Question

The molar heat of fusion of aluminum metal is \(10.79 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(293.4 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of fusion of aluminum so much smaller than the heat of vaporization? b. What quantity of heat would be required to vaporize \(1.00 \mathrm{~g}\) of aluminum at its normal boiling point? c. What quantity of heat would be evolved if \(5.00 \mathrm{~g}\) of liquid aluminum freezes at its normal freezing point? d. What quantity of heat would be required to melt 0.105 mole of aluminum at its normal melting point?

Short answer

a. The heat of fusion is smaller than the heat of vaporization because it takes less energy to break the bonds between solid molecules and convert them into a liquid than it does to completely separate the molecules in the liquid form and convert them into a gas. b. 10.86 kJ of heat is required to vaporize 1.00g of aluminum at its normal boiling point. c. 1.997 kJ of heat is evolved when 5.00g of liquid aluminum freezes at its normal freezing point. d. 1.133 kJ of heat is required to melt 0.105 mole of aluminum at its normal melting point.

Step-by-step solution

a. Explanation of the difference between heat of fusion and heat of vaporization

The heat of fusion is the amount of energy required to change a substance from a solid to a liquid state, whereas the heat of vaporization is the energy required to change it from a liquid to a gaseous state. The heat of fusion is smaller than the heat of vaporization because it takes less energy to break the bonds between solid molecules and convert them into a liquid than it does to completely separate the molecules in the liquid form and convert them into a gas.

b. Calculation of heat required to vaporize 1.00g of aluminum at its normal boiling point

First, we need to calculate the number of moles of aluminum in 1.00g. The molar mass of aluminum is 26.98g/mol. \(n = \frac{m}{M} = \frac{1.00g}{26.98g/mol} = 0.03704mol\) Now, we can calculate the heat required using the formula \(q = n \times \Delta H_v\): \(q = 0.03704mol \times 293.4 kJ/mol = 10.86 kJ\) Therefore, 10.86 kJ of heat is required to vaporize 1.00g of aluminum at its normal boiling point.

c. Calculation of heat evolved when 5.00g of liquid aluminum freezes at its normal freezing point

First, calculate the number of moles in 5.00g of aluminum: \(n = \frac{m}{M} = \frac{5.00g}{26.98g/mol} = 0.1852mol\) Now, calculate the heat evolved using the formula \(q = n \times \Delta H_f\): \(q = 0.1852mol \times 10.79 kJ/mol = 1.997 kJ\) Since the process is freezing, the value of q is negative; therefore, 1.997 kJ of heat is evolved when 5.00g of liquid aluminum freezes at its normal freezing point.

d. Calculation of heat required to melt 0.105 mole of aluminum at its normal melting point

We are given the number of moles, so we can directly use the formula \(q = n \times \Delta H_f\): \(q = 0.105mol \times 10.79 kJ/mol = 1.133 kJ\) Therefore, 1.133 kJ of heat is required to melt 0.105 mole of aluminum at its normal melting point.

Key concepts

Understanding the Heat of Fusion

The heat of fusion is a fundamental concept in thermochemistry. It represents the amount of energy needed to convert a solid into a liquid at its melting point. In the case of aluminum, the heat of fusion is relatively low compared to its heat of vaporization. This is because, when a substance transitions from solid to liquid, only some of the intermolecular forces are loosened. Therefore, less energy is required. Not all bonds need to be broken for transitioning from a solid state, as the molecules still remain relatively close together in the liquid form.
This energy is used to overcome the forces keeping the solid in its rigid structure, allowing it to become a fluid. It is important to note that during this phase transition, the temperature of the substance does not rise. This is because the energy is used entirely to change the state of the matter rather than increase kinetic energy.

Exploring the Heat of Vaporization

The heat of vaporization is crucial for understanding the energy involved in phase changes from liquid to gas. This transition requires significantly more energy than melting, which explains why the heat of vaporization for a substance like aluminum is much larger than its heat of fusion. During vaporization, the energy input is substantial as it needs to separate molecules entirely, overcoming molecular attractions such as Van der Waals forces.
In this process, molecules break free from the liquid form to become gas, which requires enough energy to completely overcome intermolecular attractions. As such, this involves an extensive input of heat energy. The entire input energy in vaporization goes towards breaking these interactions, and similar to melting, the temperature of the liquid remains constant throughout the process until the phase change is complete. This significant disparity between heats of transition reflects the difference in energy required to break bonds between molecules compared to loosening them.

Phase Transition Calculations Simplified

Phase transition calculations are performed to determine the amount of energy required or released during a change of state of a substance. To carry out these calculations, we use specific formulas with the values of heat of fusion or vaporization. For example, if tasked with determining the energy required to melt a certain amount of aluminum, we use the formula:

• \( q = n \times \Delta H_f \)

Here, \( q \) represents heat, \( n \) is the number of moles, and \( \Delta H_f \) is the heat of fusion. The molar mass of the substance is often needed to find \( n \), especially if the problem provides mass rather than moles.
Similarly, for vaporization, you would use:

• \( q = n \times \Delta H_v \)

When approaching these calculations, remember that energy values should be positive when heat is absorbed (like melting or vaporizing) and negative when heat is released (like freezing or condensing). Using this straightforward approach simplifies understanding how energy changes involve precise energy accounting during phase transitions.