Problem 37 - Chemistry The Molecular Science

Question

The chlorofluorocarbon \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) was once used as a refrigerant. Calculate what mass of this substance must evaporate to freeze \(2 \mathrm{~mol}\) water initially at \(20^{\circ} \mathrm{C}\). The vaporization enthalpy for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(289 \mathrm{~J} / \mathrm{g}\). The fusion enthalpy for solid ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\) and specific heat capacity for liquid water is \(4.184 \mathrm{~J} \mathrm{~g}^{-1}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}^{-1}\).

Step-by-Step Solution

Answer

To freeze the water, 52.1 g of \(\mathrm{CCl}_2\mathrm{F}_2\) must evaporate.

Step 1: Calculate Heat Required to Cool Water

First, we need to cool the water from \(20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\). The formula to calculate the heat \(q\) is: \( q = mc\Delta T \) where \(m\) is the mass of water, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. The molar mass of water is \(18 \mathrm{~g/mol}\), so the mass of \(2\) moles is \(36 \mathrm{~g}\). Thus, the heat required is: \[ q = (36 \mathrm{~g})\times (4.184 \mathrm{~J/g^{\circ}C) \times (20^{\circ}C) = 3012.48 \mathrm{~J} \]

Step 2: Calculate Heat Required to Freeze Water

Next, calculate the heat required to freeze \(2\) moles of water. The formula is: \( q = n\cdot \Delta H_f \), where \( n \) is the number of moles and \( \Delta H_f \) is the fusion enthalpy. \[ q = (2 \mathrm{~mol}) \times (6.02 \mathrm{~kJ/mol}) = 12.04 \mathrm{~kJ} = 12040 \mathrm{~J} \]

Step 3: Calculate Total Heat to be Absorbed

Add the heat required to cool the water and the heat required to freeze the water to get the total heat that must be absorbed. \[ \text{Total Heat} = 3012.48 \mathrm{~J} + 12040 \mathrm{~J} = 15052.48 \mathrm{~J} \]

Step 4: Calculate Mass of \(\mathrm{CCl}_2\mathrm{F}_2\) Required

The mass of \(\mathrm{CCl}_2\mathrm{F}_2\) required can be calculated using its vaporization enthalpy \( q = m\cdot \Delta H_v \). Rearrange to find \( m \): \[ m = \frac{q}{\Delta H_v} = \frac{15052.48 \mathrm{~J}}{289 \mathrm{~J/g}} = 52.1 \mathrm{~g} \]

Key Concepts

Enthalpy

Enthalpy is a central concept in thermodynamics and chemistry, often symbolized as \( H \). It represents the total heat content of a system at constant pressure. While enthalpy itself is not measured directly, the change in enthalpy (\( \Delta H \)) during a reaction or a process is crucial. It tells us how much heat is absorbed or released. For example, in the original exercise, the vaporization enthalpy of \( \mathrm{CCl}_2 \mathrm{F}_2 \) indicates how much energy is needed to convert a unit mass of this substance from liquid to gas, without changing its temperature.

In our context, due to the process of vaporization, \( \mathrm{CCl}_2 \mathrm{F}_2 \) absorbs heat, leading to necessary cooling of water. This heat absorption is critical for understanding how thermodynamic systems can control temperature changes—whether cooling water or freezing it to ice. If you need to cool or heat a substance in a physical process, you’ll often have to calculate the enthalpy change, which ties directly into phase changes and specific heat capacity.

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy required to change the temperature of a given quantity of a substance by 1 degree Celsius. The specific heat capacity \( c \) is expressed in \( J/g^{obreakspace obreakspace obreakspace obreakspace}^\circ\mathrm{C} \). In our problem, the specific heat capacity of water is \( 4.184 \, \text{J/g}{\underline{\phantom{xx}}}^{\circ}C \), which is relatively high. This means water requires a lot of energy to change its temperature, explaining its great use in cooling processes.

When dealing with specific heat capacity, you should note:

• It is substance-specific.
• The higher the specific heat, the more energy required for a temperature change.

This concept is prominent when calculating the energy required to cool substances. For water to drop from \( 20^{\circ}\mathrm{C} \) to \( 0^{\circ}\mathrm{C} \), knowing the specific heat capacity allows us to compute the exact amount of energy needed, which in this exercise facilitates the understanding of how much \( \mathrm{CCl}_2 \mathrm{F}_2 \) vaporization is required.

Phase Change

Phase changes involve the transformation of a substance from one state of matter to another—such as from liquid to solid. These processes absorb or release heat without changing temperature, known as latent heat. In the exercise, water changing from liquid to solid (freezing) involves the concept of heat of fusion, or fusion enthalpy \( \Delta H_f \). Here, \( 6.02 \, \text{kJ/mol} \) of energy is required to freeze one mole of water, reflecting how energy is transferred during phase changes.

The concept of phase change covers:

• Fusion/Melting (solid ↔ liquid)
• Vaporization/Condensation (liquid ↔ gas)
• Sublimation/Deposition (solid ↔ gas)

Each of these phase swaps involves specific heat or enthalpy values that must be calculated to understand energy dynamics. In this calculation exercise, the phase change does more than just explain the freezing process; it provides insights into why vaporization of \(\mathrm{CCl}_2 \mathrm{F}_2\) is effective for cooling. This refrigerant's high vaporization enthalpy allows it to absorb a significant amount of heat, effectively lowering the water's temperature until it can freeze.