1.

Show that in the first order reaction, time required for completion of 99.9% is 10 times that of half-life (t_(1//2)) of the reaction.

Answer»

Solution :The EQUATION for first order REACTION is
`t=(2.303)/(k)"log"([R]_(0))/([R])""…(i)`
For 99.9% completion, `[R]=0.001xx[R]_(0)`
For half-reaction,`[R]=(1)/(2)[R]_(0)`
Substituting the valuesin the above equation, we have
`t_(99.9%)=(2.303)/(k)"log"([R]_(0))/(0.001xx[R]_(0))=(2.303)/(k) log 10^(3) ""...(ii)`
`t_(50%)=(2.303)/(k)"log"([R]_(0))/(1//2xx[R]_(0))=(2.303)/(k)log2 ""...(iii)`
DIVIDING (ii) by (iii), we get
`(t_(99.9%))/(t_(50%))=(log 10^(3))/(log2)=(3)/(0.3010)=10`


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