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Q. In the following first order competing reactions;
$A +$ Reagent $\xrightarrow{k_{1}}$ Product
$B +$ Reagent $\xrightarrow{k_{2}}$ 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

Chemical Kinetics

Solution:

$t = \frac{2.303}{k_{2}}$ log $\frac{100}{50}$
and $t = \frac{2.303}{k_{1}}$ log $\frac{100}{6}$
$\frac{k_{1}}{k_{2}} = \frac{log \,100 - log\, 6}{log \,2} = 4$