A) \[59.2\text{ }kJ\,mo{{l}^{-1}}\]
B) \[39.2\text{ kJ }mo{{l}^{-1}}\]
C) \[52.9\,kJ\,mo{{l}^{-1}}\]
D) \[29.5\,kJ\,mo{{l}^{-1}}\]
Correct Answer: C
Solution :
[c] Activation energy can be calculated from the equation. \[\frac{{{\operatorname{logK}}_{2}}}{{{\operatorname{logK}}_{1}}}=\frac{-{{E}_{a}}}{2.303R}\left( \frac{1}{{{T}_{2}}}-\frac{1}{{{T}_{1}}} \right)\] Given \[\frac{\log {{K}_{2}}}{\log {{K}_{1}}}=2\text{ }{{T}_{2}}=308\text{ }K;{{T}_{1}}=298K\] \[\therefore \log 2=\frac{-{{E}_{a}}}{2.303\times 8.314}\left( \frac{1}{308}-\frac{1}{298} \right)\] \[{{E}_{a}}=52.9\,kJmo{{l}^{-1}}\]
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