Question

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?

Short answer

The short answer for this question: (a) ΔH_combustion (Propyne) = \(3ΔH_f(CO_2) + 2ΔH_f(H_2O) - ΔH_f(C_3H_4)\) ΔH_combustion (Propylene) = \(3ΔH_f(CO_2) + 3ΔH_f(H_2O) - ΔH_f(C_3H_6)\) ΔH_combustion (Propane) = \(3ΔH_f(CO_2) + 4ΔH_f(H_2O) - ΔH_f(C_3H_8)\) (b) 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 \(\frac{ΔH_{combustion}(C_3H_8)}{44g/mol}\) (c) The most efficient fuel in terms of heat evolved per unit mass can be determined by comparing the heat evolved on combustion of 1 kg of each substance, with the highest heat evolved per kg being the most efficient fuel.

Step-by-step solution

Write the balanced combustion reactions

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): \(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\)

Calculate the heat evolved per mole

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) = \(3ΔH_f(CO_2) + 4ΔH_f(H_2O) - ΔH_f(C_3H_8)\)

Find the molar masses

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

Calculate the heat evolved on combustion of 1 kg of each substance

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 \(\frac{ΔH_{combustion}(C_3H_8)}{44g/mol}\)

Determine the most efficient fuel

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.

Key concepts

Enthalpies of Formation

Enthalpies of formation are crucial to understanding combustion processes. This term refers to the change in enthalpy when one mole of a compound forms from its elements in their standard states. - Standard conditions generally mean 1 atm pressure and a specified temperature (usually 25°C). - Each compound has a unique enthalpy of formation, often represented as ΔH_f. For combustion reactions, it's essential to know the enthalpies of formation for both the reactants and products. For example: - The enthalpy of formation of carbon dioxide (CO₂) is a key component when calculating combustion enthalpies, alongside water vapor (H₂O). Understanding these enthalpies allows us to use Hess's Law to compute the heat involved in reactions, including those as energetic as combustion. By combining the known enthalpies for CO₂, H₂O, and the hydrocarbons mentioned, we can find the total energy change for combustion.

Combustion Reactions

Combustion reactions are a type of exothermic reaction where a substance reacts with oxygen to produce heat and light. These reactions are essential for energy production in various fields, from industrial processes to daily fuel usage. - Combustion of hydrocarbons, such as - Propyne (C₃H₄) - Propylene (C₃H₆) - Propane (C₃H₈) follows a similar pattern: 1. Reacting with oxygen. 2. Producing carbon dioxide and water vapor. The chemical equations must be balanced to accurately reflect the conservation of mass. For instance, the combustion of propane is represented as: - C₃H₈ + 5O₂ → 3CO₂ + 4H₂O Balanced equations are essential for calculating the exact enthalpy change of the reaction. Combustion reactions release significant energy, quantified using the enthalpies of formation. This energy release is commonly harnessed in engines and heating systems.

Hess's Law

Hess's Law is a powerful tool in thermochemistry, stating that the total enthalpy change for a reaction is the same, regardless of how many steps the reaction takes. To apply Hess's Law in combustion: - We calculate the enthalpy change (ΔH_combustion) of the overall reaction by using the standard enthalpies of formation of the products and subtracting those of the reactants. - The formula for the combustion of a hydrocarbon can be summarized as: ΔH_combustion = Σ(ΔH_f(products)) - Σ(ΔH_f(reactants)) For each hydrocarbon: - Products typically include CO₂ and H₂O. - Their known enthalpies of formation can be inserted into the formula. By using this method, we can quantify the energy change regardless of intermediate steps, enabling precise calculations needed for industrial and chemical applications.

Molar Masses

Molar mass is the mass of one mole of a substance, expressed in grams per mole. It is an essential concept when calculating energy changes in chemical reactions, such as combustion. - Calculating the molar mass involves adding up the atomic masses of all atoms present in a molecule. For example: - Propyne ( C₃H₄): - Molar mass = (3 × 12 g/mol for Carbon) + (4 × 1 g/mol for Hydrogen) = 40 g/mol - Propylene ( C₃H₆): - Molar mass = (3 × 12) + (6 × 1) = 42 g/mol - Propane ( C₃H₈): - Molar mass = (3 × 12) + (8 × 1) = 44 g/mol Understanding molar masses is fundamental when scaling reactions from moles to grams. For instance, in the problem, to convert the heat evolved in kJ/mol to kJ/kg, knowing the molar mass allows us to determine how much energy is involved when one kilogram of a substance is combusted. This is critical for comparing the efficiency of different fuels.