Chapter 19, Problem 86AE

(a)

Interpretation Introduction

To determine: The sign of standard entropy change for the given reaction.

Solution: The standard entropy change for the given reaction is positive.

Explanation:

Given

The balanced chemical equation for the given reaction is,

$\text{Cd}{\left({\text{NO}}_3\right)}_2\cdot 4{\text{H}}_2\text{O}\left(s\right)\to \text{Cd}{\left({\text{NO}}_3\right)}_2\left(s\right)+4{\text{H}}_2\text{O}\left(g\right)$

The process is spontaneous at room temperature.
The standard enthalpy change for the reaction is positive.

The relation between standard enthalpy change, standard free energy change and standard entropy change is given by the formula,


$\Delta G^0=\Delta H^0-T\Delta S^0$

Where,

• $\Delta G^0$ is the standard free energy change of the reaction.
• $\Delta H^0$ is the standard enthalpy change of the reaction.
• $T$ is the temperature of the reaction.
• $\Delta S^0$ is the standard entropy change of the reaction.

Since, the reaction is spontaneous which means the standard free energy change of the reaction is negative and the standard enthalpy change is positive. Therefore, for standard free energy to be negative, the standard entropy change of the reaction must be positive.

Conclusion:

The standard entropy change for the given reaction is positive.

(b)

Interpretation Introduction

To determine: The change in standard entropy change of the reaction at room temperature by placing the hydrated compound in the vessel containing large amount of water vapor.