A) \[52.988\,\,kJ\]
B) \[42.024\,\,kJ\]
C) \[52.898\,\,kJ\]
D) \[51.898\,\,kJ\]
Correct Answer: C
Solution :
\[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] \[\frac{{{k}_{2}}}{{{k}_{1}}}=2,\,\,{{T}_{1}}=298\,\,K,\,\,{{T}_{2}}=308\,\,K,\] \[R=8.314J\,\,{{K}^{-1}}\,\,mo{{l}^{-1}}\] \[\log 2=\frac{{{E}_{a}}}{2.303\times 8.314}\times \left[ \frac{308-298}{308\times 298} \right]\] \[0.3010=\frac{{{E}_{a}}}{2.303\times 8.314}\times \frac{10}{298\times 308}\] \[\therefore \]\[{{E}_{a}}=\frac{0.3010\times 2.303\times 8.314\times 298\times 308}{10}\] \[=52.898\,\,kJ\]
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