can see that acetylene, with a specific combustion enthalpy of approximately -48.25 kJ/g, releases more energy per gram upon combustion compared to benzene, which releases approximately -41.83 kJ/g. Therefore, acetylene is the better fuel in terms of energy release per gram when combusted in air.