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Explain:How the value of activation energy is dermine on the base of Aeehenius equation? |
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Answer» Solution :Method-1:Calculation of `E_(a)` :The Arrhenius equation :`T_(1)` and `T_(2)` The Arrhenius equation :k=A.`e^(-(E_(a))/(RT))` and In k=`-(E_(a))/(RT)` + In A Of the value of the rate constant at temperature `T_(1)` and `T_(2)` are `K_(1)` and `k_(2)` respectively then we have two following equations. In `k_(1)=(E_(a))/(RT_(1))+In A`.........(i) In `K_(2)=-(E_(a))/(RT_(2))+In A` .......(ii) Substracting equation (ii)from equation (i) In `k_(2)-In K_(1)=-(E_(a))/(RT_(2))-((E_(a))/(RT_(1)))`+In A-In A `therefore In (k_(2))/(k_(1))=+(E_(a))/(R)((T_(1)-T_(1))/(T_(1)xxT_(2)))` Taking log on both side log `(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` By using this formula we can calculate the activated energy `E_(a)` method-2: To draw plot of in `kto(1)/(T)` or log `kto(1)/(T)` Measuring the slope of the LINE,the value of `E_(a)`can be CALCULATED . In K`to(1)/(T),=-(E_(a))/(R)` log `kto(1)/(T),=-(E_(a))/(2.303R)` |
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