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Chemische Reaktionen
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====Massenwirkungsgesetz==== <u>Voraussetzung: ideales System (verdünnte Lösung)</u> Gleichgewicht: :<math>{{A}_{\rho }}:=-\sum\limits_{i}^{{}}{{}}{{\tilde{\mu }}_{i}}{{v}_{i\rho }}=0</math> Mit :<math>\begin{align} & {{{\tilde{\mu }}}_{i}}(T,p,{{x}_{i}})={{\Phi }_{i}}(T)+RT\ln p+RT\ln {{x}_{i}} \\ & \Rightarrow {{\Phi }_{i}}(T)+RT\ln p={{g}_{i}}(T,p) \\ & \Rightarrow {{{\tilde{\mu }}}_{i}}(T,p,{{x}_{i}})={{g}_{i}}(T,p)+RT\ln {{x}_{i}} \\ \end{align}</math> (Seite 118): :<math>\begin{align} & \sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}\left( {{\Phi }_{i}}(T)+RT\ln p+RT\ln {{x}_{i}} \right)=0 \\ & \Rightarrow \sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}\ln {{x}_{i}}=-\sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}\ln p-\sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}\frac{{{\Phi }_{i}}(T)}{RT} \\ \end{align}</math> Also: :<math>\prod\limits_{i}^{{}}{{}}{{x}_{i}}^{\left( {{v}_{i\rho }} \right)}={{p}^{-\left( \sum\limits_{i}^{{}}{{}}{{v}_{i\rho }} \right)}}\cdot K(T)</math> (Massenwirkungsgesetz) mit der Gleichgewichtskonstanten K(T)= <math>\exp \left( -\sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}\frac{{{\Phi }_{i}}(T)}{RT} \right)</math> mit <math>{{\Phi }_{i}}(T)+RT\ln p={{g}_{i}}(T,p)</math> erhält man: :<math>\frac{d}{dT}\left( \frac{{{\Phi }_{i}}(T)}{T} \right)=\frac{\partial }{\partial T}{{\left( \frac{{{g}_{i}}(T,p))}{T} \right)}_{p}}=\frac{1}{T}{{\left( \frac{\partial {{g}_{i}}(T,p))}{\partial T} \right)}_{p}}-\frac{{{g}_{i}}}{{{T}^{2}}}</math> Es gilt: :<math>\begin{align} & {{\left( \frac{\partial {{g}_{i}}(T,p))}{\partial T} \right)}_{p}}=-{{s}_{i}} \\ & {{h}_{i}}={{g}_{i}}+T{{s}_{i}} \\ & \Rightarrow \frac{d}{dT}\left( \frac{{{\Phi }_{i}}(T)}{T} \right)=-\frac{{{h}_{i}}(T)}{{{T}^{2}}} \\ \end{align}</math> Also: :<math>\frac{d}{dT}\left( \ln K(T) \right)=\frac{1}{R{{T}^{2}}}\sum\limits_{i}^{{}}{{}}{{v}_{i\rho }}{{h}_{i}}(T)=\frac{{{Q}_{p}}^{\left( \rho \right)}}{R{{T}^{2}}}</math> Im Normalbereich: :<math>\begin{align} & {{c}_{pi}}(T)=const \\ & \Rightarrow {{h}_{i}}(T)=\int_{{{T}_{o}}}^{T}{{}}{{c}_{pi}}(T\acute{\ })dT\acute{\ } \\ \end{align}</math> ist linear in T Also: :<math>{{Q}_{p}}^{\left( \rho \right)}={{Q}_{0}}^{\left( \rho \right)}+{{Q}_{1}}^{\left( \rho \right)}\cdot T</math> :<math>\begin{align} & \frac{d}{dT}\left( \ln K(T) \right)=\frac{{{Q}_{p}}^{\left( \rho \right)}}{R{{T}^{2}}}=\frac{{{Q}_{0}}^{\left( \rho \right)}}{R{{T}^{2}}}+\frac{{{Q}_{1}}^{\left( \rho \right)}}{RT} \\ & \Rightarrow \ln \frac{K(T)}{{{K}_{0}}}=-\frac{{{Q}_{0}}^{\left( \rho \right)}}{RT}+\frac{{{Q}_{1}}^{\left( \rho \right)}}{R}\ln T \\ & K(T)={{K}_{0}}{{T}^{\frac{{{Q}_{1}}^{\left( \rho \right)}}{R}}}\exp \left( -\frac{{{Q}_{0}}^{\left( \rho \right)}}{RT} \right) \\ \end{align}</math> mit der dominanten Temperaturabhängigkeit im Exponenten <u>'''Beispiel: Haber- Bosch- Verfahren:'''</u> :<math>3{{H}_{2}}+{{N}_{2}}\begin{matrix} \to \\ \leftarrow \\ \end{matrix}2N{{H}_{3}}</math> :<math>\begin{align} & {{\nu }_{i}}:-3\quad -1\quad \quad +2 \\ & {{Q}_{0}}<0 \\ & \frac{{{x}_{3}}^{2}}{{{x}_{1}}^{3}{{x}_{2}}}={{p}^{2}}K(T)\tilde{\ }\exp \left( \frac{\left| {{Q}_{0}} \right|}{RT} \right) \\ \end{align}</math> x3 entspricht der Ammoniakausbeute der Reaktion. * x3 soll möglichst groß werden! :<math>\frac{{{x}_{3}}^{2}}{{{x}_{1}}^{3}{{x}_{2}}}={{p}^{2}}K(T)\tilde{\ }\exp \left( \frac{\left| {{Q}_{0}} \right|}{RT} \right)</math> * wähle Druck möglichst groß, Temperatur möglichst niedrig. * Problem: niedrige Temperaturen → Reaktion langsam! '''Betrachte nur eine Reaktion '''<math>\rho </math> : Mit <math>A:=-\sum\limits_{i}^{{}}{{}}{{\tilde{\mu }}_{i}}{{v}_{i}}</math> folgt (vergl. S. 122) :<math>\exp \left( \frac{A}{RT} \right)={{e}^{-\sum\limits_{i}^{{}}{{}}{{v}_{i}}\frac{{{\Phi }_{i}}}{RT}-\sum\limits_{i}^{{}}{{}}{{v}_{i}}\ln \left( \rho {{x}_{i}} \right)}}</math> Also: :<math>\exp \left( \frac{A}{RT} \right)=K(T)\cdot {{p}^{-\sum\limits_{i}^{{}}{{}}{{v}_{i}}}}\prod\limits_{i}^{{}}{{}}{{x}_{i}}^{-{{\nu }_{i}}}</math> → hier: Arrhenius- Plot Gleichgewicht: A=0 A>0 → spontane Vorwärtsreaktion A<0 → spontane Rückwärtsreaktion! Nach dem Prinzip von <u>'''Le Chatelier - Braun'''</u> T<To erniedrigt → A>0 * Vorwärtsreaktion! → Wärmeproduktion → T steigt!
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