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  4. Following is the graph between log T50 and log a(a=. initial concentration) for a given reaction at 27° C. Hence order is

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Q. Following is the graph between $\log T_{50}$ and $\log a\left(a=\right.$ initial concentration) for a given reaction at $27^{\circ} C$. Hence order isChemistry Question Image

Chemical Kinetics

Solution:

$t_{1 / 2} \propto\left(\frac{1}{a}\right)^{n-1}$ or $t_{1 / 2}=k(a)^{1-n}$
$\log t_{1 / 2}=\log k+(1-n) \log a(y=c+m x)$
Slope $=(1-n)=\tan 45=1$
$\therefore(1-n)=1 \Rightarrow n=0$