A) 43.85 kJ/mol
B) 55.14 kJ/mol
C) 11.97 kJ/mol
D) 6.65 kJ/mol
Correct Answer: B
Solution :
[b] \[{{k}_{1(300)}}=\frac{0.693}{20};\] \[{{k}_{2(320)}}=\frac{0.693}{5}\] In \[\frac{{{k}_{2(320)}}}{{{k}_{1(300)}}}=\frac{{{E}_{a}}}{R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] \[{{E}_{a}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{({{T}_{2}}-{{T}_{1}})}\log \frac{{{k}_{2}}}{{{k}_{1}}}\] \[=\frac{2.303\times 8.314}{20\times 1000}\times 300\times 320\log 4\] \[=55.14kJ/mol\]
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