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%\titlehead{}
\title{Formelsammlung Chemie}
\author{Joachim Jakob, Kronberg-Gymnasium Aschaffenburg}
\date{\href{http://chemie-lernprogramme.de/daten/programme/js/formelsammlung/}{chemie-lernprogramme.de/daten/programme/js/formelsammlung/}}

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\begin{document}

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\maketitle
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\tableofcontents
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\newpage

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%%                         1. Avogadro-Konstante N_{A}                        %%
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\section{Avogadro Konstante $N_{A}$}

\begin{minipage}[t]{0.4\textwidth}

Die Avogadro-Konstante $N_{A}$ ist der Quotient aus der Teilchenzahl $N$ der Teilchen X und der Stoffmenge $n$ der Teilchen X. Ihre Einheit ist $mol^{-1}$. Sie gibt an, wie viele Teilchen in der Stoffmenge von einem MOL Teilchen enthalten sind. Sie ist unabhängig von der Art der Teilchen. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
           N_{A} & = & \dfrac{ N(X) }{ n(X) } \\
           N_{A} & = & 6,022 \cdot 10^{23} \dfrac{ 1 }{ mol } \\
\Rightarrow n(X) & = & \dfrac{ N(X) }{ N_{A} }
\end{eqnarray*}

\end{minipage}

\hline

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%%                           2. Molare Masse M                                %%
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\section{Molare Masse $M$}

\begin{minipage}[t]{0.4\textwidth}

Die Molare Masse $M$ der Teilchen X ist der Quotient aus der Masse $m$ der Teilchen X und der Stoffmenge $n$ der Teilchen X. Ihre Einheit ist $g/mol$. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
            M(X) & = & \dfrac{ m(X) }{ n(X) } \\
\Rightarrow n(X) & = & \dfrac{ m(X) }{ M(X) } 
\end{eqnarray*}

\end{minipage}

\hline

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%%                           3. Molares Volumen V_{M}                         %%
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\section{Molares Volumen $V_m$}

\begin{minipage}[t]{0.4\textwidth}

Das Molare Volumen $V_{m}$ eines (idealen)  Gases ist der Quotient aus dem Volumen $V$ der Gasteilchen X und der Stoffmenge $n$ der Gasteilchen X. Seine Einheit ist $l/mol$. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
           V_{m} & = & \dfrac{ V(X) }{ n(X) } \\
           V_{m} & = & 22,41 \dfrac{l}{mol} \\
\Rightarrow n(X) & = & \dfrac{ V(X) }{ V_{m} }
\end{eqnarray*}

\end{minipage}

\hline

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%%                     4. Stoffmengenkonzentration c                          %%
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\section{Stoffmengenkonzentration $c$}

\begin{minipage}[t]{0.4\textwidth}

Die Stoffmengenkonzentration $c$ ist der Quotient aus
der Stoffmenge $n$ der gelösten Teilchen X 
und dem Volumen der Lösung $V_{Lösung}$.
Ihre Einheit ist $mol/l$. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
            c(X) & = & \dfrac{ n(X) }{ V_{Lösung} } \\
\Rightarrow n(X) & = & c(X) \cdot V_{Lösung}
\end{eqnarray*}

\end{minipage}

\hline

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%%              5. Mittlere Reaktionsgeschwindigkeit v                        %%
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\section{Mittlere Reaktionsgeschwindigkeit $\overline{v}$}

\begin{minipage}[t]{0.4\textwidth}

Die Mittlere Reaktionsgeschwindigkeit $\overline{v}$ ist der Quotient aus
der Konzentrationsänderung der Produkte $\Delta c(Prod)$ und der Zeitänderung $\Delta t$. Sie beschreibt sowohl die Bildungsgeschwindigkeit der Produkte als auch die Zerfallsgeschwindigkeit der Edukte (mit umgekehrtem Vorzeichen). \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\lefteqn{ Ed \rightleftharpoons Prod } \\
\\
\overline{v} & = & + \dfrac{ \Delta c(Prod) }{ \Delta t } \\ 
             & = & - \dfrac{ \Delta c(Ed) }{ \Delta t } 
\end{eqnarray*} 

