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Display information for equation id:math.2530.51 on revision:2530

* Page found: Chemische Reaktionen (eq math.2530.51)

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Chemische Reaktionen Eq: math.2532.51

Hash: a1b4e8a75f87d9bbd29de95ad86abd24

TeX (original user input):

\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}

TeX (checked):

{\begin{aligned}&{\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{aligned}}

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\begin{aligned} \frac{d}{d T} (\ln K (T)) = \frac{{Q_{p}}^{(ρ)}}{R T^{2}} = \frac{{Q_{0}}^{(ρ)}}{R T^{2}} + \frac{{Q_{1}}^{(ρ)}}{R T} \\ \Rightarrow \ln \frac{K (T)}{K_{0}} = - \frac{{Q_{0}}^{(ρ)}}{R T} + \frac{{Q_{1}}^{(ρ)}}{R} \ln T \\ K (T) = K_{0} T^{\frac{{Q_{1}}^{(ρ)}}{R}} \exp (- \frac{{Q_{0}}^{(ρ)}}{R T}) \end{aligned}

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Identifiers

$d$

$d$

$T$

$K$

$T$

$Q_{p}$

$ρ$

$R$

$T$

$Q_{0}$

$ρ$

$R$

$T$

$Q_{1}$

$ρ$

$R$

$T$

$K$

$T$

$K_{0}$

$Q_{0}$

$ρ$

$R$

$T$

$Q_{1}$

$ρ$

$R$

$T$

$K$

$T$

$K_{0}$

$T$

$Q_{1}$

$ρ$

$R$

$Q_{0}$

$ρ$

$R$

$T$

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