A) 63%
B) 50%
C) 100%
D) 10%
Correct Answer: A
Solution :
From Arrhenius equation, \[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left( \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right)\] \[=\frac{9\times {{10}^{3}}}{2.303\times 2}\left( \frac{308-298}{308\times 298} \right)=0.2129\] \[\therefore \] \[\frac{{{k}_{2}}}{{{k}_{1}}}=\text{antilog}\,\,\text{0}\text{.2129}\,\text{=}\,\text{1}\text{.632}\] \[\therefore \] \[{{k}_{2}}=1.632\,{{k}_{1}}\] Hence, increase in rate constant \[={{k}_{2}}-{{k}_{1}}\] \[=1.632\,{{k}_{2}}-{{k}_{1}}=0.632\,{{k}_{1}}\] \[=\frac{0.632{{k}_{1}}}{{{k}_{1}}}\times 100=63.2%=63%\]
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