Minimization Of Gibbs Free Energy
Another common method to calculate phase diagrams is associated with the chemical equilibrium condition in which it is required that the chemical potentials for each component …show more content…
This set of equations is usually solve by using numerical methods for finding roots such as the
Newton-Raphson method. A detailed description of both methods is provided in this section.
The total Gibbs free energy is given as the summation of the Gibbs free energies for the different phases that are coexisting at equilibrium. These Gibbs free energies depend on temperature, pressure and composition of the components in each phase. Equation (1) describes this straightforward relation in which G r represents the Gibbs free energy of each phase and p the number of phases in equilibrium.
The problem of finding the compositions at equilibrium is thus to minimize the total Gibbs free energy at constant P and T
If a total of one mole of components is defined
Where f r represents the amount of the r phase, G r (x Br ) corresponds to the Gibbs free energy of mixing in the r phase, and x Br describes the molar fraction of the component B in the r phase. In addition, the total Gibbs free energy is subject to the following constraints:
These equality constraints arise from the conservation of mass and the restriction that summation …show more content…
In our case, we chose the regular solution model in order to calculate the Gibbs free energies of the solid and liquid solution. In addition, we also selected the solid state as the reference state for each component.
Lagrange multipliers can be also used in order to calculate the equilibrium state of the system. In this method, a Lagrange multiplier will be introduce in the expression for the total Gibbs free energy for each constraint defined in equations (3) and (4). These parameters are defined as μ i and λ r in equation (16):
It is worthy to notice that the new L function will have the same minimum value as the original function for the total Gibbs free energy since the additional terms that were added are all equal to zero. The minimum can be thus obtained from the following equations:
Now by solving for μ i from equation (19) and substituting in equation (18), we have:
Now it follows