Benutzer:Espresso robusta/Sandkasten

$\Theta ={K_{A} \over (1-P/P_{0})\cdot [1+P/P_{0}\cdot (K_{A}-1)]}\cdot {P \over P_{0}}$

$\Theta ={K_{A}\cdot P \over 1+K_{A}\cdot P}$

$\Theta ={K_{A}\cdot c \over 1+K_{A}\cdot c}$

$\Gamma =K_{A}\exp {\bigg [}{\frac {P}{K_{A}+1}}{\bigg ]}$

$\Gamma =K_{A}\cdot {\sqrt {P}}$

$\Gamma =K_{A}\cdot \ln(P-1)$

$\Gamma =K_{A}\cdot P^{q}\;\mathrm {mit} \;0\leq q\leq 1$

$\Gamma =K_{A}\cdot P^{q}\;\mathrm {mit} \;q\geq 1$

$\Theta ={\Gamma \over \Gamma _{mon}}={K_{L}\cdot c \over 1+K_{L}\cdot c}$

$\lim _{P\to P_{0}}\Theta =\infty$

$\Theta ={n \over n_{mon}}={C \over (1-P/P_{0})\cdot [1+P/P_{0}\cdot (C-1)]}\cdot {P \over P_{0}}$

$\Theta ={\Gamma \over \Gamma _{mon}}={K_{L}\cdot P \over 1+K_{L}\cdot P}$

$\Gamma =K_{F}\cdot P^{q}$

$\mathrm {M} _{x/n}^{n+}[\mathrm {(AlO_{2})} _{x}^{-}\mathrm {(SiO_{2})} _{y}]\cdot z\ \mathrm {H_{2}O}$

Zeolith A : Na₁₂[(AlO₂)₁₂(SiO₂)₁₂] · 27 H₂O = Zeolith A : ${\ce {Na12[(AlO2)12(SiO2)12]*27 H2O}}$

$\Delta _{R}G_{m}=\sum _{i=1}^{z}\nu _{i}\cdot \Delta _{B}G_{m,i}=\sum _{i=1}^{z}\nu _{i}\cdot \mu _{i}^{\circ }$

$n_{ges}=\sum _{i=1}^{z}n_{i}$

$x_{A}={{n_{A}} \over {n_{A}+n_{B}}}$

$x_{B}={{n_{B}} \over {n_{A}+n_{B}}}$

$x_{i}={{n_{i}} \over {n_{ges}}}$

$0\leq x_{i}\leq 1$

$\sum _{i=1}^{z}n_{i}\equiv 1$

$K_{eq}=\exp {\bigg [}-{\frac {\Delta _{R}F_{m}}{R\cdot T}}{\bigg ]}$

$\Delta _{R}F_{m}=-RT\cdot \mathrm {ln} \ K_{eq}$

$\Delta _{R}F_{m}=\sum _{i=1}^{z}\nu _{i}\cdot \Delta _{B}F_{m,i}$

$\Delta _{R}G_{m}=\sum _{i=1}^{z}\nu _{i}\cdot \Delta _{B}G_{m,i}$

$\mathrm {d} G=\sum _{i=1}^{z}\mu _{i}\cdot \mathrm {d} n_{i}$

$n_{i}(\xi )=n_{i,0}+\nu _{i}\cdot \xi$

$\mathrm {d} n_{i}=\nu _{i}\cdot \mathrm {d} \xi$

$\mathrm {d} G=\sum _{i=1}^{z}(\nu _{i}\cdot \mu _{i}\cdot \mathrm {d} \xi )={\Bigl (}\sum _{i=1}^{z}\nu _{i}\cdot \mu _{i}{\Bigr )}\cdot \mathrm {d} \xi$

$(\partial G/\partial \xi )_{p,T}=\sum _{i=1}^{z}\nu _{i}\cdot \mu _{i}$

$(\partial G/\partial \xi )_{p,T}=\sum _{i=1}^{z}\nu _{i}\cdot \mu _{i}^{\circ }+RT\cdot \sum _{i=1}^{z}\nu _{i}\cdot \mathrm {ln} \ a_{i}$

$\sum _{i=1}^{z}\nu _{i}\cdot \mu _{i}^{\circ }=\Delta _{R}G_{m}$

$RT\cdot \sum _{i=1}^{z}\nu _{i}\cdot \mathrm {ln} \ a_{i}$

$(\partial G/\partial \xi )_{p,T}=\Delta _{R}G_{m}+RT\cdot \sum _{i=1}^{z}\nu _{i}\cdot \mathrm {ln} \ a_{i}$

$(\partial G/\partial \xi )_{p,T}=0$

$\Delta _{R}G_{m}=-RT\cdot \sum _{i=1}^{z}\nu _{i}\cdot \mathrm {ln} \ a_{i}$

$\Delta _{R}G_{m}=-RT\cdot \sum _{i=1}^{z}\mathrm {ln} \ a_{i}^{\nu _{i}}=-RT\cdot \mathrm {ln} {\biggl (}\prod _{i=1}^{z}a_{i}^{\nu _{i}}{\biggr )}$

