Chapter 18, Problem 18.74QP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant $Kp$ value should be derived given the spontaneous reaction at ${\text{25}}^{°}\text{C}$.

Concept Introduction:

Spontaneous reaction: This reaction is said to be spontaneous it occurs without being drive by some of outside force. The two driving forces for all chemical reactions, first one is enthalpy and second one is entropy, spontaneous reactions occurs without outside intervention.

Equilibrium constant: The concentration at equilibrium always combine in the manner below where the products are in the numerator and the reactant are in the denominator to produce the (K) value regardless the initial concentration of species.

$K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Free energy: is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The free energy change ${\text{(ΔG}}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.31\text{4}\text{J/K×mol/K}\text{.mol)}\\ T=\text{Temprature}2\text{73}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

Explanation of Solution

First calculate the free energy ($\Delta G°$) value for given reaction

$\begin{array}{l}\text{Fe(s)}\text{+}{\text{2H}}^{\text{+}}\text{(aq})\to {\text{Fe}}^{\text{2+}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\\ {\text{ΔG}}_{\text{rxn}}^{°}={\displaystyle \sum n{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(Products)}}\text{-}{\displaystyle \sum {\text{mΔG}}_{\text{f}}^{\text{°}}(Reac\tan ts)}\\ {\text{ΔG}}^{°}\text{=}{\text{ΔG}}_{\text{f}}^{\text{o}}{\text{(H}}_{\text{2}}\text{)+}{\text{ΔG}}_{\text{f}}^{\text{o}}{\text{(Fe}}^{\text{2+}}\text{)-}[{\text{ΔG}}_{\text{f}}^{\text{o}}{\text{(Fe)-2ΔG}}_{\text{f}}^{\text{o}}{\text{(H}}^{\text{+}}\text{)]}\text{(}\because {\text{ΔG}}_{\text{f}}^{\text{o}}\text{Values}\text{ referred}\text{from}\text{Appendix-2}\text{)}\\ {\text{ΔG}}^{°}\text{=}\text{(1)(0)+}\text{(1)(-84}\text{.9}\text{kJ/mol)-[(1)(0)-(2)(0)]}\\ {\text{ΔG}}^{\text{0}}\text{=}\text{-84}\text{.9}\text{kJ/mol}\end{array}$

Next calculate the equilibrium constant ($Kp$) value of given reaction (a),

$\begin{array}{l}\text{The}\text{free}\text{energy}\text{equation}\text{is,}\\ {\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{-84}\text{.9}\times {10}^3\text{J/mol=}\text{-(8}\text{.314}\text{J/mol}\cdot \text{K)(298}\text{K)InK}\\ \text{Rewrite the above free}\text{energy }\Delta \text{G}°\text{equation }\\ \text{InKp}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{\text{-84}{\text{.9×10}}^{\text{3}}\text{J/mol}}{\text{-(8}\text{.314}\text{J/}\text{mol)(298K)}}\\ \text{InKp}\text{=}34.152\\ \text{Kp}=e^{34.152}\\ \text{Kp=}6.79\times {10}^{14}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant $Kp$ value should be derived given the non-spontaneous reaction at ${\text{25}}^{°}\text{C}$.

Concept Introduction:

Equilibrium constant: The concentration at equilibrium always combine in the manner below where the products are in the numerator and the reactant are in the denominator to produce the (K) value regardless the initial concentration of species.

$K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Free energy: is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The free energy change ${\text{(ΔG}}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(8.31\text{4}\text{J/K×mol/K}\text{.mol)}\\ T=\text{Temprature}2\text{73}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

Explanation of Solution

First calculate the free energy ($\Delta G°$) value for given reaction

$\begin{array}{l}\text{Cu(s)}\text{+}{\text{2H}}^{\text{+}}\text{(aq)}\to {\text{Cu}}^{\text{2+}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\\ {\text{ΔG}}_{\text{rxn}}^{°}={\displaystyle \sum n{\text{ΔG}}_{\text{f}}^{\text{°}}\text{(Products)}}\text{-}{\displaystyle \sum {\text{mΔG}}_{\text{f}}^{\text{°}}(Reac\tan ts)}\\ {\text{ΔG}}_{}^{°}={\text{ΔG}}_{\text{f}}^{\text{o}}{\text{(H}}_{\text{2}}\text{)+}{\text{ΔG}}_{\text{f}}^{\text{o}}{\text{(Cu}}^{\text{2+}}\text{)-}[{\text{ΔG}}_{\text{f}}^{\text{o}}\text{(}Cu{\text{)-2ΔG}}_{\text{f}}^{\text{o}}{\text{(H}}^{\text{+}}\text{)]}\\ {\text{ΔG}}^{°}=(1)(0)+(1)(\text{64}\text{.98kJ/mol})-(1)(0)-(2)(0)\\ {\text{ΔG}}^{°}=\text{64}\text{.98kJ/mol}\end{array}$

Next we calculate the equilibrium constant ($Kp$) value of given reaction (b),

$\begin{array}{l}{\text{ΔG}}^{°}\text{=-RTln}\text{K}\\ \text{The respactive values are substituted for above equlibrium reaction (2)}\\ 6\text{4}\text{.98}\times {\text{10}}^3\text{J/mol=}\text{-(8}\text{.314}\text{J/K}\cdot \text{mol)(298}\text{K)InK}\\ \text{Rewrite the above equation }\\ \text{InKp}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{64.98\times {10}^3\text{J/mol}}{-(8.314\text{J/K}\text{mol)}(298K)}\\ \text{InKp}\text{=}-26.231\\ \text{K=}{\text{e}}^{-26.231}\\ \text{Kp=}4.05\times {10}^{-12}\end{array}$

In conclusion the activity series is correct, the large value of $K$ for reaction (a) it is indicate that products are highly favoured, whereas the small value of $K$ for reaction $(b)$ indicates that reactants are highly favoured.