Question

The standard enthalpy of formation for liquid water is: $$ \mathrm{H}_2(g)+\mathrm{fiO}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \quad \Delta H^2{ }_f=-285.8 \mathrm{~kJ} / \mathrm{mol} $$ Which of the following could be the standard enthatpy of formation for water vapor? A. \(-480.7 \mathrm{~kJ} / \mathrm{mol}\) B. \(-285.8 \mathrm{~kJ}\) C. \(-241.8 \mathrm{~kJ} / \mathrm{mol}\) D. \(+224.6 \mathrm{~kJ} / \mathrm{mol}\)

Short answer

C. \( -241.8 \mathrm{~kJ} / \mathrm{mol} \)

Step-by-step solution

- Understand the given formation reaction

The problem gives the standard enthalpy of formation for liquid water: \[ \mathrm{H}_2 (g) + \frac{1}{2} \mathrm{O}_2 (g) \rightarrow \mathrm{H}_2\mathrm{O} (l) \quad \Delta H_f = -285.8 \mathrm{~kJ} / \mathrm{mol} \] This value represents the energy change when 1 mole of liquid water is formed from its elements in their standard states (hydrogen gas and oxygen gas).

- The heat of vaporization

To find the enthalpy of formation for water vapor, the enthalpy change when liquid water (H₂O (l)) converts to water vapor (H₂O (g)) must be considered. This enthalpy change is known as the heat of vaporization (ΔHvap). For water, ΔHvap is approximately +44 kJ/mol.

- Calculate the enthalpy of formation for water vapor

When converting liquid water to water vapor, the system absorbs heat. Therefore, the change in enthalpy of formation when water is vaporized can be calculated by adding the heat of vaporization to the enthalpy of formation of liquid water: \[ \Delta H_{f\,\mathrm{H_2O(g)}} = \Delta H_{f\,\mathrm{H_2O(l)}} + \Delta H_{\text{vap}} \] Substitute the given values:\[ \Delta H_{f\,\mathrm{H_2O(g)}} = -285.8 \mathrm{~kJ} / \mathrm{mol} + 44 \mathrm{~kJ} / \mathrm{mol} = -241.8 \mathrm{~kJ} / \mathrm{mol} \]

- Choose the correct answer

Among the given choices, the enthalpy of formation for water vapor that matches our calculated value is C. \( -241.8 \mathrm{~kJ} / \mathrm{mol} \)

Key concepts

standard enthalpy

The standard enthalpy of formation is a crucial concept in thermochemistry. It refers to the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. For instance, in the formation of liquid water, the standard enthalpy of formation is given as:
$$ \text{H}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O} (\text{l}) \text{, } \triangle H_f = -285.8 \text{kJ/mol} $$

This means that 285.8 kJ of energy is released when one mole of liquid water is formed from hydrogen gas and oxygen gas at standard conditions (298 K and 1 atm pressure).

Understanding standard enthalpy helps in predicting the energy changes during chemical reactions. It's a foundational block for calculating other thermodynamic properties.

heat of vaporization

The heat of vaporization (ΔHvap) is the amount of energy required to convert one mole of a liquid into vapor without a temperature change. For water, the heat of vaporization is approximately 44 kJ/mol.

When liquid water (H₂O(l)) changes to water vapor (H₂O(g)), it absorbs this energy, increasing the enthalpy. The equation for this phase change is:

$$ \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{O}(\text{g}) \text{ } \triangle H_{\text{vap}} = +44 \text{kJ/mol} $$

In our problem, we used the heat of vaporization to find the enthalpy of formation for water vapor, starting with the enthalpy of formation for liquid water. This is an essential step in understanding phase changes and how they affect the enthalpy of substances.

thermochemistry

Thermochemistry deals with the study of heat changes that accompany chemical reactions and phase changes. It helps in understanding how energy is absorbed or released in different processes.

One key concept is enthalpy, which is a measure of the total energy of a thermodynamic system. In thermochemistry, we often deal with changes in enthalpy (ΔH) to predict reaction energetics.

Key processes like bond formation, breaking, and phase transitions, all involve energy changes that can be quantified using enthalpy.

By understanding these energy changes, scientists and engineers can design processes and materials with desired thermal properties, ensuring safety, efficiency, and sustainability in various applications.

enthalpy change calculations

Calculating enthalpy changes is essential in predicting the outcome of chemical reactions. Here's how we approached it in the exercise:

1. Identify the given enthalpy value: The standard enthalpy of formation for liquid water was provided as \text{ΔH\text{sub}f} (\text{H₂O(l)}) = -285.8 kJ/mol.
2. Understand the process change: Converting liquid water to water vapor involves the heat of vaporization, ΔHvap = +44 kJ/mol.
3. Calculate the final enthalpy change: We added the heat of vaporization to the initial enthalpy of formation:

$$ \triangle H_{f \text{, H₂O(g)}} = \triangle H_{f \text{, H₂O(l)}} + \triangle H_{\text{vap}} $$
$$ \triangle H_{f \text{, H₂O(g)}} = -285.8 \text{kJ/mol} + 44 \text{kJ/mol} = -241.8 \text{kJ/mol} $$

This type of calculation is fundamental in thermochemistry, allowing us to predict how much energy is involved in different reactions and processes.