When reaction is completed 99.9%,`[R]_(n) = [R]_(0) - 0.999[R]_(0)`
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`k=2.303/t"log"[R]_(0)/[[R]]` `k=2.303/t"log"[R]_(0)/[[R]-0.999[R]_(0)]=2.303/t"log10^(3)` `t= 6.909//k` For half-life of the reaction `t_(1/2)=0.693//k` `t_(1/2)=0.693/kxxk/0.693 =10`
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