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Chemical Kinetics

Question
CBSEENCH12006402

Show that in first order reaction time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction.

Solution

For first order reaction                

t=2.303klogaa-x

                              t0.999 = 2.303kloga(a-0.999 a)              [ x = 0.999 a]          = 2.303kloga0.001 a           = 2.303klog1000 = 2.303k×3                ...(i)


and                             

t1/2 = 2.303kloga(a-0.5 a)        = 2.303klog 2 = 2.303k×0.3010......(ii)

Dividing (i) by (ii), we get  

t0.999t0.5 = 2.303k×3×k2.303×0.3010=10


or                                         t0.999 = 10 × t0.5