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InterviewSolution
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The time required for `10%` completion of a first order reaction at 298 K is equal to that required for its `25%` completiion at `308K.` If the value of A is `4xx10^(10)s^(-1).` calculate k at 318K and `E_(a),` |
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Answer» Calculation of activation energy `(E_(a))` For `1^(st)` order reaction : `k=(2.303)/(t)log ""([A]_(0))/([A])` `At 298K` `K_(1)=(2.303)/(t) log """"(100)/(90)-(i)` At 308 K `K_(2)=(2.303)/(t)log ""(100)/(75) -(ii)` Dividing eq (ii) by (i) `(k_(2))/(k_(1))=(log ""(100)/(75))/(log ""(100)/(90))=(0.1249)/(0.0458)=2.73` According to Arrhenius theory `log ""(k_(2))/(k_(1))=(E_(a))/(2.303 R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `log 2.73 =(E_(a))/(2.303R)((308-298)/(298xx308))` `E_(a)=(0.4361xx2.303xx(8.314J mol ^(-1))xx298xx308)/(10)` `E_(a) k =log A (-E_(a))/(2.303 xx(8.314J mol^(-1) K^(-1))xx(318K))` `log k =10.6021 -12.5870 =-1.9849` k = Antilog `(-19849)=` Antilog `(bar2.0151)=1.035xx10^(-2)s^(-1)` `E_(a)=76.640kJ mol ^(-1)` `k=1.035 xx10^(-2)s^(-1)` |
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