\end{minipage}

\hline

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%%                 6. Massenwirkungsgesetzt (MWG)                             %%
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\section{Massenwirkungsgesetz ($MWG$)}

\begin{minipage}[t]{0.4\textwidth}

Die Gleichgewichtskonstante $K_{C}$ für Reaktionen in Lösungen ist der Quotient aus dem Produkt der Konzentration $c$ der Produkte und dem der Edukte. 
Die Gleichgewichtskonstante $K_{P}$ für Gasreaktionen wird entsprechend aus den Partialdrücken $p$ der Produkte und Edukte ermittelt. \\
\\
Feststoffe werden in beiden Fällen gleich 1 (ohne Einheit) gesetzt.
Die Koeffizienten gehen als Exponenten ein.  \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\lefteqn{a \, A + b\, B \rightleftharpoons c\, C + d\, D} \\
\\
K_{C} & = & \dfrac{ c(C)^{c} \cdot c(D)^{d} }{ c(A)^{a} \cdot c(B)^{b} }  \\
      & = & \dfrac{ k_{hin} }{ k_{rück}} \\
K_{P} & = & \dfrac{ p(C)^{c} \cdot p(D)^{d} }{ p(A)^{a} \cdot p(B)^{b} }
\end{eqnarray*} 

\end{minipage}

\hline

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%%                    7. Gibbs-Helmholtz-Gleichung                            %%
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\section{Gibbs-Helmholtz-Gleichung}

\begin{minipage}[t]{0.4\textwidth}

Die Änderung der freien Enthalpie $\Delta G$ ist die Differenz aus der
Enthalpieänderung $\Delta H$ und dem Produkt aus der
absoluten Temperatur $T$ (in der Einheit Kelvin $K$) und der 
Entropieänderung $\Delta S$. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\Delta G & = & \Delta H - T \cdot \Delta S
\end{eqnarray*} 

\end{minipage}

\hline

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%%                     8. Ionenprodukt des Wasser K_{W}                       %%
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\section{Ionenprodukt des Wassers $K_{W}$}

\begin{minipage}[t]{0.4\textwidth}

Das Ionenprodukt des Wassers $K_{W}$
ist das Produkt der Konzentrationen der Oxoniumionen und der Hydroxidionen
in der Einheit $mol^{2}/l^{2}$.
Der $pK_{W}$-Wert ist der negative dekadische Logarithmus des Ionenprodukts 
$K_{W}$. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
 K_{W} & = & c(H_{3}O^{+}) \cdot c(OH^{-}) = 10^{-14} mol^{2}/l^{2} \\
pK_{W} & = & -lg \{K_{W}\} = 14 \\
 K_{W} & = & K_{S} \cdot K_{B} \\
pK_{W} & = & pK_{S} + pK_{B}
\end{eqnarray*} 

\end{minipage}

\hline

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%%                 9. Säurekonstante K_{S} und pK_{S}-Wert                    %%
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\section{Säurekonstante $K_{S}$ und $pK_{S}$-Wert}

\begin{minipage}[t]{0.4\textwidth}

Je größer die Säurekonstante
$K_{S}$, desto stärker ist die Säure.
Der $pK_{S}$-Wert ist der negative dekadische Logarithmus der Säurekonstante
$K_{S}$ in der Einheit $mol \cdot l^{-1}$.
Je kleiner die Säurekonstante
$pK_{S}$, desto stärker ist die Säure. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\lefteqn{H_{2}O + HA \rightleftharpoons H_{3}O^{+} + A^{-}} \\
\\
 K_{S} & = & \dfrac{c(H_{3}O^{+}) \cdot c(A^{-})} {c(HA) } \\
pK_{S} & = & -lg \{K_{S}\}
\end{eqnarray*} 

\end{minipage}

\hline

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%%             10. Basenkonstante K_{B} und pK_{B}-Wert                       %%
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\section{Basenkonstante $K_{B}$ und $pK_{B}$-Wert}

\begin{minipage}[t]{0.4\textwidth}

Je größer die Basenkonstante
$K_{B}$, desto stärker ist die Base.
Der $pK_{B}$-Wert
ist der negative dekadische Logarithmus der Basenkonstante
$K_{B}$ in der Einheit $mol \cdot l^{-1}$.
Je kleiner die Basenkonstante
$pK_{B}$, desto stärker ist die Base. \\

\end{minipage}
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\vspace{3mm}\vline
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\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\lefteqn{B + H_{2}O \rightleftharpoons BH^{+} + OH-} \\
\\
 K_{S} & = & \dfrac{c(BH^{+}) \cdot c(OH^{-})} {c(B) } \\
pK_{B} & = & -lg \{K_{B}\}
\end{eqnarray*} 

\end{minipage}

\hline

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%%                               11. pH-Wert                                  %%
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\section{$pH$-Wert}

\begin{minipage}[t]{0.4\textwidth}

Der $pH$-Wert ist der negative dekadische Logarithmus
des Zahlenwerts der Oxoniumionenkonzentration.
Je kleiner der
$pH$-Wert, desto saurer ist die Lösung. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth} 
  
\begin{eqnarray*}
pH & = & -lg \{c(H_{3}O^{+})\}
\end{eqnarray*} 

\end{minipage}

\hline

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%%                               12. pOH-Wert                                 %%
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\section{$pOH$-Wert}

\begin{minipage}[t]{0.4\textwidth}

Der $pOH$-Wert ist der negative dekadische Logarithmus
des Zahlenwerts der Hydroxidionenkonzentration.
Je niedriger der
$pOH$-Wert, desto alkalischer ist die Lösung. \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth} 

\begin{eqnarray*}
   pOH & = & -lg \{c(OH^{-})\} \\
pK_{W} & = & pH + pOH = 14
\end{eqnarray*} 

