• # question_answer The rate constant for the decomposition of a certain substance is $2.80\times {{10}^{-3}}{{M}^{-1}}{{s}^{-1}}$at $30{}^\circ C$and $1.38\times {{10}^{-2}}{{M}^{-1}}{{s}^{-1}}$ at $50{}^\circ C.$ The Arrhenius parameters [A] of the reaction is: $(R= 8.314 \times {{10}^{-3}}kJmo{{l}^{-1}}{{K}^{-1}}).$ 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) $3.34\times {{10}^{8}}{{\operatorname{M}}^{-1}}{{\operatorname{s}}^{-1}}$

 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 $k=A.{{e}^{-{{E}_{a}}/RT}}$ $\therefore$      $A=K.{{e}^{-{{E}_{a}}/RT}}$ $\operatorname{A}=2.80\times 1{{0}^{-3}}{{e}^{64}}{{^{.91}}^{/(8.314\times {{10}^{-3}}\times 303)}}$ $=4.34\times {{10}^{8}}{{M}^{-1}}{{s}^{-1}}$