Problem 95 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

(a) -964.7 kJ/mol for propyne, -971.9 kJ/mol for propylene, -2043.2 kJ/mol for propane. (b) -24103.9 kJ/kg for propyne, -23084.6 kJ/kg for propylene, -46322.0 kJ/kg for propane. (c) Propane is the most efficient fuel.

Step 1: Write Balanced Combustion Reactions

For propyne (C\(_3\)H\(_4\)): 2C\(_3\)H\(_4\) + 9O\(_2\) → 6CO\(_2\) + 4H\(_2\)O.For propylene (C\(_3\)H\(_6\)): 2C\(_3\)H\(_6\) + 9O\(_2\) → 6CO\(_2\) + 6H\(_2\)O.For propane (C\(_3\)H\(_8\)): C\(_3\)H\(_8\) + 5O\(_2\) → 3CO\(_2\) + 4H\(_2\)O.

Step 2: Calculate Heat Evolved (ΔH_comb)

Use the relation \( \Delta H_{comb} = \Sigma \Delta H_{f, \, products} - \Sigma \Delta H_{f, \, reactants} \). Since we are considering complete combustion, the products are CO\(_2\) and H\(_2\)O which have formation enthalpies of -393.5 kJ/mol and -241.8 kJ/mol, respectively.

Step 2a: Combustion of Propyne

Calculate \( \Delta H_{comb} \) for propyne: \(\Delta H_{comb}^{propyn} = [6(-393.5) + 4(-241.8)] - [2(+185.4)] = -1929.4 \text{kJ/mol}\). Divide by 2 for per mole: \(= -964.7 \text{kJ/mol}\).

Step 2b: Combustion of Propylene

Calculate \( \Delta H_{comb} \) for propylene: \(\Delta H_{comb}^{propyl} = [6(-393.5) + 6(-241.8)] - [2(+20.4)] = -1943.8 \text{kJ/mol}\). Divide by 2 for per mole: \(= -971.9 \text{kJ/mol}\).

Step 2c: Combustion of Propane

Calculate \( \Delta H_{comb} \) for propane: \(\Delta H_{comb}^{propan} = [3(-393.5) + 4(-241.8)] - [-103.8] = -2043.2 \text{kJ/mol}\).

Step 3: Calculate Heat Evolved for 1 kg of Each Substance

Use molar masses: C\(_3\)H\(_4\): 40.06 g/mol; C\(_3\)H\(_6\): 42.08 g/mol; C\(_3\)H\(_8\): 44.1 g/mol.Propyne: 1000 g / 40.06 g/mol × -964.7 kJ/mol = -24103.9 kJ.Propylene: 1000 g / 42.08 g/mol × -971.9 kJ/mol = -23084.6 kJ.Propane: 1000 g / 44.1 g/mol × -2043.2 kJ/mol = -46322.0 kJ.

Step 4: Compare Efficiencies

The heat evolved per unit mass is greatest for propane among the three, making it the most efficient fuel in terms of heat evolved per unit mass.

Key Concepts

Enthalpy of Formation

The enthalpy of formation is a crucial concept in chemistry, especially when examining combustion reactions. It refers to the heat change when one mole of a compound is formed from its elements in their standard states. These values are crucial as they help calculate the heat evolved or absorbed during chemical reactions.
For example, the standard enthalpies of formation for gaseous propyne, propylene, and propane are given as +185.4, +20.4, and -103.8 kJ/mol, respectively. These values indicate how much energy is absorbed or released when these compounds are formed. Positive values, like those for propyne and propylene, mean that energy is required to form the compound, while the negative value for propane indicates that energy is released.

• Components in their most stable form have an enthalpy of formation of zero.
• The combustion reaction uses these values to assess the energy efficiency of a fuel.

The lower the enthalpy of formation, the more stable the compound is energetically, making it a critical factor in evaluating the combustion potential of different fuels.

Heat Evolution

Heat evolution during a chemical reaction, particularly in combustion, involves energy being released as heat. Combustion reactions are typically exothermic, meaning they release heat to the surroundings. Determining how much heat is evolved in such a reaction uses the enthalpy of combustion (\( \Delta H_{comb} \)) of the fuels involved.
In our context, we calculated the heat evolved for combusting one mole of propyne, propylene, and propane. Each of these substances releases a different amount of heat:

• Propyne: -964.7 kJ/mol
• Propylene: -971.9 kJ/mol
• Propane: -2043.2 kJ/mol

These figures result from their respective balanced combustion reactions and the enthalpies of formation of the products (CO₂ and H₂O). High negative enthalpy values indicate a large amount of heat released, which is why tracking this metric is essential when evaluating fuels for their energy efficiency.

Chemical Thermodynamics

Chemical thermodynamics studies how heat and work are related to chemical reactions, a field that encompasses enthalpy changes during such reactions. Whenever a fuel undergoes combustion, the reaction falls under the realm of chemical thermodynamics.
In the context of combustion:

• Reactions are exothermic, liberating energy and aligning with the second law of thermodynamics which states that energy spontaneously transfers from hotter to cooler bodies.
• Understanding these energetics is vital for applications where energy efficiency and conservation are priorities.

By using the enthalpies of formation, we applied principles of chemical thermodynamics to determine how effectively different fuels convert chemical energy into heat. Such insights guide industries in choosing fuels that maximize energy production while minimizing waste, a paramount goal in energy-intensive sectors.

Fuel Efficiency

Fuel efficiency in the context of combustion reactions refers to the amount of energy produced per unit mass of fuel. It is a critical factor for industries aiming for eco-friendly and economically viable energy solutions.
In our exercise, we calculated the heat evolved from burning 1 kg of propyne, propylene, and propane. These values help determine which substance serves as the most efficient fuel:

• Propyne: -24103.9 kJ/kg
• Propylene: -23084.6 kJ/kg
• Propane: -46322.0 kJ/kg

Propane emerges as the most efficient fuel by releasing the most energy per kilogram. The ability to produce more energy from less fuel makes propane an attractive option for various applications from heating to power generation. Focusing on fuel efficiency helps both in reducing costs and in minimizing the environmental footprint of fuel consumption.