Question

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{kJ} / \mathrm{mol}\) . Use these data and Hess's law to determine the enthalpy of vaporization for water.

Short answer

Using Hess's law and the given enthalpy of combustion for \(CH_{4}(g)\) when \(H_{2}O(l)\) and \(H_{2}O(g)\) are formed, the enthalpy of vaporization for water can be determined through the difference in enthalpy changes for these reactions: \(\Delta H_{vap} = \Delta H_{2} - \Delta H_{1}\). Calculating the difference yields \(\Delta H_{vap} = 88\,kJ/mol\) for 2 moles of water. To find the enthalpy of vaporization for 1 mole of water, divide by 2, resulting in an enthalpy of vaporization for water of \(44\,kJ/mol\).

Step-by-step solution

Combustion reactions of methane

Write down the combustion reactions for \(CH_{4}\) with the formation of \(H_{2}O(l)\) and \(H_{2}O(g)\): Reaction 1 (with \(H_{2}O(l)\)): \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\) with enthalpy change, \(\Delta H_{1} = -891\,kJ/mol\) Reaction 2 (with \(H_{2}O(g)\)): \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\) with enthalpy change, \(\Delta H_{2} = -803\,kJ/mol\)

Using Hess's Law

According to Hess's Law, the enthalpy change of a process is independent of the path taken. The overall enthalpy change of the target process, \(H_{2}O(l) \rightarrow H_{2}O(g)\), can be determined by the difference between Reaction 2 and Reaction 1. Specifically, we want to find the enthalpy change for the process, \(2H_{2}O(l) \rightarrow 2H_{2}O(g)\), abbreviated as \(\Delta H_{vap}\). The relationship between the given reactions and the target reaction is: \(\Delta H_{vap} = \Delta H_{2} - \Delta H_{1}\)

Calculating the enthalpy of vaporization of water

Plug in the known enthalpy change values for reactions 1 and 2. Then, solve for the enthalpy change of the vaporization process: \(\Delta H_{vap} = (-803\,kJ/mol) - (-891\,kJ/mol) = 88\,kJ/mol\) This is the enthalpy change for the process of vaporizing 2 moles of water. To find the enthalpy of vaporization for 1 mole of water, divide this value by 2: \(\Delta H_{vap, 1\,mol} = \frac{88\,kJ/mol}{2} = 44\,kJ/mol\) Therefore, the enthalpy of vaporization for water is \(44\,kJ/mol\).

Key concepts

Enthalpy of Combustion

The enthalpy of combustion is a measure of the energy change that occurs when one mole of a substance combusts completely in the presence of oxygen. For \\(CH_4\), or methane, the enthalpy of combustion can vary depending on the final state of the products.
In the exercise, we explored two scenarios: one where the water product is in liquid form and another where it is in gaseous form.
This variance in states affects the overall enthalpy measurement:

• For the reaction forming \\(H_2O(l)\), the enthalpy is \\(-891 \ \text{kJ/mol}\).
• For the reaction forming \\(H_2O(g)\), it is \\(-803 \ \text{kJ/mol}\).

This difference is crucial as it indicates the heat energy required or released during combustion, reflecting the physical state of the water. Knowing the enthalpy of combustion allows scientists to understand the energy efficiency and heat output of fuels, which is critical in industries like energy production and environmental engineering.

Enthalpy of Vaporization

The enthalpy of vaporization refers to the heat required to convert a liquid into a gas at constant temperature and pressure. In context with this exercise, using Hess's Law enables us to find the enthalpy of vaporization for water by comparing relevant combustion reactions.
Hess's Law states that the total enthalpy change for a chemical process is the same, regardless of how many steps the process takes. Thus, if you know the enthalpy changes for related reactions, you can calculate the enthalpy change for the desired process by adding or subtracting these known values.
In the exercise, we used the formula:\[\Delta H_{vap} = \Delta H_{2} - \Delta H_{1}\]This calculation yields the enthalpy of vaporization for two moles of water. To find the value for a single mole, we simply divide by two:\[\Delta H_{vap, 1\,mol} = \frac{88\,kJ/mol}{2} = 44\,kJ/mol\]Understanding this value is important for fields such as climatology and HVAC systems where water undergoes phase changes.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. These reactions typically involve fuels, like \\(CH_4\) or methane in this scenario. During combustion:

• Fuel reacts with oxygen.
• Energy is released as heat.
• Products are formed, typically including water and carbon dioxide.

Combustion reactions can produce different products based on conditions, such as temperature, pressure, and the physical state of reactants or products. These factors affect the enthalpy change.
In our discussed example, methane combustion yields carbon dioxide and water in varying states. Such reactions are foundational for engines and power plants, as they provide the necessary energy for these systems to operate. Understanding the nuances of these reactions allows for the optimization and control of energy production processes.