Chapter 21, Problem 96A

(a)

Interpretation Introduction

Interpretation: The amount of heat lost or gained when 0.625 mol of NaOH (s) is dissolved in water needs to be determined.

Concept Introduction: A measurement in one set of units can be converted into another set of units using a conversion factor.

Explanation of Solution

The enthalpy of the dissolution reaction of NaOH is -41.6 kJ/mol. Thus, if 1 mol of NaOH is dissolved in water, 41.6 kJ of heat is lost.

The amount of heat lost for 0.625 mol of NaOH will be:

  $\begin{array}{c}\Delta H=-\left(\frac{41.6\text{ kJ}}{1mol}\right)\left(0.625\text{ mol}\right)\\ =-26\text{ kJ}\end{array}$

Thus, the amount of heat lost is 26 kJ. Here, the negative sign shows the loss of heat.

(b)

Interpretation Introduction

Interpretation: The amount of heat lost or gained when 1.17 mol of water freezes at $0{}^{\text{o}}\text{C}$ needs to be determined.

Concept Introduction: A measurement in one set of units can be converted into another set of units using a conversion factor.

Explanation of Solution

The enthalpy of the fusion of water is 6.01 kJ/mol. Thus, for 1 mol of water, 6.01 kJ of heat is required for fusion. To freeze water, the same amount of heat will be lost and the value will be -6.01 kJ.

The amount of heat lost for 1.17 mol of water will be:

  $\begin{array}{c}\Delta H=-\left(1.17\text{ mol}\right)\left(\frac{6.01\text{ kJ}}{1\text{ mol}}\right)\\ =-7.03\text{2 kJ}\end{array}$

Thus, the amount of heat lost is $7.03\text{2 kJ}$ .

(c)

Interpretation Introduction

Interpretation: The amount of heat lost or gained when 0.30 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)$ vaporizes needs to be determined.

Concept Introduction: A measurement in one set of units can be converted into another set of units using a conversion factor.

Explanation of Solution

Heat of vaporization of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)$ is 38.56 kJ/mol. Thus, 38.56 kJ of heat is gained to vaporize 1 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)$ .

The amount of heat required for 0.30 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)$ will be:

  $\begin{array}{c}\Delta H=\left(0.30\text{ mol}\right)\left(\frac{38.56\text{ kJ}}{\text{ 1 mol}}\right)\\ =11.57\text{ kJ}\end{array}$

Thus, the amount of heat gained is $11.57\text{ kJ}$

(d)

Interpretation Introduction

Interpretation: The amount of heat lost or gained when 0.66 mol of the steam condenses at $100{}^{\text{o}}\text{C}$ needs to be determined.

Concept Introduction: A measurement in one set of units can be converted into another set of units using a conversion factor.

Explanation of Solution

The heat of condensation of steam is 40.8 kJ/mol. Thus, for 1 mol of water, 40.8 kJ of heat is gained.

The amount of heat required for 0.66 mol of steam will be:

  $\begin{array}{c}\Delta H=\left(0.66\text{ mol}\right)\left(\frac{40.8\text{ kJ}}{1\text{ mol}}\right)\\ =26.93\text{ kJ}\end{array}$

Thus, the amount of heat gained is $26.93\text{ kJ}$ .