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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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what would be the expression in place of ? in the Van't Hoff equation: $ log \frac{K_2}{K_1} = \frac {\Delta H^\circ}{2.303R} [?]$

(a) $ \frac {1}{T_1} + \frac {1}{T_2} $

(b) $ \frac {1}{T_1} - \frac {1}{T_2} $

(c) $ \frac {1}{T_2} - \frac {1}{T_1} $

(d) $ \frac {1}{T_1 T_2} $

Can you answer this question?
 
 

1 Answer

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  • The Vant Hoff equation is: $ \frac{d ln K_p}{dT} = \frac {- \Delta H^\circ}{T^2} $
 
Answer: $ log \frac{K_2}{K_1} = \frac {\Delta H}{2.303R} [ \frac {T_2 - T_1}{T_1 T_2}] $
 
The Vant Hoff equation is:
$ \frac{d ln K_p}{dT} = \frac {- \Delta H^\circ}{T^2} $
 
Integrating the Vant Hoff equation we get,
 
$\Rightarrow \int \mathrm{d} ln K_p = \frac{\Delta H^\circ}{R} \int \frac {\mathrm{d} T}{T^2} T $
$\Rightarrow ln K_p = - \frac{\Delta H^\circ}{R} \frac {1}{T} + 1 $
 
If equilibrium constants at $ T_1 $ and $ T_2 $ are $ K_1 $ and $ K_2 $ respectively,
 
$ log \frac{K_2}{K_1} = \frac {\Delta H}{2.303R} [ \frac {T_2 - T_1}{T_1 T_2}] $

 

answered Nov 15, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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