\end{minipage}

\hline

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%%                      13. pH-Wert für starke Säuren                         %%
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\section{$pH$-Wert für starke Säuren}

\begin{minipage}[t]{0.4\textwidth}

Die Näherungsformel für starke Säuren ergibt sich aufgrund der vollständigen Dissoziation wie folgt aus der Ausgangskonzentration dieser Säure $c_{0(HA)}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth} 

\begin{eqnarray*}
c(H_{3}O^{+}) & = & c_{0}(HA)  \\
           pH & = & -lg\{c_{0}(HA) \} \\
\end{eqnarray*} 

\end{minipage}

\hline

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%%                     14. pH-Wert für schwache Säuren                        %%
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\section{$pH$-Wert für schwache Säuren}

\begin{minipage}[t]{0.4\textwidth}

Die Näherungsformel für schwache Säuren ergibt sich wie folgt aus der
Ausgangskonzentration dieser Säure $c_{0(HA)}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
pH & = & \dfrac{1}{2} \cdot (pK_{S} -lg\{c_{0}(HA) \}) \\
\end{eqnarray*} 

\end{minipage}

\hline

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%%                       15. pH-Wert für starke Basen                         %%
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\section{$pH$-Wert für starke Basen}

\begin{minipage}[t]{0.4\textwidth}

Die Näherungsformel für starke Basen ergibt sich aufgrund der vollständigen Dissoziation wie folgt aus der Ausgangskonzentration dieser Base $c_{0(B)}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
    c(OH^{-}) & = & c_{0}(B)  \\
c(H_{3}O^{+}) & = & pK_{W} - pOH \\
           pH & = & 14 -(-lg\{ c_{0}(B) \})
\end{eqnarray*} 

\end{minipage}

\hline

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%%                       16. pH-Wert für schwache Basen                       %%
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\section{$pH$-Wert für schwache Basen}

\begin{minipage}[t]{0.4\textwidth}

Die Näherungsformel für schwache Basen ergibt sich wie folgt aus der
Ausgangskonzentration dieser Base $c_{0(B)}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
c(H_{3}O^{+}) & = & pK_{W} - pOH \\
           pH & = & 14 - \left( \dfrac{1}{2} \cdot (pK_{B} -lg\{c_{0}(B) \}) \right)
\end{eqnarray*} 

\end{minipage}

\hline

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%%              17. Henderson-Hasselbalch-Gleichung                           %%
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\section{Henderson-Hasselbalch-Gleichung}
  
\begin{minipage}[t]{0.4\textwidth}

Der pH-Wert für Pufferlösungen einer schwachen Säure
HA und ihrer korrespondierenden Base
$A^{-}$ ergibt sich wie folgt aus den
Ausgangskonzentrationen $c_{0}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
pH & = & pK_{S} + lg \left\lbrace \dfrac{ c_{0}(A⁻) }{ c_{0}(HA) } \right\rbrace \\
%\mbox{ist gleichbedeutend mit:} \\
pH & = & pK_{S} - lg \left\lbrace \dfrac{ c_{0}(HA) }{ c_{0}(A⁻) } \right\rbrace
\end{eqnarray*} 

\end{minipage}

\hline

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%%                         18. Nernstsche Gleichung                           %%
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\section{Nernstsche Gleichung}

\begin{minipage}[t]{0.4\textwidth}

Das Redoxpotenzial Normalpotenzial $E$ für ein korrespondierendes Redoxpaar (Halbzelle/galvanisches Element: Redm/Oxm) lautet bei 25$^{\circ}C$ ausgehend vom Normalpotenzial $E^{0}$: \\

\end{minipage}
\hfill
\vspace{3mm}\vline
\hfill
\begin{minipage}[t]{0.5\textwidth}

\begin{eqnarray*}
\lefteqn{Ox.: a \, Redm \rightleftharpoons b \, Oxm^{n+} + n \, e^{-}} \\
\\
E & = & E^{0} + \dfrac{0,059 V}{n} \cdot lg \left\lbrace \dfrac{ c(Oxm)^{b} }{ c(Redm)^{a} }  \right\rbrace

\end{eqnarray*} 

\end{minipage}

\hline

\end{document}
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