Problem 45 - General Chemistry

Question

Water at \(0^{\circ} \mathrm{C}\) was placed in a dish inside a vessel maintained at low pressure by a vacuum pump. After a quantity of water had evaporated, the remainder froze. If \(9.31 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) was obtained, how much liquid water must have evaporated? The heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\) and its heat of vaporization is \(44.9 \mathrm{~kJ} / \mathrm{mol}\) at \(0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

1.25 g of water evaporated to freeze 9.31 g of water.

Step 1: Understand the Problem

The problem involves water initially at 0°C in a low-pressure environment, which causes some of the water to evaporate and the rest to freeze. We have 9.31 g of ice remaining, and we need to find out how much water evaporated. The key here is to apply the principles of heat transfer and use the latent heats provided.

Step 2: Determine Moles of Ice Formed

First, we calculate the number of moles of ice formed from 9.31 g of water. Water has a molar mass of 18 g/mol. The number of moles is calculated as: \[ n = \frac{9.31 ext{ g}}{18 ext{ g/mol}} = 0.5172 ext{ moles} \]

Step 3: Calculate the Heat Transfer for Ice Formation

The heat lost when water freezes is given by the latent heat of fusion. Using \( n = 0.5172 \) moles and latent heat of fusion \( 6.01 \) kJ/mol, the heat loss is: \[ q_{ ext{fusion}} = n imes ext{heat of fusion} = 0.5172 imes 6.01 = 3.109 ext{ kJ} \]

Step 4: Set Up the Heat Balance Equation

The heat lost by freezing (\( q_{ ext{fusion}} \)) must equal the heat gained by the evaporation process since no external heat is added. Thus: \[ q_{ ext{evaporation}} = q_{ ext{fusion}} \]

Step 5: Calculate Moles of Water Evaporated

Given the heat of vaporization of water is 44.9 kJ/mol, the moles of water evaporated \( n_e \) can be found from: \[ q_{ ext{evaporation}} = n_e imes ext{heat of vaporization} \] \[ 3.109 = n_e imes 44.9 \] Solving for \( n_e \), we get: \[ n_e = \frac{3.109}{44.9} = 0.0692 ext{ moles} \]

Step 6: Convert Moles of Evaporated Water to Grams

Convert the moles of evaporated water to grams using the molar mass of water (18 g/mol): \[ ext{mass} = n_e imes ext{molar mass} = 0.0692 imes 18 = 1.2456 ext{ g} \]

Step 7: Conclusion

Therefore, approximately 1.25 grams of water must have evaporated to allow 9.31 grams to freeze.

Key Concepts

Heat of Fusion

The heat of fusion refers to the amount of energy required to change a substance from a solid to a liquid at its melting point. For water at 0°C, this heat is 6.01 kJ/mol.
When a substance freezes, it releases this amount of energy per mole, as it transitions from a liquid to a solid.
In this problem, 9.31 grams of water freeze into ice.

• First, we calculate the number of moles of water, using its molar mass, which is 18 g/mol. This value helps us understand the total energy released during the freezing process.
• The latent heat of fusion shows us how much energy per mole is released. For water, it's 6.01 kJ/mol, and it's crucial for calculating the total heat lost during this phase transformation.

Understanding the heat of fusion is essential in explaining why the remaining water froze. It provides insight into how the energy transfer works without adding external energy sources.

Heat of Vaporization

The heat of vaporization is the energy required to convert a liquid into a gas at its boiling point. At 0°C for water, the heat of vaporization is 44.9 kJ/mol.
In a low-pressure environment, like in this exercise, some water may evaporate even at 0°C. This evaporation requires a significant amount of energy, which is drawn from the remaining water, causing some to freeze.

• We calculated the moles of evaporated water by considering the energy required for evaporation. Given the heat of vaporization is 44.9 kJ/mol, it's clear that even a small amount of evaporation requires considerable energy.
• This energy comes from the latent heat released as some water freezes, illustrating the interplay between evaporation and freezing in this scenario.

The heat of vaporization not only quantifies the energy needed for vaporization but also provides context for the energy balancing in this freezing and evaporation process.

Heat Transfer

Heat transfer is a vital concept in this exercise as it defines how energy is transferred between the water undergoing evaporation and the part that freezes.
In a closed system like the one described, energy transfer dictates that the energy isn't lost; it merely changes form.

• The energy released by the freezing of water (computed using the heat of fusion) equals the energy absorbed for the evaporation of water (using the heat of vaporization).
• This balance of heat shows there is no need for an external heat source, as the system self-regulates by transferring energy from the freezing process to the evaporation process and vice versa.

This concept highlights that even at seemingly constant temperatures, significant internal changes can occur due to the transfer of heat, facilitating two different phase changes simultaneously.