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 enthalpy change during the combustion of propyne, propylene, and propane are given by: \[\Delta H_{comb, \,C_3H_4} = -[3\Delta H_f(CO_2) + 2\Delta H_f(H_2O)] + \Delta H_f(C_3H_4)\] \[\Delta H_{comb, \,C_3H_6} = -[3\Delta H_f(CO_2) + 3\Delta H_f(H_2O)] + \Delta H_f(C_3H_6)\] \[\Delta H_{comb, \,C_3H_8} = -[3\Delta H_f(CO_2) + 4\Delta H_f(H_2O)] + \Delta H_f(C_3H_8)\] Using the provided standard enthalpies of formation and the balanced combustion reactions, we can calculate the heat evolved during combustion for each substance. Then, using the molar mass of each substance, we can find the heat evolved per unit mass. Finally, by comparing the heat evolved per unit mass for the three substances, we can determine which one is the most efficient fuel in terms of heat evolved per unit mass.

Step-by-step solution

Write the combustion reactions for each substance

Before working with the enthalpy values, we need to write down the balanced combustion reactions for propyne, propylene, and propane. For propyne \((\mathrm{C}_{3}\mathrm{H}_{4}):\) \[C_3H_4(g) + \frac{5}{2}O_2(g) \rightarrow 3CO_2(g) + 2H_2O(g)\] For propylene \((\mathrm{C}_{3}\mathrm{H}_{6}):\) \[C_3H_6(g) + \frac{9}{2}O_2(g) \rightarrow 3CO_2(g) + 3H_2O(g)\] For propane \((\mathrm{C}_{3}\mathrm{H}_{8}):\) \[C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)\]

Calculate the enthalpy change for each combustion reaction

Using the given standard enthalpies of formation for each substance and the balanced reactions, we can now calculate the enthalpy change during combustion. For propyne: \[\Delta H_{comb, C_3H_4} = -[3\Delta H_f(CO_2) + 2\Delta H_f(H_2O)] + \Delta H_f(C_3H_4)\] For propylene: \[\Delta H_{comb, C_3H_6} = -[3\Delta H_f(CO_2) + 3\Delta H_f(H_2O)] + \Delta H_f(C_3H_6)\] For propane: \[\Delta H_{comb, C_3H_8} = -[3\Delta H_f(CO_2) + 4\Delta H_f(H_2O)] + \Delta H_f(C_3H_8)\] Here, we need to use the standard enthalpy of formation values: \[\Delta H_f(C_3H_4) = +185.4 \, \text{kJ/mol}\] \[\Delta H_f(C_3H_6) = +20.4 \, \text{kJ/mol}\] \[\Delta H_f(C_3H_8) = -103.8 \, \text{kJ/mol}\] \[\Delta H_f(CO_2) = -393.5 \, \text{kJ/mol}\] \[\Delta H_f(H_2O) = -241.8 \, \text{kJ/mol}\] Calculate the enthalpy change for each combustion reaction.

Calculate the heat evolved per unit mass for each substance

Now, we'll calculate the heat evolved per unit mass for each substance. We will need the molar mass of each substance to do this: Molar mass of propyne \((\mathrm{C}_{3}\mathrm{H}_{4}) = 3(12.01) + 4(1.01) = 40.07 \, \mathrm{g/mol}\) Molar mass of propylene \((\mathrm{C}_{3}\mathrm{H}_{6}) = 3(12.01) + 6(1.01) = 42.08 \, \mathrm{g/mol}\) Molar mass of propane \((\mathrm{C}_{3}\mathrm{H}_{8}) = 3(12.01) + 8(1.01) = 44.10 \, \mathrm{g/mol}\) Now, for each substance, divide the heat evolved during combustion by the molar mass to get the heat evolved per unit mass.

Determine the most efficient fuel

Compare the heat evolved per unit mass for each substance. The most efficient fuel will be the one with the highest heat evolved per unit mass.

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance, often a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and heat. These reactions are exothermic, meaning they release energy in the form of heat.

For example, in the combustion of propyne, a complex reaction occurs as represented by the equation: \[C_3H_4(g) + \frac{5}{2}O_2(g) \rightarrow 3CO_2(g) + 2H_2O(g)\] The coefficients in front of each chemical represent the amount of each substance needed to balance the reaction. Ensuring reactions are balanced is crucial to accurately calculate the energy change involved. Remember, the amount of energy released during combustion is directly related to the kind and amount of fuel burned. When it comes to fuels, hydrocarbons like propyne, propylene, and propane, which vary in their carbon and hydrogen content, are common subjects of study for their energy release patterns.

Enthalpy Change

Enthalpy change, represented by \(\Delta H\), is a measure of the total heat content of a system. It represents the heat absorbed or released during a reaction at constant pressure. For combustion reactions, the enthalpy change is typically negative, indicating that heat is released to the surroundings, which is why they are considered exothermic reactions.

The enthalpy change for a combustion reaction can be calculated using the enthalpies of formation for the reactants and products. The standard enthalpy of formation of a substance is the heat change when one mole of the substance is formed from its elements at standard conditions. By adding the enthalpies of formation of the reactants and subtracting those of the products, the enthalpy change of the reaction is derived, as demonstrated in the step-by-step solution provided in the example problem.

Heat Evolved per Mole

Heat evolved per mole quantifies the amount of heat released when one mole of a substance undergoes combustion. It's a particularly useful metric for comparing different fuels. Calculating this value involves using enthalpy changes for the reactions, which in turn are based on the standard enthalpies of formation. By comparing the enthalpy change before and after the reaction for each substance, we determine the total heat released per mole during the combustion process.

For instance, using the balanced equations for the combustion of propyne, propylene, and propane, the heat evolved per mole can be determined by plugging in known values of the enthalpies of formation and applying stoichiometry to find the net energy released in each reaction.

Heat Evolved per Unit Mass

While knowing the heat released per mole is valuable, practical applications like fuel efficiency require understanding the heat evolved per unit mass. The heat evolved per unit mass allows us to compare different fuels in terms of energy release by weight, which is crucial for uses where weight is a constraint, such as in transportation.

To calculate this, it's necessary to first determine the molar mass of each substance. Then, divide the total heat evolved in the reaction (from the enthalpy change calculation) by the molar mass to find how much heat is released per gram. This enables us to evaluate the efficiency of different fuels, not in terms of moles, but in a more tangible sense: how much energy can be extracted from a kilogram of the fuel being burned. By performing this comparison, as shown in the exercise example, we can identify the most efficient fuel in terms of heat output for a given mass – a critical factor for energy resource selection.