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 ?

Explanation

We know from first -order reaction,
`k = (2.303)/t log. a/(a-x)`
`1.5 xx10^(-6) =(2.303)/(10xx60xx60) log. a/(a-x) " "(t=10 xx60xx60s)`
` :. " " log a/(a-x) =(1.5 xx10^(-6) xx10xx 60 xx60)/(2.303)`
`= 0.02344`
or `log. (a-x)/a = 0.02344 = T.9765`
` :. (a-x)/a =0.9473`
Percentage of reactant remained = `(a-x)/a xx100 =94.73 %`
Further `t_(1/2) =(0.6932)/k = (0.6932)/(1.5xx10^(-6))=0.462 xx10^(-6)s`