ISITEKS: TETAPAN KESEIMBANGAN KIMIA DITURUNKAN DARI SIKLUS K

Siklus K disingkat menjadi sikla

$\large \bg_black K(I,n)=\oint Idn=\Delta I.\Delta n$

$\large \bg_black K(n,I)=\oint ndI=-\Delta n\Delta I$

$\large \bg_black K(F,n)=\oint Fdn=\iint(\frac{\partial F}{\partial I})dIdn$

$\large \bg_black K(F,G)=\oint FdG=\iint(\frac{\partial F}{\partial n})dndG$

DEFINISI

$\large \bg_black K(I,n)\equiv \sum_{lk\ sikla}I\ dn$

$\large \bg_black K(n,I)\equiv \sum_{lk\ sikla}\ n\ dI$

$\large \bg_black K(F,G(I))\equiv \sum_{lk\ sikla} F\ dG(I)$

$\large \bg_black K(F,n)\equiv \sum_{lk\ sikla}F\ dn$

DALIL
Pada siklus infinitesimal (takberhingga kecil)

$\large \bg_black a)\ \ K(I,n)=dIdn$

$\large \bg_black b)\ \ K(n,I)=-dndI$

$\large \bg_black c)\ \ K(F,G(I))=-\frac{\partial F}{dn}\ dndG(I)$

$\large \bg_black d)\ \ K(F,n)=\frac{\partial F}{\partial I}dIdn$

Dari

$\LARGE \bg_black aA\rightleftharpoons bB$

$\LARGE \bg_black \frac{dn{A}}{dn{B}}=-\frac{a}{b}$

sikla A
sikla B

$\bg_black p_{A}V=n_{A}RT$

$\bg_black p_{B}V=n_{B}RT$

$\bg_black K_{A}(p_{A},V)+K_{B}(p_{B},V)=0$

Jadikan

$\LARGE \bg_black \frac{p_{B}^{b}}{p_{A}^{a}}=tetap |$

Jawab :

$\large \bg_black K_{A}(p_{A},V)=-K_{A}(V,p_{A})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K(G,y)=\frac{\partial G}{dx}dxdy$

   $\LARGE \bg_black =-K_{A}(\frac{n_{A}RT}{p_{A}},p_{A})$

   $\large \bg_black =-.-\frac{\partial n_{A}RT/p_{A}}{dn_{A}}\ dn_{A}\ dp_{A}$

   $\large \bg_black \frac{RT}{p_{A}}\ \frac{\partial n_{A}}{dn_{A}}\ dn_{A}dp_{A}}$

$\large \bg_black =dn_{A}\ RT\ \frac{dp_{A}}{p_{A}}=dn_{A}\ RT\ dlnp_{A}$

$\large \bg_black K_{B}(p_{B},V)=-K_{B}(V,p_{B})$

   $\large \bg_black =-K_{B}(\frac{n_{B}RT}{p_{B}},p_{B})$

   $\large \bg_black =-.-\frac{\partial n_{B}RT}{dn_{B}}dn_{B}dp_{B}$

$\large \bg_black =\frac{RT}{p_{B}}\ \frac{\partial n_{B}}{dn_{B}}\ dn_{B}\ dp_{B}$

$\large \bg_black =dn_{B}\ RT\ \frac{dp_{B}}{p_{B}}=dn_{B}\ RT\ dlnp_{B}$

   $\large \bg_black K_{A}(p_{A},V)+K_{B}(p_{B},V)=0$

$\large \bg_black dn_{A}\ RT\ dlnp_{A}+dn_{B}\ RT\ dlnp_{B}=0$

   $\large \bg_black RT\ dn_{B}(\frac{dn_{A}}{dn_{B}}\ dlnp_{A}+dlnp_{B})=0$

$\large \bg_black -\frac{a}{b}\ dln\ p_{A}+dln\ p_{B}=0$

   $\large \bg_black - a\ dln\ p_{A}+b\ dln\ p_{B}=0$

   $\large \bg_black -d\ a\ lnp_{A}+d\ b\ lnp_{B}=0$

   $\large \bg_black -d\ ln\ p_{A}^{a}\ +\ d\ ln\ p_{B}^{b}=0$

ISITEKS

Kamis, 24 Januari 2013

TETAPAN KESEIMBANGAN KIMIA DITURUNKAN DARI SIKLUS K

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