# CORRECT OPTION?

$Forareactionofnthorder\phantom{\rule{0ex}{0ex}}{t}_{1/2}\alpha \frac{1}{({C}_{o}{)}^{n-1}}\phantom{\rule{0ex}{0ex}}thus\phantom{\rule{0ex}{0ex}}{t}_{1/2}\alpha \frac{1}{({C}_{o}{)}^{2}}\phantom{\rule{0ex}{0ex}}Henceorder=3\phantom{\rule{0ex}{0ex}}$

The correct option is 4

Regards

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