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At 407 K, the rate contant of a chemical...

At 407 K, the rate contant of a chemical reaction is `9.5xx10^(-5) s^(-1)` and at 420 K the rate constant is `1.9xx10^(-4)s^(-1)`. Calculate the frequency equation is

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`"log"_(10) =(E_(2))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given `k_(1)=9.5xx10^(-5)s^(-1),k_(2)=1.9xx10^(-4)s^(-1),`
R=8.314J `"mol"^(-1)K^(-1),`
`T_(1)-407 K and T_(2)=420K`
Substituting the values in Arrhenius equation
`"log"_(10)(1.9xx10^(-4))/(9.5xx10^(-5))=(E_(a))/(2.303xx8.314)[(420-407)/(420xx407)]`
Applying now log `k_(1)` = log `A-(E_(a))/(2.3030RT_(1))`
log `9.5xx10^(-5)="log"A-(75782.3)/(2.303xx8.314xx407)`
or log `(A)/(9.5xx10 ^(-5))=(757.82.3)/(2.3030xx8.314xx407)=9.7246`
`A=5.04xx10^(5)s^(-1)`
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