<p>The Te–O phase diagram and the TeO<sub>2</sub> thermodynamic properties are of interest for many industrial fields: nuclear applications, steel making industry and chalcogenide glass processes. Both, the thermodynamic properties and phase diagram of this relevant binary system were reviewed and assessed using the Calphad method. From this assessment, the standard Gibbs free energy and corresponding heat capacity of the binary oxides are calculated as:<Equation ID="Equa"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11669_2025_1175_Article_Equa.gif" Format="GIF" Height="51" Rendition="HTML" Resolution="72" Type="Linedraw" Width="514" /> </MediaObject> <EquationSource Format="TEX">\(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{TeO}}_{2} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 109.646 + 0.1034 \cdot T - 0.0064 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{p}}_{{{\text{TeO}}_{2} }} }} \left( {{\text{J}} \cdot {\text{K}} \cdot {\text{mol}}^{ - 1} } \right) = 19.59 + 0.0101 \cdot {\text{T}} - 1.6 \cdot 10 ^{- 6}.T^{2} - 186517 \cdot T^{ - 2} \hfill \\ \end{gathered}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi mathvariant="normal">Δ</mi> <mtext>f</mtext> </msub> <msubsup> <mtext>G</mtext> <mrow> <msub> <mtext>TeO</mtext> <mn>2</mn> </msub> </mrow> <mo>∘</mo> </msubsup> <mfenced close=")" open="("> <mrow> <mtext>KJ</mtext> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mo>-</mo> <mn>109.646</mn> <mo>+</mo> <mn>0.1034</mn> <mo>·</mo> <mi>T</mi> <mo>-</mo> <mn>0.0064</mn> <mo>·</mo> <mi>T</mi> <mo>·</mo> <mo>ln</mo> <mrow> <mo stretchy="false">(</mo> <mi>T</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <msub> <mtext>C</mtext> <msub> <mtext>p</mtext> <msub> <mtext>TeO</mtext> <mn>2</mn> </msub> </msub> </msub> <mfenced close=")" open="("> <mrow> <mtext>J</mtext> <mo>·</mo> <mtext>K</mtext> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mn>19.59</mn> <mo>+</mo> <mn>0.0101</mn> <mo>·</mo> <mtext>T</mtext> <mo>-</mo> <mn>1.6</mn> <mo>·</mo> <msup> <mn>10</mn> <mrow> <mo>-</mo> <mn>6</mn> </mrow> </msup> <mo>.</mo> <msup> <mi>T</mi> <mn>2</mn> </msup> <mo>-</mo> <mn>186517</mn> <mo>·</mo> <msup> <mi>T</mi> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow /> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation><Equation ID="Equb"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11669_2025_1175_Article_Equb.gif" Format="GIF" Height="51" Rendition="HTML" Resolution="72" Type="Linedraw" Width="467" /> </MediaObject> <EquationSource Format="TEX">\(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{Te}}_{2} {\text{O}}_{5} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 104.81 + 0.1175 \cdot T - 0.0077 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{p}}_{{{\text{Te}}_{2} {\text{O}}_{5} }} }} \left( {{\text{J}} \cdot {\text{K}}^{ - 1} \cdot {\text{mol}}^{ - 1} } \right) = 19.86 + 0.0091 \cdot T - 285181 \cdot T^{ - 2} \hfill \\ \end{gathered}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi mathvariant="normal">Δ</mi> <mtext>f</mtext> </msub> <msubsup> <mtext>G</mtext> <mrow> <mrow> <msub> <mtext>Te</mtext> <mn>2</mn> </msub> <msub> <mtext>O</mtext> <mn>5</mn> </msub> </mrow> </mrow> <mo>∘</mo> </msubsup> <mfenced close=")" open="("> <mrow> <mtext>KJ</mtext> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mo>-</mo> <mn>104.81</mn> <mo>+</mo> <mn>0.1175</mn> <mo>·</mo> <mi>T</mi> <mo>-</mo> <mn>0.0077</mn> <mo>·</mo> <mi>T</mi> <mo>·</mo> <mo>ln</mo> <mrow> <mo stretchy="false">(</mo> <mi>T</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <msub> <mtext>C</mtext> <msub> <mtext>p</mtext> <mrow> <msub> <mtext>Te</mtext> <mn>2</mn> </msub> <msub> <mtext>O</mtext> <mn>5</mn> </msub> </mrow> </msub> </msub> <mfenced close=")" open="("> <mrow> <mtext>J</mtext> <mo>·</mo> <msup> <mrow> <mtext>K</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mn>19.86</mn> <mo>+</mo> <mn>0.0091</mn> <mo>·</mo> <mi>T</mi> <mo>-</mo> <mn>285181</mn> <mo>·</mo> <msup> <mi>T</mi> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow /> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation><Equation ID="Equc"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="11669_2025_1175_Article_Equc.gif" Format="GIF" Height="48" Rendition="HTML" Resolution="72" Type="Linedraw" Width="558" /> </MediaObject> <EquationSource Format="TEX">\(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{TeO}}_{3} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 92.79 + 0.111 \cdot T - 0.00776 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{PTeO}}_{3} }} \left( {{\text{J}} \cdot {\text{K}}^{ - 1} \cdot {\text{mol}}^{ - 1} } \right) = 17.475 + 0.013 \cdot {\text{T}} - 2.085 \cdot 10^{ - 6} \cdot T^{2} - 280000 \cdot T^{ - 2} \hfill \\ \end{gathered}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi mathvariant="normal">Δ</mi> <mtext>f</mtext> </msub> <msubsup> <mtext>G</mtext> <mrow> <msub> <mtext>TeO</mtext> <mn>3</mn> </msub> </mrow> <mo>∘</mo> </msubsup> <mfenced close=")" open="("> <mrow> <mtext>KJ</mtext> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mo>-</mo> <mn>92.79</mn> <mo>+</mo> <mn>0.111</mn> <mo>·</mo> <mi>T</mi> <mo>-</mo> <mn>0.00776</mn> <mo>·</mo> <mi>T</mi> <mo>·</mo> <mo>ln</mo> <mrow> <mo stretchy="false">(</mo> <mi>T</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <msub> <mtext>C</mtext> <msub> <mtext>PTeO</mtext> <mn>3</mn> </msub> </msub> <mfenced close=")" open="("> <mrow> <mtext>J</mtext> <mo>·</mo> <msup> <mrow> <mtext>K</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>·</mo> <msup> <mrow> <mtext>mol</mtext> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </mfenced> <mo>=</mo> <mn>17.475</mn> <mo>+</mo> <mn>0.013</mn> <mo>·</mo> <mtext>T</mtext> <mo>-</mo> <mn>2.085</mn> <mo>·</mo> <msup> <mn>10</mn> <mrow> <mo>-</mo> <mn>6</mn> </mrow> </msup> <mo>·</mo> <msup> <mi>T</mi> <mn>2</mn> </msup> <mo>-</mo> <mn>280000</mn> <mo>·</mo> <msup> <mi>T</mi> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow /> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation></p>

