What is the relationship between the half-lif | vedclass

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(B) For a reaction of order $n$ (where $n 
eq 1$),the integrated rate law is given by: $k = \frac{1}{(n-1)t} \left( \frac{1}{[R]^{n-1}} - \frac{1}{[R

Vedclass · Accepted Answer

$t_{1/2} \propto [R]_0^{1-n}$

Vedclass · Answer

(B) For a reaction of order $n$ (where $n 
eq 1$),the integrated rate law is given by: $k = \frac{1}{(n-1)t} \left( \frac{1}{[R]^{n-1}} - \frac{1}{[R]_0^{n-1}} ight)$ At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$. Substituting these values: $k = \frac{1}{(n-1)t_{1/2}} \left( \frac{1}{([R]_0/2)^{n-1}} - \frac{1}{[R]_0^{n-1}} ight)$ $k = \frac{1}{(n-1)t_{1/2}} \left( \frac{2^{n-1} - 1}{[R]_0^{n-1}} ight)$ Rearranging for $t_{1/2}$: $t_{1/2} = \frac{2^{n-1} - 1}{k(n-1)} 	imes \frac{1}{[R]_0^{n-1}}$ Therefore,$t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$ or $t_{1/2} \propto [R]_0^{1-n}$.

What is the relationship between the half-life period $(t_{1/2})$ and the initial concentration $([R]_0)$ for a reaction of order $n$?

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