Question

The enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) 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

The enthalpy of vaporization for water can be determined using Hess's law and the given enthalpy of combustion values for methane when water is formed as a liquid and a gas. By subtracting the enthalpy of combustion for the liquid reaction from the enthalpy of combustion for the gas reaction, we find that the enthalpy of vaporization for water is \(ΔH₃ = 88 \: \text{kJ/mol}\).

Step-by-step solution

Write down the given information

The given information is the enthalpy of combustion of methane when water is formed as a liquid and as a gas: 1. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH₁ = -891 kJ/mol 2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔH₂ = -803 kJ/mol

Write the reaction for the enthalpy of vaporization of water

The enthalpy of vaporization describes the conversion of one mole of water from the liquid phase to the gas phase: 3. H₂O(l) → H₂O(g), ΔH₃ = ΔH(vap)

Apply Hess's law

Hess's law states that the total enthalpy change in a reaction is equal to the sum of individual enthalpy changes in the reactions forming the desired reaction. We need to use reaction 1 and 2 to obtain reaction 3. To obtain reaction 3, we can visualize subtracting reaction 1 from reaction 2 (eliminating the CH₄ and O₂ molecules): ΔH₃ = ΔH₂ - ΔH₁ = (-803 kJ/mol) - (-891 kJ/mol)

Calculate ΔH₃

Now we can calculate the enthalpy of vaporization for water (ΔH₃): ΔH₃ = 891 kJ/mol - 803 kJ/mol = 88 kJ/mol Hence, the enthalpy of vaporization for water is 88 kJ/mol.

Key concepts

Enthalpy of Combustion

The enthalpy of combustion measures the energy released when a substance combusts completely in the presence of oxygen. It's vital because it tells us how much energy is available from fuels like methane. The enthalpy of combustion is typically expressed in kilojoules per mole (kJ/mol). In our example, the enthalpy of combustion for methane when water is produced as a liquid is \(-891 \text{ kJ/mol}\). This means that when methane combusts, it releases 891 kJ of energy for every mole of methane consumed. If water is instead formed as a gas, the enthalpy of combustion is \(-803 \text{ kJ/mol}\). These differences arise because forming water in different states (liquid vs. gas) involves different amounts of energy.

Enthalpy of Vaporization

The enthalpy of vaporization represents the energy required to convert a liquid into a gas at constant pressure. It's significant because it helps us understand the phase transition and energy requirements for boiling or vaporization. The reaction for the enthalpy of vaporization for water can be written as: \( \text{H}_2\text{O (l)} \to \text{H}_2\text{O (g)} \), with an enthalpy value \( \Delta H_{\text{vap}} \). In the context of Hess's Law, we use differences in combustion enthalpies to indirectly compute this value. The enthalpy of vaporization for water is calculated from the differences in combustion enthalpies: \( \Delta H_{\text{vap}} = -803 \text{ kJ/mol} - (-891 \text{ kJ/mol}) = 88 \text{ kJ/mol} \). This tells us that 88 kJ of energy is required to turn one mole of liquid water into vapor.

Thermochemical Equations

Thermochemical equations are chemical equations that show both the reactants and products of a reaction along with the enthalpy change. They are essential for understanding energy changes in chemical reactions. These equations help us visualize the energy shifts as substances react under specified conditions. In our exercise, two thermochemical equations describe the combustion of methane:

• \( \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}, \Delta H = -891 \text{ kJ/mol} \)
• \( \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}, \Delta H = -803 \text{ kJ/mol} \)

These equations give us a descriptive picture of how energy is released based on the phase of water formed. When combined using Hess's Law, they help calculate the enthalpy of vaporization, providing insights into energy transitions between liquid and gaseous states of water.