For a reaction A→PHalf life measured for two different values of initial concentrations 5×10−3 M and 25×10−4 M are 1 and 8 hrs respectively. If initial concentration is adjusted to 1.25×10−3 M, what is the half-life? (in hrs)

For a ‘n’ th order reactiont12∝1an−1 where a= initial conc.(5×10−325×10−4)n−1=81⇒n=4 t121=(5×10−31.25×10−3)n−1t12=64 hr

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For a reaction $A \to P$
Half life measured for two different values of initial concentrations $5 \times 10^{- 3} M a n d 25 \times 10^{- 4} M$ are 1 and 8 hrs respectively. If initial concentration is adjusted to $1.25 \times 10^{- 3} M,$ what is the half-life? (in hrs)

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answer is 64.

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Detailed Solution

For a ‘n’ th order reaction
$t_{\frac{1}{2}} \propto \frac{1}{a^{n - 1}} w h e r e a = initial conc .$
${(\frac{5 \times 10^{- 3}}{25 \times 10^{- 4}})}^{n - 1} = \frac{8}{1} \Rightarrow n = 4$
$\frac{t_{\frac{1}{2}}}{1} = {(\frac{5 \times 10^{- 3}}{1.25 \times 10^{- 3}})}^{n - 1}$
$t_{\frac{1}{2}} = 64 h r$

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