Chapter 21, Problem 21.135MP

Interpretation Introduction

(a)

Interpretation:

Determine the larger entropy $\Delta S°$ from the following two reactions.

${\text{Ni(H}}_{\text{2}}{\text{O)}}_6^{2+}(aq)+{\text{6NH}}_{\text{3}}(aq)\rightleftharpoons {\text{Ni(NH}}_{\text{3}}{\text{)}}_6^{2+}(aq)+6{\text{H}}_{\text{2}}\text{O}(l)$ with $K_f=2.0\times {10}^8$ …………… (1)

${\text{Ni(H}}_{\text{2}}{\text{O)}}_6^{2+}(aq)+3en(aq)\rightleftharpoons \text{Ni(en}{)}_3^{2+}(aq)+6{\text{H}}_{\text{2}}\text{O}(l)$ with $K_f=4\times {10}^{17}$ ……………… (2)

Concept introduction:

The entropy is used to define the randomness and disorder of molecules on the basis of their different arrangements.

Interpretation Introduction

(b)

Interpretation:

The greater stability of $Ni{(en)}_3^{2+}$ in terms of relative values of $\Delta S°$ for the two reactions needs to be determined.

Concept introduction:

The following formula will be used −

$\Delta G°=\Delta H°-T\Delta S°$

Here, $\Delta G°$ is standard Gibbs free energy change, $\Delta H°$ is standard enthalpy change, $\Delta S°$ is standard entropy change and T is temperature.

Interpretation Introduction

(c)

Interpretation:

The value of $\Delta S{°}_2-\Delta S{°}_1$ is to be determined, where it is assumed that $\Delta H{°}_2-\Delta H{°}_1$ is zero.

Concept introduction:

The following formula will be used −

$\Delta G°=\Delta H°-T\Delta S°$

Here, $\Delta G°$ is standard Gibbs free energy change, $\Delta H°$ is standard enthalpy change, $\Delta S°$ is standard entropy change and T is temperature.