The standard enthalpies of formation of gaseous propyne $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\( propylene \)\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\( and propane \)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\( are \)+185.4,+20.4,\( and \)-103.8 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and $\mathrm{H}_{2} \mathrm{O}(g) .\( (b) Calculate the heat evolved on combustion of \)1 \mathrm{~kg}\( of each substance. \)(\mathbf{c})$ Which is the most efficient fuel in terms of heat evolved per unit mass?

For each substance, we need to write the balanced combustion equation, which involves the complete reaction of the substance with oxygen (O₂) to form CO₂ and H₂O as products. (a) Propyne (C₃H₄): \(C_3H_4 + 4O_2 \rightarrow 3CO_2 + 2H_2O\) (b) Propylene (C₃H₆): \(C_3H_6 + 4.5O_2 \rightarrow 3CO_2 + 3H_2O\) (c) Propane (C₃H₈): \(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\)

Now we will calculate the heat evolved per mole during the combustion of each substance. We can do this by applying Hess's law and using the standard enthalpies of formation: ΔH_combustion = Σ(ΔH_f(products)) - Σ(ΔH_f(reactants)) For each substance, the heat evolved per mole is as follows: (a) Propyne: ΔH_combustion (Propyne) = \(3ΔH_f(CO_2) + 2ΔH_f(H_2O) - ΔH_f(C_3H_4) - 4ΔH_f(O_2)\) Since the standard enthalpy of formation for O₂ is zero, we can simplify the equation as: ΔH_combustion (Propyne) = \(3ΔH_f(CO_2) + 2ΔH_f(H_2O) - ΔH_f(C_3H_4)\) (b) Propylene: ΔH_combustion (Propylene) = \(3ΔH_f(CO_2) + 3ΔH_f(H_2O) - ΔH_f(C_3H_6) - 4.5ΔH_f(O_2)\) ΔH_combustion (Propylene) = \(3ΔH_f(CO_2) + 3ΔH_f(H_2O) - ΔH_f(C_3H_6)\) (c) Propane: ΔH_combustion (Propane) = \(3ΔH_f(CO_2) + 4ΔH_f(H_2O) - ΔH_f(C_3H_8) - 5ΔH_f(O_2)\) ΔH_combustion (Propane) = \(3ΔH_f(CO_2) + 4ΔH_f(H_2O) - ΔH_f(C_3H_8)\)

We need to find the molar masses of each substance to calculate the heat evolved on combustion of 1 kg of each substance. Molar mass (C₃H₄) = 3 × 12 + 4 × 1 = 36 + 4 = 40 g/mol Molar mass (C₃H₆) = 3 × 12 + 6 × 1 = 36 + 6 = 42 g/mol Molar mass (C₃H₈) = 3 × 12 + 8 × 1 = 36 + 8 = 44 g/mol

To find the heat evolved on combustion of 1 kg of each substance, we divide 1000 g by the molar mass and multiply it by the heat evolved per mole of the substance. Heat evolved (C₃H₄) = 1000 g x \(\frac{ΔH_{combustion}(C_3H_4)}{40g/mol}\) Heat evolved (C₃H₆) = 1000 g x \(\frac{ΔH_{combustion}(C_3H_6)}{42g/mol}\) Heat evolved (C₃H₈) = 1000 g x \(\frac{ΔH_{combustion}(C_3H_8)}{44g/mol}\)

To compare which fuel is the most efficient in terms of heat evolved per unit mass, we can compare the heat evolved on combustion of 1 kg of each substance. The fuel with the highest heat evolved per kg is the most efficient fuel.