MCQOPTIONS
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MCQOPTIONS
| 1. |
Activation energy is given by the formula [DCE 1999] |
| A. | \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] |
| B. | \[\log \frac{{{K}_{1}}}{{{K}_{2}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] |
| C. | \[\log \frac{{{K}_{1}}}{{{K}_{2}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]\] |
| D. | None of these |
| Answer» B. \[\log \frac{{{K}_{1}}}{{{K}_{2}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] | |