Problem 138 - General Chemistry: Principles and Modern Applications

Question

The following data are given for \(\mathrm{CCl}_{4}\). Normal melting point, \(-23^{\circ} \mathrm{C} ;\) normal boiling point, \(77^{\circ} \mathrm{C} ;\) density of liquid \(1.59 \mathrm{g} / \mathrm{mL} ; \Delta H_{\text {fus }}=3.28 \mathrm{kJ} \mathrm{mol}^{-1} ;\) vapor pressure at \(25^{\circ} \mathrm{C}, 110\) Torr. (a) What phases-solid, liquid, and/or gas-are present if \(3.50 \mathrm{g} \mathrm{CCl}_{4}\) is placed in a closed \(8.21 \mathrm{L}\) container at \(25^{\circ} \mathrm{C} ?\) (b) How much heat is required to vaporize 2.00 L of \(\mathrm{CCl}_{4}(\mathrm{l})\) at its normal boiling point?

Step-by-Step Solution

Answer

a) The CCl4 would be in the liquid phase. b) The heat required to vaporize the CCl4 would be 67.50 kJ.

Step 1: Determine phase(s) of CCl4

Given that the melting point of \(\mathrm{CCl}_{4}\) is \(-23^{\circ} \mathrm{C}\) and the boiling point is \(77^{\circ} \mathrm{C}\), if \(\mathrm{CCl}_{4}\) is placed in a closed \(8.21 \mathrm{L}\) container at \(25^{\circ} \mathrm{C}\), it would be in liquid state because \(25^{\circ} \mathrm{C}\) is between the melting and boiling points.

Step 2: Calculate molar volume of CCl4

First, use the given density to convert the volume of CCl4 to be vaporized into mass. The density of \(\mathrm{CCl}_{4}\) is given as \(1.59 \mathrm{g}/\mathrm{mL}\). Therefore, \(2.00 \mathrm{L} = 2000 \mathrm{mL} = 2000 \mathrm{mL} \times 1.59 \mathrm{g}/\mathrm{mL} = 3180 \mathrm{g}\). Then, calculate the molar volume using the molar mass of CCl4 (154.5 g/mol): molar volume = \(3180 \mathrm{g} / 154.5 \mathrm{g/mol} = 20.58 \mathrm{mol}\)

Step 3: Find the heat of vaporization

The heat of vaporization can be found by multiplying the number of moles by the heat of fusion (change in enthalpy during phase transition). The given heat of fusion of CCl4 is \(3.28 kJ/mol\). So, the heat of vaporization is \(20.58 \mathrm{mol} \times 3.28 \mathrm{kJ/mol} = 67.50 \mathrm{kJ}\)

Step 4: Transfer heat to vaporize

The heat required to vaporize 2.00 L of liquid CCl4 at its normal boiling point is simply equal to the heat of vaporization, which has been calculated to be 67.50 kJ.

Key Concepts

Vaporization

Vaporization is a phase transition where a substance changes from its liquid state to its gaseous state. It is an essential concept in thermodynamics and occurs at the boiling point of the substance.
This process requires energy to overcome intermolecular forces between the liquid molecules. The required energy is called the heat of vaporization. This energy input does not change the temperature of the substance but transforms its state from liquid to gas.
There are two main types of vaporization:

• Evaporation: Occurs at the surface of a liquid at any temperature below its boiling point.
• Boiling: Takes place throughout the entire body of the liquid at a specific temperature called the boiling point.

For example, when \(\mathrm{CCl}_4\), with a normal boiling point of \(77^{\circ} \mathrm{C}\), is heated to this temperature, it converts from liquid to gas, requiring significant energy input to overcome its intermolecular forces.

Molar Volume

Molar volume is an essential concept in understanding the amount of space that one mole of a substance occupies. It is typically expressed in liters per mole (L/mol). For most liquids and solids, molar volume can be calculated by dividing the mass of the substance by its density and then by the molar mass of the substance.
This value gives insight into the structure and the packing of molecules in a particular phase. For example, knowing the density of \(\mathrm{CCl}_4\) (which is \(1.59 \text{g/mL}\)), you can calculate its molar volume by converting the volume of liquid to grams and then converting the grams to moles.
To find the molar volume of \(\mathrm{CCl}_4\), we used the formula:

• Molar Volume = \(\frac{\text{Mass}}{\text{Molar Mass}}\)

For \(\mathrm{CCl}_4\), with a calculated mass of \(3180 \mathrm{g}\), this results in a molar volume:

• \(3180 \mathrm{g} \div 154.5 \mathrm{g/mol} = 20.58 \mathrm{mol}\)

Heat of Fusion

The heat of fusion refers to the energy required to change a substance from solid to liquid at its melting point, without changing its temperature. It is a crucial measure of a substance's internal bonding strength.
For \(\mathrm{CCl}_4\), the heat of fusion is given as \(3.28\, \mathrm{kJ/mol}\). This value reflects the amount of energy required for each mole of carbon tetrachloride to transition from a solid to a liquid.
Though we focused on vaporization above for the boiling point calculation, the heat of fusion allows us to understand the enthalpy changes during melting:

• Phase transition occurs at the melting point, which for \(\mathrm{CCl}_4\), is \(-23^{\circ} \mathrm{C}\).
• It considers only the solid-to-liquid transformation, not the liquid-to-gas seen in vaporization.

Even though the exercise emphasized the energy for vaporization, understanding the heat of fusion gives a complete cycle of transitioning phases for any given substance.