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

Short answer

The heat evolved per mole during the combustion of propyne, propylene, and propane are -2420.4 kJ/mol, -2220.9 kJ/mol, and -2044.1 kJ/mol, respectively. On combustion of 1 kg of each substance, the heat evolved are -60340.06 kJ for propyne, -52725.10 kJ for propylene, and -46337.65 kJ for propane. Propyne is the most efficient fuel in terms of heat evolved per unit mass, as it releases the most heat per kg.

Step-by-step solution

Write the balanced chemical equations for combustion

For each of the three gaseous hydrocarbons, write the balanced chemical equation for their combustion reaction, which involves converting the hydrocarbon into CO2(g) and H2O(g): 1. Combustion of propyne: \(C_3H_4(g) + \dfrac{5}{2}O_2(g) \rightarrow 3CO_2(g) + 2H_2O(g)\) 2. Combustion of propylene: \(C_3H_6(g) + \dfrac{9}{2}O_2(g) \rightarrow 3CO_2(g) + 3H_2O(g)\) 3. Combustion of propane: \(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)\)

Calculate the heat evolved per mole for each combustion

We know that the standard enthalpies of formation are additive. Using Hess's Law, we can find the heat evolved in each reaction by subtracting the standard enthalpies of formation of the reactants from the standard enthalpies of formation of the products. 1. Heat Evolved in the combustion of propyne: ΔH = 3(ΔfH°(CO2)) + 2(ΔfH°(H2O)) - ΔfH°(C3H4) = 3(-393.5) + 2(-241.8) - 185.4 ΔH = -2420.4 kJ/mol 2. Heat Evolved in the combustion of propylene: ΔH = 3(ΔfH°(CO2)) + 3(ΔfH°(H2O)) - ΔfH°(C3H6) = 3(-393.5) + 3(-241.8) - 20.4 ΔH = -2220.9 kJ/mol 3. Heat Evolved in the combustion of propane: ΔH = 3(ΔfH°(CO2)) + 4(ΔfH°(H2O)) - ΔfH°(C3H8) = 3(-393.5) + 4(-241.8) + 103.8 ΔH = -2044.1 kJ/mol

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

To do this, we first need to find the molar mass of each hydrocarbon and then multiply the heat evolved per mole by the number of moles in 1 kg. 1. Heat evolved on combustion of 1 kg of propyne: Molar mass of \(C_3H_4\) = (3 x 12.01) + (4 x 1.01) = 40.06 g/mol Number of moles in 1 kg of propyne = 1000 g / 40.06 g/mol = 24.94 mol Heat evolved = -2420.4 kJ/mol x 24.94 mol = -60340.06 kJ 2. Heat evolved on combustion of 1 kg of propylene: Molar mass of \(C_3H_6\) = (3 x 12.01) + (6 x 1.01) = 42.08 g/mol Number of moles in 1 kg of propylene = 1000 g / 42.08 g/mol = 23.76 mol Heat evolved = -2220.9 kJ/mol x 23.76 mol = -52725.10 kJ 3. Heat evolved on combustion of 1 kg of propane: Molar mass of \(C_3H_8\) = (3 x 12.01) + (8 x 1.01) = 44.10 g/mol Number of moles in 1 kg of propane = 1000 g / 44.10 g/mol = 22.68 mol Heat evolved = -2044.1 kJ/mol x 22.68 mol = -46337.65 kJ

Determine the most efficient fuel

To find the most efficient fuel in terms of heat evolved per unit mass, compare the heat evolved on combustion of 1 kg of each substance: 1. Propyne: -60340.06 kJ/kg 2. Propylene: -52725.10 kJ/kg 3. Propane: -46337.65 kJ/kg Since the heat evolved is more negative for propyne, it releases the most heat per unit mass, making it the most efficient fuel amongst the three given hydrocarbons.

Key concepts

Combustion Reaction

Combustion reactions are a type of chemical reaction where a substance, typically a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy in the form of heat. This is the fundamental process by which fuels like gasoline, natural gas, and propane release energy that can then be harnessed for various uses like heating, powering engines, or generating electricity.

These reactions are exothermic, meaning they release energy. In many cases, the substance that undergoes combustion initially requires some energy input, like a spark, to start the process, but once begun, the reaction releases more energy than was put in, resulting in a net release of energy. This is what makes fuels so useful; their ability to produce large amounts of heat upon burning can be translated into work or other forms of energy through mechanical systems or generators.

To ensure complete combustion and maximum energy output, an adequate supply of oxygen is important. This allows the fuel to burn fully, converting all the carbon to CO₂ and hydrogen to H₂O, maximizing heat output and reducing emissions of unburned hydrocarbons and other pollutants.

Hess's Law

Hess's Law is a statement in chemistry that describes the total enthalpy change in a chemical reaction and is independent of the path taken from reactants to products. This principle is immensely useful in calculating the enthalpy changes of reactions that are difficult to measure experimentally.

Essentially, Hess's Law allows us to use the enthalpies of formation, which are measured under standard conditions for various compounds, to determine the enthalpy change (ΔH) for a given reaction. This is done by summing the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of the reactants.

In the context of combustion reactions, Hess's Law enables us to predict the amount of energy released when a fuel burns. We can write the overall reaction for the combustion process and then use known values for the enthalpies of formation of the reactants and products to determine the total heat evolved, even if we can't measure the reaction directly. This method is systematic and follows the laws of thermodynamics, providing us with a clear framework to calculate energy changes in chemical reactions.

Heat Evolved per Mole

The 'heat evolved per mole' is a term that describes the amount of heat energy released when one mole of a substance undergoes a chemical reaction. For combustion reactions, this value is particularly important because it directly relates to the amount of energy a fuel can provide.

In calculating the heat evolved per mole, it's important to consider the stoichiometry of the reaction, the enthalpies of formation of products and reactants, and the conditions under which combustion takes place.

Understanding the heat evolved per mole helps when comparing different fuels. It allows us to assess which fuel releases more energy per mole and is thus more efficient. The standard units for measuring this heat energy are kilojoules per mole (kJ/mol). In applications such as powering an engine or heating a home, a higher heat output per mole generally indicates a more effective fuel. Calculation of this value is a combination of careful measurements, use of standard enthalpy values, and a clear understanding of chemical reaction equations.