InterviewSolution
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InterviewSolution
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From the following data for the reaction between A and B Calculate the following (i) The order of the reaction with respect to A and with respect to B (ii) The rate constant at 300K (iii) The energy of activation and (iv) The pre-exponential factor. |
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Answer» (i) Let the rat law be : Rate= `k[A]^(x)[B]^(y)` `"From expt".(1),5.0xx10^(-4)=k[2.5xx10^(-4)]^(x)[3.0xx10^(-5)]^(y)" "....(i)` `"From expt".(2),4.0xx10^(-4)=k[5.0xx10^(-4)]^(x)[6.0xx10^(-5)]^(y)" "....(ii)` Dividing by eq. (i), `(4.0xx10^(-3))/(5.0xx10^(-4))=2^(x)*2^(y)=8` `"From expt".(3),1.6xx10^(-2)=k[1.0xx10^(-3)]^(x)[6.0xx10^(-5)]^(y)" "....(iii)` `"From expt".(1),5xx10^(-4)=k[2.5xx10^(-4)]^(2)[3.0xx10^(-5)]` or `k=(5xx10^(-4))/([2.5xx10^(-4)]^(2)[3.0xx10^(-5)])=2.67xx10^(8)L^(2)"mol"^(-2)s^(-1)` (iii) From Arrhenius eqauation, `log_(10).(2.0xx10^(-3))/(5.0xx10^(4))=(E_(a))/(2.3030xx8.314)xx(20)/(300xx320)` `E_(a)=(2.303xx8.314xx300xx320)/(20)xxlog_(10)4` (iv) Applying `log_(10) k=log_(10) A-(E_(a))/(2.3030RT)` `log_(10).(A)/(k)=(55333)/(2.3030xx8.314xx300)=9.633` or `(A)/(k)=4.29xx10^(9)` or `A=4.29xx10^(9)xx2.67xx10^(8)` `=1.145xx10^(18)` |
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