Chapter 14, Problem 22A

(a)

Interpretation Introduction

Interpretation:

The reason for the heat of fusion of aluminum to be much smaller than the heat of vaporization is to be determined.

Concept introduction:

The heat of fusion is the amount of heat energy required to change the state of $\text{1}$ mole of solid to liquid.

The heat of vaporization is the amount of heat energy required to change the state of $\text{1}$ mole of liquid to vapor.

Answer to Problem 22A

From the solid to liquid to gas, the distance between molecules increases and the force of attraction decreases. In liquid state, the aluminum molecules are relatively close together than the gaseous state, hence to separate the aluminum molecules enough to form a gas requires large quantity of energy.

Explanation of Solution

In a solid state, aluminum molecules are packed in a crystalline form. Heat of fusion converts the solid aluminum into liquid, in liquid state the aluminum molecules are enough close together, so most of the intermolecular forces are still present.

Whereas, heat of vaporization converts liquid aluminum into gas, in gaseous state the aluminum molecules are far apart, so all of the intermolecular forces must be overcome and require large amount of energy.

As the distance between molecules increases and the force of attraction decreases, the amount of energy required increases.

(b)

Interpretation Introduction

Interpretation:

The amount of heat required to vaporize aluminum at its normal boiling point is to be determined.

Concept introduction:

The amount of heat required to change the state of $\text{1}$ mole of liquid to vapor at its normal boiling point is called as heat of vaporization.

Answer to Problem 22A

The amount of heat required to vaporize $\text{1}\text{.00 g}$ of aluminum is $\text{10}\text{.9 kJ}$ .

Explanation of Solution

Given Information:

Heat of vaporization $\text{= 293}\text{.4 }\frac{\text{kJ}}{\text{mol}}$

Mass of aluminum $\text{= 1}\text{.00 g}$

Calculation:

According to the heat of vaporization, $\text{293}\text{.4 kJ}$ of heat is required to vaporize $\text{1}$ mole of aluminum.

Moles of $\text{1}\text{.00 g}$ aluminum (molar mass $\text{= 23}\text{.98 }\frac{\text{kJ}}{\text{mol}}$ ) are calculated as,

Moles = mass / molar mass

  $\text{= }\frac{\text{1}\text{.00 g}}{\text{23}\text{.98 }\frac{\text{g}}{\text{mol}}}$

  $\text{= 0}\text{.037064 mol}$

So, $\text{1}$ mole of aluminum required $\text{293}\text{.4 kJ}$ of heat, hence the amount of heat required for $\text{0}\text{.037064 mol}$ of aluminum is calculated as,

  $\text{0}\text{.037064 mol}$ Aluminum x $\text{293}\text{.4 }\frac{\text{kJ}}{\text{1 mol}}\text{ = 10}\text{.87 kJ}$

(c)

Interpretation Introduction

Interpretation:

The amount of heat evolved to freeze aluminum at its normal freezing point is to be determined.

Concept introduction:

The amount of heat evolved to change the state of $\text{1}$ mole of liquid to solid at its normal freezing point is called as heat of fusion.

Answer to Problem 22A

The amount of heat evolved to freeze $\text{5}\text{.00 g}$ of aluminum is $\text{-2}\text{.00 kJ}$ .

Explanation of Solution

Given Information:

Heat of fusion $\text{= 10}\text{.79 }\frac{\text{kJ}}{\text{mol}}$

Mass of aluminum $\text{= 5}\text{.00 g}$

Calculation:

According to the heat of fusion, $\text{10}\text{.79 kJ}$ of heat is required to freeze $\text{1}$ mole of aluminum.

Moles of $\text{5}\text{.00 g}$ aluminum (molar mass $\text{= 23}\text{.98 }\frac{\text{kJ}}{\text{mol}}$ ) are calculated as,

Moles = mass / molar massM

  $\text{= }\frac{\text{5}\text{.00 g}}{\text{23}\text{.98 }\frac{\text{g}}{\text{mol}}}$

  $\text{= 0}\text{.1853 mol}$

So, $\text{1}$ mole of aluminum required $\text{10}\text{.79 kJ}$ of heat, hence the amount of heat required for $\text{0}\text{.1853 mol}$ of aluminum is calculated as,

  $\text{0}\text{.1853 mol}$ Aluminum x $\text{10}\text{.79 }\frac{\text{kJ}}{\text{1 mol}}\text{ = 1}\text{.999 kJ}$

As the heat is evolved during this process, hence the amount of heat would have negative sign.

Therefore, the heat evolved during the freezing process of liquid aluminum is $\text{-2}\text{.00 kJ}$ .

(d)

Interpretation Introduction

Interpretation:

The amount of heat evolved to freeze aluminum at its normal melting point is to be determined.

Concept introduction:

The amount of heat required to change the state of 1 mole of solid to vapor at its normal melting point is called as heat of fusion.

Answer to Problem 22A

The amount of heat required to melt $\text{0}\text{.105 mol}$ of aluminum is $\text{1}\text{.13 kJ}$ .

Explanation of Solution

Given Information:

Heat of fusion $\text{= 10}\text{.79 }\frac{\text{kJ}}{\text{mol}}$

Mole of aluminum $\text{= 0}\text{.105 mol}$

According to the heat of fusion, $\text{10}\text{.79 kJ}$ of heat is required to melt $\text{1}$ mole of aluminum.

So, $\text{1}$ mole of aluminum required $\text{10}\text{.79 kJ}$ of heat, hence the amount of heat required for $\text{0}\text{.105 mol}$ of aluminum is calculated as,

  $\text{0}\text{.105 mol}$ Aluminum x $\text{10}\text{.79 }\frac{\text{kJ}}{\text{1mol}}\text{ = 1}\text{.132 kJ}$

Hence the amount of heat required to melt $\text{0}\text{.105 mol}$ aluminum is $\text{1}\text{.13 kJ}$ .