Question

Thestandard 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. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

Short answer

The heat evolved per mole on combustion of propyne, propylene, and propane, are -2083.9 kJ/mol, -2058.9 kJ/mol, and -2219.0 kJ/mol respectively. For 1 kg of each substance, the heat evolved is -52,097.5 kJ (propyne), -49,014.1 kJ (propylene), and -50,424.7 kJ (propane). Propyne is the most efficient fuel in terms of heat evolved per unit mass.

Step-by-step solution

Write combustion reactions

Write a balanced combustion reaction for each substance. Combustion reactions involve the reaction of the fuel with molecular oxygen (O2) to form carbon dioxide (CO2) and water (H2O). For propyne (C3H4): C3H4(g) + 4O2(g) -> 3CO2(g) + 2H2O(g) For propylene (C3H6): C3H6(g) + 9/2O2(g) -> 3CO2(g) + 3H2O(g) For propane (C3H8): C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)

Calculate the enthalpy change (∆H) for each reaction

Using the balanced combustion reactions and the given standard enthalpies of formation, calculate the enthalpy change for the combustion reactions according to the formula: ∆H_combustion = ∆H_products - ∆H_reactants Standard enthalpies of formation (∆Hf°) values: C3H4(g) = +185.4 kJ/mol C3H6(g) = +20.4 kJ/mol C3H8(g) = -103.8 kJ/mol CO2(g) = -393.5 kJ/mol H2O(g) = -241.8 kJ/mol For propyne: ∆H_combustion = [3(-393.5) + 2(-241.8)] - [+185.4 + 4(0)] = -2083.9 kJ/mol For propylene: ∆H_combustion = [3(-393.5) + 3(-241.8)] - [+20.4 + 0] = -2058.9 kJ/mol For propane: ∆H_combustion = [3(-393.5) + 4(-241.8)] - [-103.8 + 5(0)] = -2219.0 kJ/mol

Calculate the heat evolved for 1 kg of each substance

To calculate the heat evolved for 1 kg (1000 g) of each substance, first, find the molar mass of each substance. Then, determine the moles of each substance in 1 kg and multiply this value by the enthalpy change for combustion. Molar masses: C3H4 = 40 g/mol C3H6 = 42 g/mol C3H8 = 44 g/mol For propyne: Moles in 1 kg: 1000 / 40 = 25 mol Heat evolved: 25 mol × -2083.9 kJ/mol = -52,097.5 kJ For propylene: Moles in 1 kg: 1000 / 42 ≈ 23.81 mol Heat evolved: 23.81 mol × -2058.9 kJ/mol = -49,014.1 kJ For propane: Moles in 1 kg: 1000 / 44 ≈ 22.73 mol Heat evolved: 22.73 mol × -2219.0 kJ/mol = -50,424.7 kJ

Compare the heat evolved per unit mass

We have the heat evolved for 1 kg of each substance: Propyne: -52,097.5 kJ Propylene: -49,014.1 kJ Propane: -50,424.7 kJ The highest amount of heat evolved per unit mass is obtained from propyne. Therefore, propyne is the most efficient fuel in terms of heat evolved per unit mass.

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation is a key concept in thermodynamics and chemistry. It refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. Standard states typically refer to the pure substances at a pressure of 1 atmosphere and a specified temperature, commonly 25°C (298 K).

For example, the formation of water from hydrogen gas and oxygen gas involves a specific enthalpy change, which is the standard enthalpy of formation of water. These values are essential for calculating the enthalpy changes in chemical reactions, such as combustion. By using the standard enthalpy of formation values for the reactants and products, we can determine the overall enthalpy change for the reaction—a procedure extensively utilized in the exercise you are working on.

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance combines with oxygen to produce heat and light. In this context, organic compounds like propyne, propylene, and propane are combusted, resulting in products such as carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)).

Combustion reactions are exothermic, which means they release heat. This release of energy makes them important for generating energy and heat. The balanced chemical equations for combustion reactions provide the basis to calculate heat energy changes. Knowing exactly how these substances react with oxygen helps in determining how much heat they can produce, which is vital in efficiency analysis.

Heat Evolved per Unit Mass

Calculating the heat evolved per unit mass is crucial to understanding how efficient a fuel is, specifically when comparing different substances. This involves determining how much heat is released when a specific mass of a substance combusts. To accomplish this, it is necessary to know the enthalpy change per mole of the substance and then apply it to a determined mass, such as 1 kilogram.

This exercise compares propyne, propylene, and propane. The computations show the heat energy released from 1 kilogram of each substance. Through such comparison, you can identify which substance produces the most energy per mass unit. This metric is highly valuable in industrial applications where cost efficiency is linked to the energy output from fuel.

Molar Mass Calculation

Molar mass calculation is a fundamental skill in chemistry that involves adding the atomic masses of all atoms in a molecule. This calculation is necessary to convert between grams and moles, allowing the determination of how many moles are in a given mass. For instance, when you know the mass of propyne, propylene, or propane, you can compute the number of moles by dividing the mass by the molar mass.

This conversion is pivotal when comparing heat evolved per unit mass because it links the enthalpy change calculated per mole to a definite quantity of substance. Understanding molar mass calculations ensures accuracy in chemical computations and energy comparisons across different fuels or substances in various chemical contexts.