1.

75% of a first order reaction was completed in 32 min . When would 50% of the reaction completed

Answer»

24 min
16 min
8 min
64 min

Solution :The first order RATE constant is given as
`k = (2.303)/(t) "log" (a_(0))/(a_(0) - x) ` _______(i)
also , half life `t_(1//2) =(2.303 "log"2)/(K)` ______(ii)
EQUATING K from equation (i) and (ii)
`therefore (2.303"log"2)/(t_(1//2)) = (2.303)/(32) "log" (100)/(100-75)` or , `("log"2)/(t_(1//2)) = (1)/(32)` log 4 .
or `("log"2)/(t_(1//2)) = (1)/(32) xx 2` log 2
`therefore t_(1//2) = 16 ` MINUTES


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