Chapter 13, Problem 56E

Interpretation Introduction

(a)

Interpretation:

The reason for different values of $\text{ΔrG° }$ for following two equations should be explained.

${\text{Mg(OH)}}_{\text{2}}{\text{(s)+2H}}^{\text{+}}\text{(aq)}\stackrel{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l) ΔrG° }=-95.5\text{ kJ/mole}$

$\frac{1}{2}Mg{(OH)}_2(s)+H^{+}(aq)\stackrel{}{\rightleftharpoons }\frac{1}{2}M{g}^{2+}(aq)+H_{2}O(l)\text{ ΔrG° }=-47.8\text{ kJ/mole}$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔG = ΔH - TΔS }$

With the help of this equation, one can predict the change in $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$. ${\text{K}}_{\text{p}}$ or $\text{Kc}$ are the equilibrium constants for the reaction which are ratio of gaseous and aqueous products and reactant molecules.

$\text{A + B }\rightleftharpoons \text{C + D }$

$\text{K = Kp = }\frac{\text{[PC] [PD]}}{\text{[PA] [PB]}}$

Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.

The relation between equilibrium constant and $\Delta \text{rG}°$ can be written as:

$\text{ΔrG° = - 2}\text{.303 RT log K}$

Here,

• R = 8.314 J / mol .K
• T = temperature in Kelvin

Interpretation Introduction

(b)

Interpretation:

Whether K for following two equations will be different or same should be explained.

${\text{Mg(OH)}}_{\text{2}}{\text{(s)+2H}}^{\text{+}}\text{(aq)}\stackrel{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l) ΔrG° }=-95.5\text{ kJ/mole}$

$\frac{1}{2}Mg{(OH)}_2(s)+H^{+}(aq)\stackrel{}{\rightleftharpoons }\frac{1}{2}M{g}^{2+}(aq)+H_{2}O(l)\text{ ΔrG° }=-47.8\text{ kJ/mole}$

Concept introduction:

The Gibb’s equation of thermodynamic purposed a relation between $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$ with temperature. The mathematical expression of Gibb’s equation can be written as:

$\text{ΔG = ΔH - TΔS }$

With the help of this equation, one can predict the change in $\text{ΔS}$, $\text{ΔH}$ and $\text{ΔG}$. ${\text{K}}_{\text{p}}$ or $\text{Kc}$ are the equilibrium constants for the reaction which are ratio of gaseous and aqueous products and reactant molecules.

$\text{A + B }\rightleftharpoons \text{C + D }$

$\text{K = Kp = }\frac{\text{[PC] [PD]}}{\text{[PA] [PB]}}$

Here, ‘a’ represents the active mass and ‘P’ represents the partial pressure.

The relation between equilibrium constant and $\Delta \text{rG}°$ can be written as:

$\text{ΔrG° = - 2}\text{.303 RT log K}$

Here,

• R = 8.314 J / mol .K
• T = temperature in Kelvin