Chapter 15, Problem 15.116QP

Interpretation Introduction

 Interpretation:

To calculate the free energy $(\text{ΔG})$ values for given aqueous phase equilibrium reaction at ${\text{25}}^{\text{0}}\text{C}$.

Concept Introduction:

Gibbs free energy (G): The thermodynamic quantity to the $({\text{ΔG}}^0)$ enthalpy of a system process and minutes the product of entropy and the absolute temperature. The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy $(\text{ΔH)}$ of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation $\text{G=H-(TS) or (ΔG=ΔH-TΔS)}$.

To find: Calculate the $(\text{ΔG})$ values for given the aqueous phase equilibrium reaction (a and b) Calculate and analyze the $(\text{ΔG})$ values with respective concentrations.

Answer to Problem 15.116QP

The aqueous phase equilibrium concentrations (free energy $\text{ΔG}$) of given the four statements of reactions (a-d) are shown below.

$\begin{array}{l}{\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{OH}}^{\text{-}}\text{(aq)}\\ \text{a)}\text{.}{\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-7}}M,[O{H}^{-}]=\text{1}\text{.0}\times {\text{10}}^{\text{-7}}M\\ \Delta \text{G}=8.0\times {\text{10}}^4\text{J/mol}\\ \text{b)}\text{.}{\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-3}}M,[O{H}^{-}]=\text{1}\text{.0}\times {\text{10}}^{\text{-4}}M\\ \Delta \text{G}=4.0\times {\text{10}}^4\text{J/mol}\\ \text{c)}\text{.}{\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-12}}M,[O{H}^{-}]=2.\text{0}\times {\text{10}}^{\text{-8}}M\\ \Delta \text{G}=-3.2\times {\text{10}}^4\text{J/mol}\\ \text{d)}\text{.}{\text{[H}}^{\text{+}}\text{]=}3.5\text{M},[O{H}^{-}]=4.\text{8}\times {\text{10}}^{\text{-4}}M\\ \Delta \text{G}=6.4\times {\text{10}}^4\text{J/mol}\end{array}$

Explanation of Solution

(a) and (b)

Let us consider the following standard free energy equation

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\\ \text{ΔG}=\text{Free energy;}{\text{ΔG}}^{\text{0}}=\text{Standard state free energy}\\ \text{R}\text{=}\text{Gas}\text{Constant-}(8.314\text{J/K}\text{.mol});T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{);}\ln =(-ve(\log )\text{State Function})\\ \text{Here }\\ \text{ΔG=}{\text{ΔG}}^{\text{ο}}+RT\ln Q(or)-------[2]\\ \text{ΔG=}{\text{ΔG}}^{\text{ο}}+RT\ln {\text{[H}}^{\text{+}}\text{]}[O{H}^{-}]\end{array}$

The reactions (a and b) concentrations at 25⁰C. Science the reaction is at equilibrium $(\text{ΔG=0})$, this is advantageous, because it allows us to calculate $({\text{ΔG}}^{\text{0}})$ and the recall that at equilibrium Q=K, the modified equation (2) showed above.

$\begin{array}{l}\text{Let}\text{us consider following reaction (a}\text{and b)}\\ \text{Reaction(a)}\\ {\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{OH}}^{\text{-}}\text{(aq)}\\ \Delta {\text{G}}^{\text{0}}=-RT\ln K_w\\ {\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-7}}M,[O{H}^{-}]=\text{1}\text{.0}\times {\text{10}}^{\text{-7}}M\\ \Delta \text{G}=\Delta {\text{G}}^{\text{0}}+RT\ln ([H^{+}][O{H}^{-}])-------[3]\\ Givenvalues(R,\text{T})aresubstitutedequation(3)\\ \Delta {\text{G}}^{\text{0}}=-(8.314J/K\cdot mol)(298\text{K})ln(1.0\times {10}^{-7})(1.0\times {10}^{-7})\\ =-2477.572\ln (1.0\times {10}^{-14})[\because Here\text{In}1.0\times {10}^{-7}=-29.9336]\\ \Delta {\text{G}}^{\text{0}}=8.0\times {10}^{4}J/mol\\ \text{Reaction}\text{(b)}\\ \text{b)}\text{.}{\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-3}}M,[O{H}^{-}]=\text{1}\text{.0}\times {\text{10}}^{\text{-4}}M\\ \Delta {\text{G}}^{\text{0}}=8.0\times {10}^{4}J/mol\\ \Delta {\text{G}}^{\text{0}}=-RT\mathrm{lnQ}\\ \Delta \text{G}=\Delta {\text{G}}^{\text{0}}+RT\ln ([H^{+}][O{H}^{-}])-----[4]\\ \text{Given values are substituted equatuion (4)}\\ \Delta \text{G}=8.0\times {10}^{4}J/mol+(8.314J/K\cdot mol)(298\text{K})ln(1.0\times {10}^{-3})(1.0\times {10}^{-4})\\ \Delta \text{G}=4.0\times {10}^{4}J/mol\end{array}$

The standard free energy values are calculated given the aqueous phase equilibrium reactions (a and b).

To find: Calculate the $(\text{ΔG})$ values for given the aqueous phase equilibrium reaction (c and d).

Calculate and analyze the $(\text{ΔG})$ values with respective concentrations.

$\begin{array}{l}\text{Let us consider the following reaction (c and d)}\\ \text{Reaction(c)}\\ {\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{OH}}^{\text{-}}\text{(aq)}\\ \Delta {\text{G}}^{\text{0}}=-RT\ln K_w\\ {\text{[H}}^{\text{+}}\text{]=}\text{1}\text{.0}\times {\text{10}}^{\text{-12}}M,[O{H}^{-}]=2.\text{0}\times {\text{10}}^{\text{-8}}M\\ \Delta \text{G}=-3.2\times {\text{10}}^4\text{J/mol}\\ \Delta \text{G}=\Delta {\text{G}}^{\text{0}}+RT\ln ([H^{+}][O{H}^{-}])-------[4]\\ Givenvalues(R,\text{T})aresubstitutedequation(3)\\ \Delta \text{G}=(8.0\times {10}^{4}J/mol)+(8.314J/K\cdot mol)(298\text{K})ln(\text{1}\text{.0}\times {\text{10}}^{\text{-12}})(2.\text{0}\times {\text{10}}^{\text{-8}})\\ \Delta \text{G}=-3.2\times {10}^{4}J/mol\\ \text{Reaction}\text{(d)}\\ {\text{[H}}^{\text{+}}\text{]=}3.5\text{M},[O{H}^{-}]=4.\text{8}\times {\text{10}}^{\text{-4}}M\\ \Delta \text{G}=6.4\times {\text{10}}^4\text{J/mol}\\ \Delta {\text{G}}^{\text{0}}=-RT\mathrm{lnQ}\\ \Delta \text{G}=\Delta {\text{G}}^{\text{0}}+RT\ln ([H^{+}][O{H}^{-}])-----[5]\\ \text{Given values are substituted equatuion (5)}\\ \Delta \text{G}=8.0\times {10}^{4}J/mol+(8.314J/K\cdot mol)(298\text{K})ln(3.5)(4.\text{8}\times {\text{10}}^{\text{-4}})\\ \Delta \text{G}=6.4\times {10}^{4}J/mol\\ \end{array}$

The standard free energy values are calculated given the aqueous phase equilibrium reactions (c and d).

Conclusion

The free energy values are calculated $(\text{ΔG})$ given the different aqueous phase equilibrium reactions.