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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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We know that equilibrium constant, K changes with temperature. At 300K, equilibrium constant is 25 and at 400K it is 10. Hence, backward reaction will have energy of activation:

(a) equal to that of forward reaction

(b) less than that of forward reaction

(c) greater than that of forward reaction

(d) given values are not sufficient to explain given statement

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • Van't Hoff equation: $log\frac{K_2}{K_1} = \frac{\Delta H}{2.303 \times R} [\frac{T_2 - T_1}{T_1 T_2}]$
  • When $\Delta H = -ve$, then the reaction is exothermic
  • When $\Delta H = +ve$, then the reaction is endothermic$
 
Answer: greater than that of forward reaction
 
According to the Van't Hoff equation,
$log\frac{K_2}{K_1} = \frac{\Delta H}{2.303 \times R} [\frac{T_2 - T_1}{T_1 T_2}]$
 
By putting the values provided,
$\Delta H = \frac{2.303 \times R T_1 T_2}{(T_2-T_1)} log \frac{K_2}{K_1}$
$= \frac{2.303 \times 8.314 \times 300 \times 400}{400-300} log \frac{10}{25}$
$= \frac{2.303 \times 8.314 \times 300 \times 400}{100} log \frac{10}{25}$ < 0
 
As $\Delta H$ is -ve, the reaction is exothermic
 
Hence, backward reaction will have energy of activation greater than that of forward reaction.
 

 

answered Nov 20, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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