A) \[+1\]
B) \[-1\]
C) \[+1000\]
D) \[-1000\]
Correct Answer: D
Solution :
\[k=A{{e}^{-{{E}_{a}}/RT}}\] \[k'=A{{e}^{-E{{'}_{a}}/RT}}\] where k = rate constant for non-catalysed reaction, and k? = rate constant for catalysed reaction. \[{{E}_{a}}\] = activation energy for non-catalysed reaction and \[E{{'}_{a}}\] = activation energy for catalysed reaction) \[\frac{k'}{k}={{e}^{{{E}_{a}}-E{{'}_{a}}}}\] Also given \[\operatorname{k}'=k+1.718\,\,k=2.718\,\,k\] \[\therefore \,\,\,\,2.718={{e}^{\frac{{{E}_{a}}-E{{'}_{a}}}{RT}}}\] \[{{\log }_{e}}\,2.718=\frac{{{E}_{a}}-E{{'}_{a}}}{8.314\times {{10}^{-3}}\times 500}\] \[{{E}_{a}}-E_{a}^{'}=4.11\] \[E_{a}^{'}=\,\,4.15\] \[\therefore \,\,\,\,\,{{E}_{a}}=8.3\,\,kJ/mo{{l}^{-1}}\] \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[\therefore \,\,\,\,\,\,\,\,{{\log }_{e}}k={{\log }_{e}}A-\frac{Ea}{RT}\] This is an equation for straight line with slope \[=-\frac{Ea}{R}=-\frac{8.3}{8.3\times {{10}^{-3}}}\] \[=\,\,-1000\]You need to login to perform this action.
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