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The first order reaction has k=1.5xx10^(...

The first order reaction has `k=1.5xx10^(-6)` per second at `200^(@)C` . If the reaction is allowed to ruf for 10 hours at the same temperature, what percentage of the initial concentration would have changed into the product? What is the half life period of this reaction ?

Text Solution

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(i) Give, `= 1.5 xx 10^(-6) s^(-1)`
`:.t_(1//2)= 0.693/k = 0.693/(1.5 xx 10^(-6)) = 462000s = 128.33h`
(ii) Let initial conc (a) = 100, t = 10h = 36000 s, x = (?)
`k = 2.303/t log""(100)/((100-x))`
` :. 1.5 xx 10^(-6)= 2.303/(3.6 xx 10^4) log""(100)/((100-x))`
`:. x = 5.2`
Thus, initial concentration changed into product is 5.2 %
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