1.

If the rate constant for a first order reactions K ,the time (t) required for the completion of 99% of the reaction is given by :

Answer»

t=2.303 /k
t=0.693/k
t=6.909 /K
t=4.606/k

Solution :When 99% REACTION is completed,
`[R]_(t)=((100-99)/(100))[R]_(0)`
`[R]_(t)=((100-99)/(100))[R]_(0)`
`t=(2.303)/(K)` log `([R]_(o))/([R]_(t))`
`therefore t=(2xx303)/(K)` log` ([R]_(o))/((100-99)/(100))[R]_(o)`
`t=(2.303)/(K)` log `100=(2.303)/(K)`log`10^(2)`
`t=(2.303)/(K).2"log"10`
`t=(2.303xx2)/(K)[therefore log 10=1]`
`t=(4.606)/(K)`


Discussion

No Comment Found

Related InterviewSolutions