1.

20%of a first order reaction was found to be completed at 10 a.m at 11.30 a.m on the samedat, 20% of the reaction was foundto be remaining . The half life period in minutes of the reaction is

Answer»

90
45
60
30

Solution :a = initial concentration
Set - I Given,
(a - X) = conc.At time t.
Amount LEFT` = (a - x) at t_(10_(A.M)) = a`
`(for t_(11.30 A.M))= 100 - 20 = 80`(initial)
= ( a for set - 2)
Set - 2 ` (a - x) at t_(11.30_(A.M))`
= 100 - 80 = 20 (Final)
( t = 11.30 (A.M) - 10 (A.M)= 1.50` H xx 60 = 90 min`)
`:. K = (2.303)/(t) "log"(a)/(a -x)`
`K = (2.303)/(90) "log"(80)/(20)`
`K =(2.303)/(90) log 4 `
` = (2 .303 xx 0 . 6020)/(90)`
K = 0 015 min
`:.t_(1//2) = (0.693)/(K)`
`t_(1//2) = (0.6930)/(0.015) = 46 .2 ~~ 45 min`


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