A) 1842.4 R
B) 921.27 R
C) 460.6 R
D) 230.3 R
Correct Answer: B
Solution :
Activation energy can be calculated with the help of Arrhenius equation. \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] Where, \[{{K}_{1}}=\]Rate constant at \[{{T}_{1}}\]temperature \[{{K}_{2}}=\]Rate constant at \[{{T}_{2}}\] temperature \[{{E}_{a}}=\]Activation Energy R = Gas constant \[{{T}_{1}},{{T}_{2}}=\]Temperature in Kelvin \[\log \frac{10}{1}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{400-200}{400\times 200} \right]\] \[1=\frac{{{E}_{a}}}{2.303R}\left[ \frac{200}{400\times 200} \right]\] \[{{E}_{a}}=2.303\times R\times 400\] \[{{E}_{a}}=921.2R\]
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