logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

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

1 Answer

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
 

Related questions

...