A - B- K1-0-693 sec-1 C - D- K2-0-693 -min-1- If t1 and t2 are half lives of two reactions- then-

The correct option is C t_2=60 t_1 t_1/2=0.693K t_1t_2=K_2K_1 K_1=0.693 sec^ 1 and K_2=0.69360sec^ 1 As, 1 min=60 sec t_ ...

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Standard XII

Chemistry

First Order Reaction

A → B, K1=0.6...

Question

$A \to B, K_{1} = 0.693 s e c^{- 1}$
$C \to D, K_{2} = 0.693, m i n^{- 1}$ . If $t_{1}$ and $t_{2}$ are half lives of two reactions, then:

t_{1} = t_{2}

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t_{1} = 60 t_{2}

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t_{2} = 60 t_{1}

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t_{2} = 2.303 t_{1}

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Solution

The correct option is C $t_{2} = 60 t_{1}$
$t_{1 / 2} = \frac{0.693}{K}$
$\frac{t_{1}}{t_{2}} = \frac{K_{2}}{K_{1}}$
$K_{1} = 0.693 s e c^{- 1}$ and $K_{2} = \frac{0.693}{60} s e c^{- 1}$
As, $1 m i n = 60 s e c$
$\frac{t_{1}}{t_{2}} = \frac{0.693 / 60}{0.693}$
$t_{2} = 60 t_{1}$

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A→B,K1=0.693sec−1C→D,K2=0.693,min−1. If t1 and t2 are half lives of two reactions, then:

The correct option is C t2=60t1t1/2=0.693Kt1t2=K2K1K1=0.693 sec−1 and K2=0.69360sec−1As, 1 min=60 sect1t2=0.693/600.693t2=60t1

$A \to B, K_{1} = 0.693 s e c^{- 1}$
$C \to D, K_{2} = 0.693, m i n^{- 1}$ . If $t_{1}$ and $t_{2}$ are half lives of two reactions, then:

The correct option is C $t_{2} = 60 t_{1}$
$t_{1 / 2} = \frac{0.693}{K}$
$\frac{t_{1}}{t_{2}} = \frac{K_{2}}{K_{1}}$
$K_{1} = 0.693 s e c^{- 1}$ and $K_{2} = \frac{0.693}{60} s e c^{- 1}$
As, $1 m i n = 60 s e c$
$\frac{t_{1}}{t_{2}} = \frac{0.693 / 60}{0.693}$
$t_{2} = 60 t_{1}$