In the following first order competing reactions; A + Reagent xrightarrowk1 Product B + Reagent xrightarrowk2 Product The ratio of k1/ k2 if only 50 % of B will have been reacted when 94 % of A has been reacted is

Question

In the following first order competing reactions; A + Reagent xrightarrowk1 Product B + Reagent xrightarrowk2 Product The ratio of k1/ k2 if only 50 % of B will have been reacted when 94 % of A has been reacted is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t = (2.303/k2) log (100/50) and t = (2.303/k1) log (100/6) (k1/k2) = (log 100 - log 6/log 2) = 4

Q. In the following first order competing reactions;
$A +$ Reagent $k_{1}$ Product
$B +$ Reagent $k_{2}$ Product
The ratio of $k_{1} / k_{2}$ if only $50%$ of $B$ will have been reacted when $94%$ of $A$ has been reacted is

$t = \frac{2.303}{k _{2}}$ log $\frac{100}{50}$
and $t = \frac{2.303}{k _{1}}$ log $\frac{100}{6}$
$\frac{k _{1}}{k _{2}} = \frac{l o g 100 - l o g 6}{l o g 2} = 4$

Q. In the following first order competing reactions; A+ Reagent k1​​ Product B+ Reagent k2​​ Product The ratio of k1​/k2​ if only 50% of B will have been reacted when 94% of A has been reacted is

t=k2​2.303​ log 50100​ and t=k1​2.303​ log 6100​ k2​k1​​=log2log100−log6​=4

Q. In the following first order competing reactions;
$A +$ Reagent $k_{1}$ Product
$B +$ Reagent $k_{2}$ Product
The ratio of $k_{1} / k_{2}$ if only $50%$ of $B$ will have been reacted when $94%$ of $A$ has been reacted is

$t = \frac{2.303}{k _{2}}$ log $\frac{100}{50}$
and $t = \frac{2.303}{k _{1}}$ log $\frac{100}{6}$
$\frac{k _{1}}{k _{2}} = \frac{l o g 100 - l o g 6}{l o g 2} = 4$