A) \[43.84\text{ }kJmo{{l}^{-1}}\]
B) \[34.84\text{ }kJmo{{l}^{-1}}\]
C) \[84.00\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[30.00\text{ }kJmo{{l}^{-1}}\]
Correct Answer: A
Solution :
According to Arrhenius, \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[{{k}_{27}}=\frac{2.303}{30}\log \frac{100}{50}\] \[=\frac{2.303}{30}\log 2=2.3\times {{10}^{-2}}{{\min }^{-1}}\] \[{{k}_{{{47}^{o}}}}=\frac{2.303}{10}\log \frac{100}{50}\] \[=0.0693\,{{\min }^{-1}}\]
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