1.

Show that in a first order reaction,time required for completion of 99.9 % is 10 times of half-life (t_((t)/(2))) of the reaction.

Answer»

Solution :When reaction is completed 99.9 % `[R]_(n)[R]_(0)-0.999[R]_(0)`
`K=(2.303)/(t)` log `([R]_(0))/([R])`
`k=(2.303)/(t)`=log`([R]_(0))/([R]_(0)-0.999[R]_(0))`
`=(2.303)/(t)=log 10^(3)`
t=6.909 /K
For HALF -life of the reaction
`t((t)/(2))=(0.693)/(k)`
`(t)/(t(1)/(2))=(6.909)/(k)XX(6.909)/(k)=10`


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