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Half- life $(t_1)$ of the first order reaction and half-life $(t_2)$ of the second order reaction are equal. Hence ratio of the rate of the rate at te start of the reaction.

$(A)\;1 \\ (B)\;2 \\(C)\;0.693 \\(D)\;1.44 $
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For I order $K_1=\large\frac{0.693}{t 1/2}$
For II order $K_2=\large \frac{1}{t 1/2 \alpha}$
$K_1= \large\frac{0.693}{T_1}$
$K_2 = \large\frac{1}{T_2 \alpha}$
If $T_1=T_2$ then $\large\frac{K_1}{K_2}$$=0.693\;\alpha$
Initially $r_1=K_1[a]^1, r_2= K_2[ \alpha]^2$
$\therefore \large\frac{r_1}{r_2}=\frac{K_1}{K_2 \alpha}=\frac{K_2 \times 0.693 \times \alpha}{K_2 \times \alpha}$
$\qquad=0.693$
Hence c is the correct answer.
answered Apr 3, 2014 by meena.p
 

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