Question

The standard heat of combustion of propane is \(-2220.1 \mathrm{~kJ} / \mathrm{mol}\). The standard heat of vaporization of liquid water is \(44 \mathrm{~kJ} / \mathrm{mol}\). What is the \(\Delta H^{\text {o }}\) of the reaction: \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ?\) (a) \(-2220.1 \mathrm{~kJ}\) (b) \(-2044.1 \mathrm{~kJ}\) (c) \(-2396.1 \mathrm{~kJ}\) (d) \(-2176.1 \mathrm{~kJ}\)

Short answer

-2396.1 kJ

Step-by-step solution

Identify the heat of combustion of propane

The standard heat of combustion of propane (C3H8) is given as -2220.1 kJ/mol. This value refers to the heat released when one mole of propane combusts completely in the presence of oxygen to form carbon dioxide (CO2) and water (H2O).

Identify the products of the reaction

In the given reaction, the combustion of propane results in the formation of 3 moles of CO2 and 4 moles of H2O. However, the state of H2O in the products is gaseous (H2O(g)), not liquid.

Consider the heat of vaporization of water

Since the standard heat of vaporization of liquid water is specified, and the products have water in the gaseous state, we need to account for the energy required to vaporize water from the liquid to gaseous state. This is 44 kJ/mol for water, and there are 4 moles of water produced in the reaction.

Calculate the total heat of vaporization for the water produced

Multiply the heat of vaporization of water (44 kJ/mol) by the number of moles of water produced (4 moles): \(4 \text{ moles} \times 44 \frac{kJ}{mol} = 176 kJ\).

Adjust the heat of combustion of propane for the gaseous water

To find the \(\Delta H^\circ\) of the reaction, we need to subtract the energy used to vaporize water from the heat of combustion of propane. This is because the heat of combustion was given for liquid water, not gaseous:\(-2220.1 kJ - 176 kJ = -2396.1 kJ\).

Choose the correct answer

The \(\Delta H^\circ\) for the combustion reaction of propane with the formation of gaseous water is -2396.1 kJ. The correct answer is therefore option (c).

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), represents the total heat content change of a system when a process occurs at constant pressure. It's a measure of the energy exchanged, particularly in exothermic or endothermic reactions. For example, during the combustion of propane, energy is released in the form of heat to the surroundings, indicating an exothermic reaction. This release of energy is represented by a negative enthalpy change, demonstrating that the system has lost heat energy.

In thermodynamics exercises, students are encouraged to understand that the sign of the enthalpy change is crucial. A negative sign corresponds to an exothermic process where heat is released, while a positive sign indicates an endothermic process where heat is absorbed.

For the reaction given, since we are concerned with the standard heat of combustion, this will relate directly to the enthalpy change of the reaction. It is crucial for students to acknowledge that the heat of combustion given (-2220.1 kJ/mol) already incorporates a phase change of water from liquid to gas within the reaction context.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a substance from its liquid phase to its gaseous phase at constant temperature and pressure. It's an important concept in thermochemistry, particularly when dealing with phase changes within a reaction or process.

When calculating energy changes in reactions, such as the given combustion of propane, recognizing and accounting for the heat of vaporization can be essential. In the step-by-step solution, we see the need to incorporate the additional energy (44 kJ/mol) required to convert water from liquid to its gaseous form, which is crucial because the product of the reaction is gaseous, not liquid water.

This kind of detail is a common place where students might make mistakes. Educators often stress the importance of paying attention to the physical states of reactants and products in thermochemical equations and adjusting calculations accordingly.

Thermochemistry

Thermochemistry deals with the study of energy changes associated with chemical reactions and changes of state. It's a subdivision of thermodynamics and involves principles like conservation of energy and the first law of thermodynamics.

In the context of the combustion reaction provided, thermochemistry allows us to understand and quantify the energy released as heat when propane burns in oxygen to form carbon dioxide and water. Calculating the enthalpy changes forms the core of thermochemistry in such reactions.

When approaching thermochemical problems, it's helpful for students to remember that the total energy of the system and the surroundings remains constant. Thus, the energy released or absorbed in a reaction can be tracked and must balance according to the principles of conservation of energy.

Combustion Reactions

Combustion reactions are exothermic chemical reactions where a substance combines with oxygen, releasing a large amount of energy in the form of heat and light. Typical examples include the burning of fuel like propane in a Bunsen burner or gasoline in an engine.

In such reactions, the compound being burned, commonly a hydrocarbon, reacts with oxygen gas to form carbon dioxide and water as the primary products. Understanding combustion reactions is fundamental not only in chemistry but also in fields like environmental science and energy engineering due to their associated energy outputs and potential environmental impacts.

In educational settings, exercises involving combustion reactions, such as the one provided, are an excellent way for students to apply various concepts from thermochemistry and learn about energy conservation in chemical processes. Recognizing the type of reaction and the states of the products is essential for correctly calculating the associated enthalpy changes.