Question

For the evaporation of water, \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g),\) at \(373 \mathrm{~K}, \Delta H=+40.7 \mathrm{~kJ} / \mathrm{mol}\) (a) How many kilojoules are needed to vaporize \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}(l) ?\) (b) How many kilojoules are released when \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed?

Short answer

(a) 22.6 kJ needed for vaporization; (b) 22.6 kJ released during condensation.

Step-by-step solution

Calculate the Moles of Water

First, we need to calculate the number of moles of water. The molar mass of water (\(\mathrm{H}_2\mathrm{O}\)) is approximately 18.02 g/mol. Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\) to find the moles of water. For a water mass of 10.0 g:\[\text{moles of water} = \frac{10.0 \text{ g}}{18.02 \text{ g/mol}} \approx 0.555 \text{ mol}\]

Calculate Heat for Vaporization

Using the enthalpy change (\(\Delta H\)) of vaporization, calculate the energy required to vaporize the water:\[\text{Energy} = \text{moles} \times \Delta H = 0.555 \text{ mol} \times 40.7 \text{ kJ/mol} \approx 22.6 \text{ kJ}\]

Calculate Heat for Condensation

The process of condensation is the reverse of vaporization, so the same amount of energy will be released when water condenses, which means:\[\text{Energy released for condensation} = 22.6 \text{ kJ}\]

Key concepts

Moles Calculation

Understanding how many moles of a substance are present in a given mass is crucial for various chemical calculations. To begin, you need to know the molar mass of the substance in question, which is the mass of one mole of that substance in grams. Water (\(\mathrm{H}_2\mathrm{O}\)) has a molar mass of approximately 18.02 g/mol.
To find the number of moles, use the formula:

• \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)

Using the example with 10.0 g of water:

• Divide 10.0 g by 18.02 g/mol, resulting in approximately 0.555 moles of water.

This conversion is fundamental because chemical reactions are balanced in terms of moles, not grams. Knowing the moles lets you link mass changes to energy changes in reactions.

Heat Calculations

Once you've determined the moles of water, the next step is to calculate the heat energy needed for a phase change, such as vaporization. This involves using the enthalpy of vaporization (\(\Delta H\)), which tells you the amount of heat needed to convert a mole of liquid into gas at constant temperature and pressure.
The formula for calculating the energy required is:

• \(\text{Energy} = \text{moles} \times \Delta H\)

For our water example, you multiply the moles of water (0.555 mol) by the enthalpy of vaporization (40.7 kJ/mol), resulting in approximately 22.6 kJ. This quantitative step shows how much energy is consumed or released during a physical phase change and is critical for understanding energy flow in chemical processes.

Phase Change

Phase changes are transformations from one state of matter to another, such as from liquid to gas or vice versa. In this exercise, we're focusing on the evaporation (liquid to gas) and condensation (gas to liquid) of water.
During evaporation, water absorbs energy, which is used to overcome intermolecular forces, allowing molecules to escape as vapor. Conversely, during condensation, water molecules release the same amount of energy back to the surroundings. This is why the energy value remains the same: 22.6 kJ is needed to vaporize the water, and 22.6 kJ is released upon condensation.
Understanding phase changes helps you grasp why certain processes need energy input or release energy. This keeps in mind the conservation of energy principle and highlights the importance of enthalpy changes in chemical and physical transformations.