InterviewSolution
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InterviewSolution
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The energy of activation for a reaction is `100 KJ mol^(-1)`. The peresence of a catalyst lowers the energy of activation by `75%`. What will be the effect on the rate of reaction at `20^(@)C`, other things being equal? |
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Answer» The arrhenius equation is , `k=Ae^(-E_(a)//R)` In absence of catalyast, `k_(1)=Ae^(-100//RT)` In presence of catalyst, `k_(2)=Ax^(-25//RT)` So, `(k_(2))/(k_(1))=^(75//RT)or2.303 "log" (k_(2))/(k_(1))=(75)/(RT)` `or 2.303 "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293)` `or "log" (k_(2))/(k_(1))=(75)/(8.314xx10^(-3)xx293xx2.303)` or `(k_(2))/(k_(1))=2.314xx10^(13)` As the things beings equal in presence or absence of a catalyst, `(k_(2))/(k_(1))` must be =`("rate in presence of catalyst")/("rate in absence of catalyst")` `i.e., (r_(2))/(r_(1))=(k_(2))/(k_(1))=2.34xx10^(13)` |
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