Sistem eutektik
Perhitungan eutektik

Komposisi dan suhu eutektik dapat dihitung dari entalpi dan entropi fusi masing-masing komponen.
Entalpi bebas Gibbs G tergantung pada perbedaan masing-masing dengan rumus (
 G=H-TS\Rightarrow {\left\{{\begin{array}{l}H=G+TS\\\\{\left({\frac {\partial G}{\partial T}}\right)_{P}=-S}\end{array}}\right.}\Rightarrow H=G-T\left({\frac {\partial G}{\partial T}}\right)_{P}.
Dengan demikian, turunan G / T pada tekanan konstan dihitung dengan persamaan
\left({\frac {\partial G/T}{\partial T}}\right)_{P}={\frac {1}{T}}\left({\frac {\partial G}{\partial T}}\right)_{P}-{\frac {1}{T^{2}}}G=-{\frac {1}{T^{2}}}\left({G-T\left({\frac {\partial G}{\partial T}}\right)_{P}}\right)=-{\frac {H}{T^{2}}}
Potensi kimia  \ mu_{i}
dihitung jika kita menganggap kegiatan sama dengan konsentrasi.
\mu _{i}=\mu _{i}^{\circ }+RT\ln {\frac {a_{i}}{a}}\approx \mu _{i}^{\circ }+RT\ln x_{i}
Pada kesetimbangan,  \mu _{i}=0, thus  \mu _{i}^{\circ } didapat dengan:
\mu _{i}=\mu _{i}^{\circ }+RT\ln x_{i}=0\Rightarrow \mu _{i}^{\circ }=-RT\ln x_{i}.
Dengan menggabungkan formula di atas, didapat persamaan:
 {\begin{array}{l}\left({\frac {\partial \mu _{i}/T}{\partial T}}\right)_{P}={\frac {\partial }{\partial T}}\left({R\ln x_{i}}\right)\Rightarrow R\ln x_{i}=-{\frac {H_{i}^{\circ }}{T}}+K\\\\\end{array}}
Integrasi konstanta K dapat ditentukan untuk komponen murni dengan suhu leleh  T^{\circ } dan entalpi fusi  H^{\circ } Eq. x_{i}=1\Rightarrow T=T_{i}^{\circ }\Rightarrow K={\frac {H_{i}^{\circ }}{T_{i}^{\circ }}}
Kita mendapatkan hubungan yang menentukan fraksi molar sebagai fungsi suhu untuk masing-masing komponen.
 R\ln x_{i}=-{\frac {H_{i}^{\circ }}{T}}+{\frac {H_{i}^{\circ }}{T_{i}^{\circ }}}
Campuran komponen n digambarkan oleh sistem
{\begin{array}{l}\left\{{\begin{array}{*{20}c}{\ln x_{i}+{\frac {H_{i}^{\circ }}{RT}}-{\frac {H_{i}^{\circ }}{RT_{i}^{\circ }}}=0}\\{\sum \limits _{i=1}^{n}{x_{i}=1}}\\\end{array}}\right.\\\\\end{array}}
 {\begin{array}{l}\left\{{\begin{array}{*{20}c}{\forall i<n\Rightarrow \ln x_{i}+{\frac {H_{i}^{\circ }}{RT}}-{\frac {H_{i}^{\circ }}{RT_{i}^{\circ }}}=0}\\{\ln \left({1-\sum \limits _{i=1}^{n-1}{x_{i}}}\right)+{\frac {H_{n}^{\circ }}{RT}}-{\frac {H_{n}^{\circ }}{RT_{n}^{\circ }}}=0}\\\end{array}}\right.\\\\\end{array}}
yang dapat diselesaikan dengan
{\begin{array}{c}\left[{\begin{array}{*{20}c}{\Delta x_{1}}\\{\Delta x_{2}}\\{\Delta x_{3}}\\\vdots \\{\Delta x_{n-1}}\\{\Delta T}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{1/x_{1}}&0&0&0&0&{-{\frac {H_{1}^{\circ }}{RT^{2}}}}\\0&{1/x_{2}}&0&0&0&{-{\frac {H_{2}^{\circ }}{RT^{2}}}}\\0&0&{1/x_{3}}&0&0&{-{\frac {H_{3}^{\circ }}{RT^{2}}}}\\0&0&0&\ddots &0&{-{\frac {H_{4}^{\circ }}{RT^{2}}}}\\0&0&0&0&{1/x_{n-1}}&{-{\frac {H_{n-1}^{\circ }}{RT^{2}}}}\\{\frac {-1}{1-\sum \limits _{1=1}^{n-1}{x_{i}}}}&{\frac {-1}{1-\sum \limits _{1=1}^{n-1}{x_{i}}}}&{\frac {-1}{1-\sum \limits _{1=1}^{n-1}{x_{i}}}}&{\frac {-1}{1-\sum \limits _{1=1}^{n-1}{x_{i}}}}&{\frac {-1}{1-\sum \limits _{1=1}^{n-1}{x_{i}}}}&{-{\frac {H_{n}^{\circ }}{RT^{2}}}}\\\end{array}}\right]^{-1}.\left[{\begin{array}{*{20}c}{\ln x_{1}+{\frac {H_{1}^{\circ }}{RT}}-{\frac {H_{1}^{\circ }}{RT_{1}^{\circ }}}}\\{\ln x_{2}+{\frac {H_{2}^{\circ }}{RT}}-{\frac {H_{2}^{\circ }}{RT_{2}^{\circ }}}}\\{\ln x_{3}+{\frac {H_{3}^{\circ }}{RT}}-{\frac {H_{3}^{\circ }}{RT_{3}^{\circ }}}}\\\vdots \\{\ln x_{n-1}+{\frac {H_{n-1}^{\circ }}{RT}}-{\frac {H_{n-1}^{\circ }}{RT_{n-1i}^{\circ }}}}\\{\ln \left({1-\sum \limits _{i=1}^{n-1}{x_{i}}}\right)+{\frac {H_{n}^{\circ }}{RT}}-{\frac {H_{n}^{\circ }}{RT_{n}^{\circ }}}}\\\end{array}}\right]\end{array}}