Question

A follows parallel path $I$ order reactions giving $B$ and $C$ as: If initial concentration of $A$ is 0.25 $M$, the concentration of $C$ after 5 hour of reaction is :

$[Given:{\lambda }_1=1.5\times {10}^{-5}{s}^{-1},{\lambda }_2=5\times {10}^{-6}{s}^{-1}]$

• A. $7.56\times {10}^{-3}M$
• B. $6.62\times {10}^{-3}M$
• C. $9.36\times {10}^{-3}M$
• D. None of these

Answer 1

The correct option is A $7.56\times {10}^{-3}M$

${\lambda }_A={\lambda }_1+{\lambda }_2=1.5\times {10}^{-5}+5\times {10}^{-6}$
$=20\times {10}^{-6}{s}^{-1}$
Also, $2.303\log \frac{[A{]}_0}{[A{]}_t}=\lambda \times t$
$\therefore 2.303\log \frac{0.25}{[A{]}_t}=20\times {10}^{-6}\times 5\times 60\times 60$
$\therefore [A{]}_t=0.1744M$
$\therefore \left[A\right]decomposed=[A{]}_0-[A{]}_t=0.25-0.1744$
$=0.0756M$
$FractionofCformed=\left[\frac{\lambda }{{\lambda }_1+{\lambda }_2}\right]\times [A{]}_{decomposed}\times \frac{2}{5}$
$=0.0756\times \frac{5\times {10}^{-6}}{20\times {10}^{-6}}\times \frac{2}{5}$
$=7.56\times {10}^{-3}M$
Note that, 5 mole of $A$ are used to give 2 mole of $C$.