Chapter 17, Problem 62A

(a)

Interpretation Introduction

Interpretation:

The quantity of heat lost or gained is to be calculated.

Concept Introduction :

The molar heat of solidification or freezing of a substance is the heat released by one mole of that substance as it is converted from a liquid state to a solid state. The negative sign of heat represents an exothermic reaction, which means heat is released during the reaction.

Answer to Problem 62A

The quantity of heat lost is $21.04\text{ kJ}$

Explanation of Solution

Molar heat of freezing of water at $\text{0}°\text{C}$ is $\text{ -6}\text{.01 kJ}$ .

Thus, to freeze $\text{1 mole}$ of water at $\text{0}°\text{C}$ , heat released is = $\text{6}\text{.01 kJ}$

To freeze $\text{3}\text{.50 mole}$ of water at $\text{0}°\text{C}$ , heat released is = $\frac{\text{6}\text{.01 kJ}}{\text{1 mol}}\times 3.50\text{ mol}=21.04\text{ kJ}$

(b)

Interpretation Introduction

Interpretation:

The quantity of heat lost or gained is to be calculated.

Concept Introduction :

The molar heat of condensation of a substance is the heat released by one mole of that substance as it is converted from a gaseous state to a liquid state. The negative sign of heat represents an exothermic reaction, which means heat is released during the reaction.

Answer to Problem 62A

The quantity of heat lost is $18.0\text{ kJ}$

Explanation of Solution

Molar heat of condensation of steam at $\text{100°C}$ is $\text{-40}\text{.7 kJ}$ .

Thus, to condense $\text{1 mole}$ of steam at $\text{100°C}$ , heat released is = $\text{40}\text{.7 kJ}$

To freeze $0.44\text{ mol}$ of steam at $\text{100°C}$ , heat released is = $\frac{\text{40}\text{.7 kJ}}{\text{1 mol}}\times 0.44\text{ mol}=18.0\text{ kJ}$

(c)

Interpretation Introduction

Interpretation:

The quantity of heat lost or gained is to be calculated.

Concept Introduction :

The molar heat of the solution of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. The negative sign of heat represents an exothermic reaction, which means heat is released during the reaction.

Answer to Problem 62A

The quantity of heat lost is $55.64\text{ kJ}$

Explanation of Solution

Molar heat of solution of $\text{NaOH}$ is $\text{-44}\text{.51 kJ}$ .

Thus, to dissolve $\text{1 mole}$ of $\text{NaOH}$ , heat released is = $\text{44}\text{.51 kJ}$

To dissolve $1.25\text{ mol}$ of $\text{NaOH}$ , heat released is = $\frac{\text{44}\text{.51 kJ}}{\text{1 mol}}\times 1.25\text{ mol}=55.64\text{ kJ}$

(d)

Interpretation Introduction

Interpretation:

The quantity of heat lost or gained is to be calculated.

Concept Introduction:

The molar heat of the vaporization of a substance is the heat released by one mole of that substance as it is converted from a liquid state to a gaseous state. A positive sign of heat represents an endothermic reaction, which means heat is absorbed during the reaction.

Answer to Problem 62A

The quantity of heat gained is $5.784\text{ kJ}$

Explanation of Solution

Molar heat of vaporization of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O)}$ is $\text{38}\text{.56 kJ}$ .

Thus, to vaporize $\text{1 mole}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ , heat absorbed is = $\text{38}\text{.56 kJ}$

To dissolve $0.15\text{ mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ , heat absorbed is = $\frac{\text{38}\text{.56 kJ}}{\text{1 mol}}\times 0.15\text{ mol}=5.784\text{ kJ}$