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The rate constant for the first order de...

The rate constant for the first order decompoistion of a certain reaction is described by the equation
`log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)`
(a) What is the energy of activation for the reaction?
(b) At what temperature will its half-life periof be `256 min`?

Text Solution

Verified by Experts

(a) Given, ln `K=14.34-(1.25xx10^(4))/(T)`
Also we have ln `K= ln A-(E_(a))/(RT)`
Therefore, comparing these two `E_(a)/R=1.25xx10^(4)`
`:. E_(a)=1.25xx10^(4)xxR cal mol^(-1)`
`=1.25xx10^(4)xx1.987 cal mol^(-1)`
`E_(a)=24.83 kcal mol^(-1)`
(b) ln `K=14.34-(1.25xx10^(4))/(500)`,
(since `T=500K`)
`:. K=2.35xx10^(-5) sec^(-1)`
(c ) `K=0.693/(256xx60), ( :' K=0.693/t_(1//2))`
`:. ln" (0.693)/(256xx60)=14.34-(1.25xx10^(4))/(T)`
or `-10.00=14.34-(1.25xx10^(4))/(T)`
`:. T=513 K`
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