The standard molar Gibbs energy of formation of $YbPt_3$ and $LuPt_3$ intermetallics

The standard molar Gibbs energies of formation of $YbPt_3$ and $LuPt_3$ intermetallic compounds have been measured in the temperature range 880 K to 1100 K using the solid-state cells: $Ta\mid\hspace{2mm}Yb+YbF_2\mid\hspace{2mm}CaF_2\mid\hspace{2mm}YbPt_3Pt+YbF_2\mid\hspace{2mm}Ta,$ and $Ta\mid\hspace{2mm}Lu+LuF_3\mid\hspace{2mm}CaF_2\mid\hspace{2mm}LuPt_3+Pt+LuF_3\mid\hspace{2mm}Ta$, The trifluoride of Yb is not stable in equilibrium with Yb or $YbPt_3$. The results can be expressed by the equations: $\Delta_fG_m\hspace{1mm}^°(YbPt_3)/(J·mol^{-1})=-322100+0.39(T/K)\pm400$, $\Delta_fG_m \hspace{1mm}^°(LuPt_3)/(J·mol^{-1})=-366800+3.82(T/K)\pm 400$. The standard molar Gibbs energy of formation of $LuPt_3 is -41.1 kJ – mol^{-1}$more negative than that for $YbPt_3$ at 1000 K. Ytterbium is divalent in the pure metal and trivalent in the intermetallic $YbPt_3$. The energy required for the promotion of divalent Yb to the trivalent state is responsible for the less negative $\Delta_f\hspace{2mm}G_m\hspace{1mm} ^o $ of $YbPt_3$. The enthalpies of formation of the two intermetallics are in reasonable agreement with Miedema's model. Because of the extraordinary stability of these compounds it is possible to reduce oxides of Yb and Lu with hydrogen in the presence of platinum at $T/K>1473$. The equilibrium chemical potential of oxygen corresponding to the reduction of $Yb_2O_3$ and $Lu_2O_3$ by hydrogen in the presence of platinum is presented in the form of an Ellingham diagram