$K_{eq}=\prod _{i=1}^{z}a_{i}^{\nu _{i}}$

$\prod _{i=1}^{z}a_{i}^{\nu _{i}}$

$\Delta _{R}G_{m}=-RT\cdot \mathrm {ln} \ K_{eq}$

$K_{eq}=\exp {\bigg [}-{\frac {\Delta _{R}G_{m}}{R\cdot T}}{\bigg ]}$

$\mathrm {Dimension} (K_{c})=(N\cdot L^{-3})^{-6}=N^{-6}\cdot L^{18}$

$K_{c}={\frac {c(\mathrm {N} )^{\nu _{\mathrm {N} }}\cdot c(\mathrm {O} )^{\nu _{\mathrm {O} }}\cdot \ldots \cdot c(\mathrm {Z} )^{\nu _{\mathrm {Z} }}}{c(\mathrm {A} )^{|\nu _{\mathrm {A} }|}\cdot c(\mathrm {B} )^{|\nu _{\mathrm {B} }|}\cdot \ldots \cdot c(\mathrm {M} )^{|\nu _{\mathrm {M} }|}}}$

$\mathrm {Dimension} (K_{c})=(N_{D}/L_{D}^{3})^{\nu _{ges}}$

$K_{eq}={\frac {{a_{\text{N}}}^{\nu _{\text{N}}}\cdot {a_{\text{O}}}^{\nu _{\text{O}}}\cdot \ldots \cdot {a_{\text{Z}}^{\nu _{\mathrm {Z} }}}}{{a_{\text{A}}}^{|\nu _{\text{A}}|}\cdot {a_{\text{B}}}^{|\nu _{\text{B}}|}\cdot \ldots \cdot {a_{\text{M}}^{|\nu _{\mathrm {M} }|}}}}$

$a_{i}=f_{i}\cdot x_{i}$

$T(r)=B\cdot \mathrm {exp} (-L\cdot r^{2})$

$r=0\ \mathrm {m} \ \mathrm {und} \ T(r)=473.15\ \mathrm {K} \ \Rightarrow \ B=473.15\ \mathrm {K}$

$r=0.2\ \mathrm {m} \ \mathrm {und} \ T(r)=293.15\ \mathrm {K}$

$293.15\ \mathrm {K} =473.15\ \mathrm {K} \cdot \mathrm {exp} (-L\cdot 0.04\ \mathrm {m} ^{2})$

${\frac {293.15\ \mathrm {K} }{473.15\ \mathrm {K} }}=\mathrm {exp} (-L\cdot 0.04\ \mathrm {m} ^{2})$

$-L\cdot 0.04\ \mathrm {m} ^{2}=\mathrm {ln} {\frac {293.15\ \mathrm {K} }{473.15\ \mathrm {K} }}$

$L=-{\frac {1}{0.04\ \mathrm {m} ^{2}}}\cdot \mathrm {ln} {\frac {293.15\ \mathrm {K} }{473.15\ \mathrm {K} }}=11.97\ \mathrm {m^{-2}}$

$\operatorname {d} \!W={p_{E}}\cdot \operatorname {d} \!{V_{m}}$

$W=\textstyle \int _{V_{m,A}}^{V_{m,E}}\displaystyle \operatorname {d} \!{V_{m}}=p_{E}\cdot \Delta V_{m}$

$W=p_{E}\cdot \Delta V\cdot n$

${\operatorname {d} \!Q_{p} \over \operatorname {d} \!t}={\operatorname {d} \!T \over \operatorname {d} \!t}\cdot {\operatorname {d} \!Q_{p} \over \operatorname {d} \!T}$

$C_{p}={\operatorname {d} \!Q_{p} \over \operatorname {d} \!T}={\biggl (}{\partial H \over \partial T}{\biggr )}_{p}$

$\operatorname {d} Q_{p}=\operatorname {d} H=C_{p}\cdot \operatorname {d} T$

$\Delta H=\int _{T_{A}}^{T_{E}}C_{p}(T)\operatorname {d} T$

$\operatorname {d} S={\operatorname {d} Q_{p} \over T}={C_{p}(T)\cdot \operatorname {d} T \over T}$

$\Delta S=\int _{T_{A}}^{T_{E}}{C_{p}(T) \over T}\operatorname {d} T$

$\Delta S=C_{p}\cdot \int _{T_{A}}^{T_{E}}{1 \over T}\operatorname {d} T=C_{p}\cdot \operatorname {ln} {T_{E} \over T_{A}}$

$\mathrm {Steigung} ={\frac {\sum \nolimits _{i=1}^{n}[(T_{i}-{\overline {T}})(C_{p,s,i}-{\overline {C}}_{p,s})]}{\sum \nolimits _{i=1}^{n}(T_{i}-{\overline {T}})^{2}}}$

$\mathrm {Steigung} ={\frac {\sum \nolimits _{i=1}^{n}[(x_{i}-{\overline {x}})(y_{i}-{\overline {y}})]}{\sum \nolimits _{i=1}^{n}(x_{i}-{\overline {x}})^{2}}}$

