GIBBS FREE ENERGY CHANGE OF MIXING FOR AN IDEAL SOLUTION

Q: What is an example of Gibbs free energy?

In A reaction between Potassium chloride and water, Solid KCl adds to the water, and solid KCl dissociates itself in water without giving outside energy to the system.

WHAT IS GIBBS FREE ENERGY?

According to thermodynamics, substances with available energy that tend to complete the reaction are called Gibbs free energy. It is denoted as ∆G.

WHAT IS THE UNIT OF GIBBS FREE ENERGY?

Unit of ∆G = kJ/mol

If the value of ∆G is Negative, the reaction is spontaneous.

EXAMPLE OF GIBBS FREE ENERGY

For example, in A reaction between Potassium chloride and water, Solid KCl adds to the water, and solid KCl dissociates itself in water without giving outside energy to the system.

If the value of ∆G is Positive, it means the reaction is non-spontaneous.

If the value of ∆G = 0, the reaction is an Equilibrium condition.

The solution was made by mixing n_A moles of liquid A and n_B moles of liquid B,

n_AA +n_BB → solution

For thermodynamics, the free energy (G) of the solution at a given temperature and pressure is,

\[\displaystyle G={{n}_{A}}{{{\bar{G}}}_{A}}+{{n}_{B}}{{{\bar{G}}}_{B}}-----(1)\]

Here,

\[\displaystyle {{{\bar{G}}}_{A}}=\text{Partial molar free energy of constituents A}\]

\[\displaystyle {{{\bar{G}}}_{B}}=\text{Partial molar free energy of constituents B}\]

\[\displaystyle G_{A}^{0}=\text{Free energies per mole of the constituents A and}\]

\[\displaystyle \text{G}_{B}^{0}=\text{Free energies per mole of the constituents B}\]

So, change in free energy of the system on mixing is known as free energy change of mixing, ∆G_mix is,

\[\displaystyle \Delta {{\text{G}}_{{mix}}}=G-\left( {{{n}_{A}}G_{A}^{0}+{{n}_{B}}G_{B}^{0}} \right)------(2)\]

Now, substituting the value of G in equation (1), we get,

\[\displaystyle \Delta {{\text{G}}_{{mix}}}=\left( {{{n}_{A}}{{{\bar{G}}}_{A}}+{{n}_{B}}{{{\bar{G}}}_{B}}} \right)-\left( {{{n}_{A}}G_{A}^{0}+{{n}_{B}}G_{B}^{0}} \right)\]

\[\displaystyle ={{n}_{A}}\left( {{{{\bar{G}}}_{A}}-G_{A}^{0}} \right)+{{n}_{B}}\left( {{{{\bar{G}}}_{B}}-G_{B}^{0}} \right)-----(3)\]

\[\displaystyle A\text{ quantities }\left( {{{{\bar{G}}}_{A}}-G_{A}^{0}} \right)=\text{Change in partial molar free enegy components A}\]

\[\displaystyle \left( {{{{\bar{G}}}_{B}}-G_{B}^{0}} \right)=\text{Change in partial molar free enegy components B}\]

The thermodynamic equation given for the state is the chemical potential µ_i of component i so,

\[\displaystyle {{\mu }_{i}}=\mu _{i}^{0}+RT\ln {{a}_{i}}-------(4)\]

Here, T = Temperature, a = activity of the component

\[\displaystyle \mu _{i}^{0}=\text{chemical potential of substance in its standard state}\]

Above equation (4), write for constituents A and B, written as,

\[\displaystyle {{{\bar{G}}}_{A}}=G_{A}^{0}+RT\ln {{a}_{A}}\]

\[\displaystyle {{{\bar{G}}}_{A}}-G_{A}^{0}=RT\ln {{a}_{A}}------(5)\]

\[\displaystyle {{{\bar{G}}}_{B}}=G_{B}^{0}+RT\ln {{a}_{B}}\]

\[\displaystyle {{{\bar{G}}}_{B}}-G_{B}^{0}=RT\ln {{a}_{B}}------(6)\]

Now, put the value of an equation (5) and (6) into equation (3), we obtain,

\[\displaystyle \Delta {{G}_{{mix}}}={{n}_{A}}RT\ln {{a}_{A}}+{{n}_{B}}RT\ln {{a}_{B}}-----(7)\]

The activity of each component should be equal to the mole fraction. If the solution is ideal so,

a_A =X_A and a_B = X_B

Now, the free energy from the mixing of an ideal solution is expressed as,

\[\displaystyle \Delta {{G}_{{mix}}}={{n}_{A}}RT\ln {{x}_{A}}+{{n}_{B}}RT\ln {{x}_{B}}------(8)\]

If more than two components of the solution, the equation is written as,

\[\displaystyle \Delta {{G}_{{mix}}}={{n}_{A}}RT\ln {{x}_{A}}+{{n}_{B}}RT\ln {{x}_{B}}+{{n}_{c}}RT\ln {{x}_{c}}+.........\]

\[\displaystyle =RT\Sigma {{n}_{i}}\ln {{x}_{i}}------(9)\]

Dividing both the side equation (8) n_A + n_B, we obtain,

\[\displaystyle \frac{{\Delta {{G}_{{mix}}}}}{{{{n}_{A}}+{{n}_{B}}}}=\frac{{{{n}_{A}}}}{{{{n}_{A}}+{{n}_{B}}}}RT\ln {{x}_{A}}+\frac{{{{n}_{B}}}}{{{{n}_{A}}+{{n}_{B}}}}RT\ln {{x}_{B}}\]

\[\displaystyle ={{x}_{A}}RT\ln {{x}_{A}}+{{x}_{B}}RT\ln {{x}_{B}}-----(10)\]

If n_A +n_B = 1 so, equation (10) written as,

\[\displaystyle \Delta {{G}_{{mix}}}=RT\left( {{{x}_{A}}\ln {{x}_{A}}+{{x}_{B}}\ln {{x}_{B}}} \right)-----(11)\]

For more than two components of the solution,

\[\displaystyle \Delta {{G}_{{mix}}}=RT\left( {{{x}_{A}}\ln {{x}_{A}}+{{x}_{B}}\ln {{x}_{B}}+{{x}_{c}}RT\ln {{x}_{c}}+.........} \right)\]

\[\displaystyle =RT\Sigma {{x}_{i}}\ln {{x}_{i}}-----(12)\]

X_i= mole fraction, and X_i << 1