A→BK1=0.693S−1 C→DK2=0.693Min−1 If t1 and t2 are half lives of reactions, then

t1 For A→B⇒0.6930.693=1t2For C→D⇒0.6930.693×60=160 t1t2=k2k1 t1t2=160 t2=60t1

A→BK1=0.693S−1 C→DK2=0.693Min−1 If t1 and t2 are half lives of reactions, then

t1 For A→B⇒0.6930.693=1t2For C→D⇒0.6930.693×60=160 t1t2=k2k1 t1t2=160 t2=60t1

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$A \to B$ $K_{1} = 0.693 S^{- 1}$
$C \to D$ $K_{2} = 0.693 M i n^{- 1}$
If t₁ and t₂ are half lives of reactions, then

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$t_{1} = t_{2}$

$t_{2} = 60 t_{1}$

$t_{2} = 2.303 t_{1}$

$t_{1} = 6 t_{2}$

answer is B.

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Detailed Solution

$t_{1}$ For $A \to B$ $\Rightarrow \frac{0.693}{0.693} = 1$

$t_{2}$ For $C \to D \Rightarrow \frac{0.693}{0.693 \times 60} = \frac{1}{60}$

$\frac{t_{1}}{t_{2}} = \frac{k_{2}}{k_{1}}$ $\frac{t_{1}}{t_{2}} = \frac{1}{60}$ $t_{2} = 60 t_{1}$

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A→BK1=0.693S−1 C→DK2=0.693Min−1 If t1 and t2 are half lives of reactions, then

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$A \to B$ $K_{1} = 0.693 S^{- 1}$
$C \to D$ $K_{2} = 0.693 M i n^{- 1}$
If t₁ and t₂ are half lives of reactions, then