A) \[8.68\times {{10}^{8}}{{\operatorname{M}}^{-1}}{{\operatorname{s}}^{-1}}\]
B) \[2.16\times {{10}^{7}}{{\operatorname{M}}^{-1}}{{\operatorname{s}}^{-1}}\]
C) \[4.34\times {{10}^{8}}{{\operatorname{M}}^{-1}}{{\operatorname{s}}^{-1}}\]
D) \[3.34\times {{10}^{8}}{{\operatorname{M}}^{-1}}{{\operatorname{s}}^{-1}}\]
Correct Answer: C
Solution :
| E and A can be evaluated using Arrhenius equation, since energy of activation \[{{E}_{a}}\] and pre-exponential factor, A are Arrhenius parameters. given rate constant |
| \[{{k}_{1}},=2.80\times {{10}^{-3}}{{M}^{-1}}{{s}^{-1}}at30{}^\circ C(303K)\] |
| \[{{k}_{2}}=1.38\times {{10}^{-2}}{{M}^{-1}}{{s}^{-1}}at50{}^\circ C(323K)\] |
| \[{{E}_{a}}=\frac{2.303R{{T}_{1}}{{T}_{2}}}{\left( {{T}_{1}}-{{T}_{2}} \right)}{{\log }_{10}}\frac{{{k}_{2}}}{{{k}_{1}}}\] |
| \[=\frac{2.303\times 8.314\times {{10}^{-3}}\times 303\times 323}{\left( 323-303 \right)}{{\log }_{10}}\] |
| \[\left( \frac{1.38\times {{10}^{-2}}}{2.80\times {{10}^{-3}}} \right)\] |
| \[{{E}_{a}}=64.91kJ\,mo{{l}^{-1}}\] |
| also \[k=A.{{e}^{-{{E}_{a}}/RT}}\] |
| \[\therefore \] \[A=K.{{e}^{-{{E}_{a}}/RT}}\] |
| \[\operatorname{A}=2.80\times 1{{0}^{-3}}{{e}^{64}}{{^{.91}}^{/(8.314\times {{10}^{-3}}\times 303)}}\] |
| \[=4.34\times {{10}^{8}}{{M}^{-1}}{{s}^{-1}}\] |
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