A)
B) \[{{K}_{p}}<K_{p}^{'}\]
C) \[{{K}_{p}}=K_{p}^{'}\]
D) \[{{K}_{p}}=\frac{1}{K_{p}^{'}}\]
Correct Answer: A
Solution :
\[{{K}_{p}}>K_{p}^{'}\]| 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 \[\Delta H=-ve\] |
| \[\therefore \] \[{{K}_{p}}>K{{'}_{p}}\] |
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