Chapter 15.4, Problem 10PPB

K_sp for Co(OH)₂ at 25°C is 3.3 ×10^-16 Using this and data from Appendix 2, calculate the value of ΔG_f^° for Co(OH)₂(s).

Interpretation Introduction

Interpretation:

Calculate the standard free energy values $({\text{ΔG}}_{\text{f}}^{\text{0}})$ for given the calcium hydroxide $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ equilibrium (Ksp) reaction at $25°C$.

Concept Introduction:

Standard free energy $({\text{ΔG}}_{\text{f}}^{\text{0}})$: This thermodynamic function associated with a chemical reaction that can be used to do work. The standard free energy of formation of a compound change of Gibbs free energy the formation of 1 mole of substance from its component elements, at their standard states (For example the obtained product is more stable at $25°C$).

Solubility of equilibrium (Ksp): Equilibrium constant, the value of $\text{Kw}$ varies with temperature. Its value is usually taken to be $\text{1}{\text{.00 x 10}}^{\text{-14}}{\text{mol}}^{\text{2}}{\text{dm}}^{\text{-6}}$ at room temperature. In fact, this is its value at a bit less than $25°C$.

Entropy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Gibbs free energy (G): The thermodynamic quantity to the ( $\text{ΔH}$) 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.

Answer to Problem 10PPB

The respective entropy $({\text{ΔG}}^0)$ values are given the statement of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ equilibrium reaction is shown below.

$\begin{array}{l}{\text{Ca(OH)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Ca}}^{2+}(aq)+2{\text{OH}}^{2-}(aq)\\ \text{The molar solubility product and Gibbs free energy is determined}\\ \text{Ksp}\text{=}[C{a}^{2+}]{[}^2\\ {\text{ΔG}}^0=55.9\text{KJ/mol}\end{array}$

Explanation of Solution

To find: To understand and derived the second law thermodynamic entropy $({\text{ΔG}}^0)$ equation of given $AgCl$ dissociation reaction.

We consider the fallowing entropy equation $({\text{ΔG}}^0)$

$\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{Ksp---------[1]}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\text{(or)}\\ \text{Ksp solubility of reactant species}\\ \ln =(-ve(\log )\text{State Function})\end{array}$

The equilibrium chemistry should explained for more details in different type of reactions (like forward and backward reactions), at the same rate constant that is no net reactions is occurring in either directions of reactants and products are constant. The above equation Q=K & $\text{ΔG}=0$, above equation larger K is the more negative ${\text{ΔG}}^{\text{0}}$ is. The most of chemists followed by this equation is one of the most important in the thermodynamics because it enables us to find the equilibrium (K) constant of a reaction. Here the indicate (Ksp) is autoionization water (or) ionization of respective equilibrium reaction.

To find: Calculate the entropy values $({\text{ΔG}}^0)$ for given the statement.

Let we consider the above chemical equilibrium process $({\text{ΔG}}^0)$.

$\begin{array}{l}{\text{Ca(OH)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Ca}}^{2+}(aq)+2{\text{OH}}^{2-}(aq)-------[DessociationReaction]\\ \text{Ksp}\text{=}[C{a}^{2+}]{[}^2(or)\frac{1}{[C{a}^{2+}]{[2O{H}^{-}]}^2}-------[2]\\ {\text{ ΔG}}^{\text{0}}\text{=-RTln}\text{K---------[1]}\text{(or)}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{Ksp}\\ Here\text{R}\text{=}\text{0}\text{.08314}\text{J/K}\text{.}\text{mol (or) (-8}\text{.314}\times {\text{10}}^{\text{-3}}\text{J/K}\text{.}\text{mol)}\\ T={273}^0\text{C}+{25}^0\text{C}=298\text{K}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{Ksp}=3.3\times {10}^{-16}\end{array}$

The given statement of values is substitute in above equation (2)

$\begin{array}{l}{\text{ΔG}}_{\text{f}}^0\text{=-RTln}\text{K}\\ {\text{ΔG}}_{\text{f}}^0=-\left(\frac{\text{8}\text{.314}\times {10}^{-3}\text{kJ}}{\text{K×mol}}\right)(298K)\ln (3.3\times {10}^{-16})\\ {\text{ΔG}}_{\text{f}}^0=-454.4\text{kJ/mol }\end{array}$

Given the $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ equilibrium reaction has heterogeneous, the single reactant molecule produce a two aqueous molecules, hence we derived a products species only because reactant will not appeared in this equilibrium. Moreover the equilibrium constant (Kp) value is large (K) values when compare to the standard free energy values, so equilibrium shifted into right side. Finally this reaction has large activation energy and converted into rate of reaction is extremely slow; the calculated standard free energy values are showed above.

Conclusion

The standard free energy values $({\text{ΔG}}_{\text{f}}^{\text{0}})$ are derived given $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ equilibrium reaction.