1.

If 60% of a first order reaction was completed in 60 minutes 50% of the same reaction would be completed in approximately :

Answer»

45 minutes
60 minutes
40 minutes
50 minutes .

Solution :(A) `k = (2.303)/(t) "log"([A]_(0))/([A])`
`:. K = (2.303)/(60) "log" ([A]_(0))/(0.4[A]_(0))`
`= (2.303)/(60) "log" (10)/(4)`
`= (2.303)/(60) ` (log 10 -log 4)
Now `t_(1//2) = (0.693)/(k) = (0.693xx60)/(2.303xx0.4)`
`= 45.14 ~~45 ` min .


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