A) \[63 %\]
B) \[50 %\]
C) \[100 %\]
D) \[10 %\]
Correct Answer: A
Solution :
\[2.303\,\,\log \,\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{9.0\times {{10}^{3}}}{2.303\times 2}\left[ \frac{308-298}{308\times 298} \right]\] \[\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{9.0\times {{10}^{3}}}{2.303\times 2}\left[ \frac{308-298}{308\times 298} \right]\] \[\frac{{{k}_{2}}}{{{k}_{1}}}=1.63;\,\,{{k}_{2}}=1.63\,{{k}_{1}};\,\,\frac{1.63{{k}_{1}}-{{k}_{1}}}{{{k}_{1}}}\times 100=63\,%\]
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