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Tellurium Oxides: Thermodynamics and Phase Relations in the Te–O System

  • S. Gossé

摘要

The Te–O phase diagram and the TeO2 thermodynamic properties are of interest for many industrial fields: nuclear applications, steel making industry and chalcogenide glass processes. Both, the thermodynamic properties and phase diagram of this relevant binary system were reviewed and assessed using the Calphad method. From this assessment, the standard Gibbs free energy and corresponding heat capacity of the binary oxides are calculated as: \(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{TeO}}_{2} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 109.646 + 0.1034 \cdot T - 0.0064 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{p}}_{{{\text{TeO}}_{2} }} }} \left( {{\text{J}} \cdot {\text{K}} \cdot {\text{mol}}^{ - 1} } \right) = 19.59 + 0.0101 \cdot {\text{T}} - 1.6 \cdot 10 ^{- 6}.T^{2} - 186517 \cdot T^{ - 2} \hfill \\ \end{gathered}\) Δ f G TeO 2 KJ · mol - 1 = - 109.646 + 0.1034 · T - 0.0064 · T · ln ( T ) C p TeO 2 J · K · mol - 1 = 19.59 + 0.0101 · T - 1.6 · 10 - 6 . T 2 - 186517 · T - 2 \(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{Te}}_{2} {\text{O}}_{5} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 104.81 + 0.1175 \cdot T - 0.0077 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{p}}_{{{\text{Te}}_{2} {\text{O}}_{5} }} }} \left( {{\text{J}} \cdot {\text{K}}^{ - 1} \cdot {\text{mol}}^{ - 1} } \right) = 19.86 + 0.0091 \cdot T - 285181 \cdot T^{ - 2} \hfill \\ \end{gathered}\) Δ f G Te 2 O 5 KJ · mol - 1 = - 104.81 + 0.1175 · T - 0.0077 · T · ln ( T ) C p Te 2 O 5 J · K - 1 · mol - 1 = 19.86 + 0.0091 · T - 285181 · T - 2 \(\begin{gathered} \Delta_{{\text{f}}} {\text{G}}_{{{\text{TeO}}_{3} }}^{ \circ } \left( {{\text{KJ}} \cdot {\text{mol}}^{ - 1} } \right) = - 92.79 + 0.111 \cdot T - 0.00776 \cdot T \cdot \ln (T) \hfill \\ {\text{C}}_{{{\text{PTeO}}_{3} }} \left( {{\text{J}} \cdot {\text{K}}^{ - 1} \cdot {\text{mol}}^{ - 1} } \right) = 17.475 + 0.013 \cdot {\text{T}} - 2.085 \cdot 10^{ - 6} \cdot T^{2} - 280000 \cdot T^{ - 2} \hfill \\ \end{gathered}\) Δ f G TeO 3 KJ · mol - 1 = - 92.79 + 0.111 · T - 0.00776 · T · ln ( T ) C PTeO 3 J · K - 1 · mol - 1 = 17.475 + 0.013 · T - 2.085 · 10 - 6 · T 2 - 280000 · T - 2