A) \[51.8\,kJ\,mo{{l}^{-1}}\]
B) \[82.1\,kJ\,mo{{l}^{-1}}\]
C) \[23.8\,kJ\,mo{{l}^{-1}}\]
D) \[62.1\,kJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
According to Arrhenius equation log\[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[\frac{{{k}_{2}}}{{{k}_{1}}}=2;\] \[{{T}_{1}}=295\,K;{{T}_{2}}=305\,K\] \[R=8.314\,\,J{{K}^{-1}}mo{{l}^{-1}}\] \[\therefore \] \[\log 2=\frac{{{E}_{a}}}{2.303\times (8.314\,\,J{{K}^{-1}}\,mo{{l}^{-1}})}\] \[\left[ \frac{1}{295K}-\frac{1}{305K} \right]\] Or \[0.3010=\frac{{{E}_{a}}}{2.303\times (8.314\,J\,mo{{l}^{-1}})}\times \frac{(10)}{295\times 305}\] \[{{E}_{a}}=\frac{0.3010\times 2.303\times 8.314\times 295\times 305}{10}\] \[=51855\,J\,\,mo{{l}^{-1}}=51.855\,kJ\,mo{{l}^{-1}}\]
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