$\mathrm {Achsenabschnitt} ={\overline {C}}_{p,s}-\mathrm {Steigung} \cdot {\overline {T}}$

$\mathrm {Achsenabschnitt} ={\overline {y}}-\mathrm {Steigung} \cdot {\overline {x}}$

$C_{p,m,basis}(T)={\frac {C_{p,m}(T_{E})-C_{p,m}(T_{onset})}{T_{E}-T_{onset}}}\cdot (T-T_{onset})+C_{p,m}(T_{onset})$

${\biggl (}{\partial \mu ^{*} \over \partial T}{\biggr )}_{n,p}={\biggl (}{\partial G_{M} \over \partial T}{\biggr )}_{n,p}=-S_{M}$

$\left({\frac {\partial G_{\mathrm {sys} }}{\partial \xi }}\right)_{T,p}=\sum _{i=1}^{z}(\nu _{i}\cdot \mu _{i})$

$\sum _{i=1}^{z}(\nu _{i}\cdot \mu _{i})=0\mathrm {\ J/mol}$

$r={\frac {1}{\nu _{\mathrm {A} }}}\cdot {\frac {\mathrm {d} [\mathrm {A} ]}{\mathrm {d} t}}=k_{G}\cdot [\mathrm {A} ]$

Das Zeitgesetz erster Ordnung wird erhalten, indem zunächst die Variablen getrennt werden:

${\frac {1}{\nu _{\mathrm {A} }}}\cdot {\frac {1}{[\mathrm {A} ]}}\cdot \mathrm {d} [\mathrm {A} ]=k_{G}\cdot {\mathrm {d} t}$

Im nächsten Schritt berechnet man die bestimmten Integrale:

${\frac {1}{\nu _{\mathrm {A} }}}\cdot \int _{[A]_{0}}^{[A](t)}{\frac {1}{[\mathrm {A} ]}}\cdot \mathrm {d} [\mathrm {A} ]=\int _{0}^{t}k_{G}\cdot {\mathrm {d} t}$

Man erhält zunächst unter Berücksichtigung der Tatsache, dass $\ln[A]$ die Stammfunktion zu $1/[A]$ ist:

${1 \over \nu _{A}}\cdot \{\ln {[A](t)}-\ln {[A]_{0}}\}=k_{G}\cdot t$

Anwendung der Logarithmusregeln ergibt:

${\frac {1}{\nu _{\mathrm {A} }}}\cdot \ln {\frac {[\mathrm {A} ](t)}{[\mathrm {A} ]_{0}}}=k_{G}\cdot t$

Umformen ergibt:

$\ln {\frac {[\mathrm {A} ](t)}{[\mathrm {A} ]_{0}}}=\nu _{\mathrm {A} }\cdot k_{G}\cdot t$

Entlogarithmieren ergibt das Zeitgesetz für Reaktionen erster Ordnung:

${\frac {[\mathrm {A} ](t)}{[\mathrm {A} ]_{0}}}=\mathrm {e} ^{\nu _{\mathrm {A} }\cdot k_{G}\cdot t}$

Beziehungsweise:

$[\mathrm {A} ](t)={[\mathrm {A} ]_{0}}\cdot \mathrm {e} ^{\nu _{\mathrm {A} }\cdot k_{G}\cdot t}$

Sofern, wie es bei Reaktionen erster Ordnung häufig der Fall ist, gilt

$\nu _{\mathrm {A} }=-1$,

wird das Zeitgesetz für Reaktionen erster Ordnung:

$[\mathrm {A} ](t)={[\mathrm {A} ]_{0}}\cdot \mathrm {e} ^{-k_{G}\cdot t}$

$[A](t_{1/2})=1/2\cdot [A]_{0}$

${1 \over {\nu _{A}}}\cdot \ln {\Biggl (}{1 \over 2}\cdot {[A]_{0} \over {[A]_{0}}}{\Biggr )}=k_{G}\cdot t_{1/2}$

${1 \over \nu _{A}}\cdot \ln {1 \over 2}=k_{G}\cdot t_{1/2}$

$t_{1/2}={1 \over k_{G}}\cdot {1 \over \nu _{A}}\cdot -\ln 2$

$t_{1/2}={1 \over k_{G}}\cdot \ln 2$

$K_{eq}={\frac {a{\bigl (}R_{1}COOR_{2}{\bigr )}\cdot a{\bigl (}H_{2}O{\bigr )}}{a{\bigl (}R_{1}COOH{\bigr )}\cdot a{\bigl (}R_{2}OH{\bigr )}}}$

$K_{eq}={a_{\text{A}}}^{\nu _{\text{A}}}\cdot {a_{\text{B}}}^{\nu _{\text{B}}}\cdot \ldots \cdot {{a_{\text{M}}}^{\nu _{\mathrm {M} }}\cdot a_{\text{N}}}^{\nu _{\text{N}}}\cdot {a_{\text{O}}}^{\nu _{\text{O}}}\cdot \ldots \cdot {{a_{\text{Z}}}^{\nu _{\mathrm {Z} }}}$

$K_{eq}=\prod _{i=1}^{z}a_{i}^{{\nu }_{i}}$