A) \[{{K}_{p}}>K_{p}^{'}\]
B) \[{{K}_{p}}<K_{p}^{'}\]
C) \[{{K}_{p}}=K_{p}^{'}\]
D) \[{{K}_{p}}=\frac{1}{K_{p}^{'}}\]
Correct Answer: A
Solution :
The equilibrium constant at two different temperatures for a thermodynamic process is given by \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{\Delta {{H}^{{}^\circ }}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] Here, \[{{K}_{1}}\] and \[{{K}_{2}}\] are replaced by \[{{K}_{p}}\] and\[K{{'}_{p}}.\] Therefore, \[\log \frac{K{{'}_{p}}}{{{K}_{p}}}=\frac{\Delta H{}^\circ }{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] For exothermic reaction, \[{{T}_{2}}>{{T}_{1}}\] and \[\Delta H=-ve\] \[\therefore \] \[{{K}_{p}}>K{{'}_{p}}\]
You need to login to perform this action.
You will be redirected in
3 sec
