Question

A follows parallel path Ist order reactions giving $B$ and $C$ as shown:

If initial concentration of $A$ is $0.25M$, calculate the concentration of $C$ after 5 hour of reaction. Given, ${\lambda }_1=1.5\times 10^{-5}{s}^{-1},{\lambda }_2=5\times 10^{-6}{s}^{-1}$

• A. $7.55\times 10^{-3}M$
• B. $1.89\times 10^{-2}M$
• C. $5.53\times 10^{-3}M$
• D. $3.51\times 10^{-3}M$

Answer 1

Answer: (B)

${\lambda }_A={\lambda }_1+{\lambda }_2=1.5\times 10^{-5}+5\times 10^{-6}$
$=20\times 10^{-6}{s}^{-1}$

Also, $2.303\log \frac{[A{]}_0}{[A{]}_t}=\lambda \times t$
$\therefore 2.303\log \frac{0.25}{[A{]}_t}=20\times 10^{-6}\times 5\times 60\times 60$
$\Rightarrow \log \frac{0.25}{[A{]}_t}=0.1563$
$\therefore [A{]}_t=0.1744M=0.25-[B]-[C]$
$\therefore [B]+[C]=0.25-0.1744=0.0756M$
Fraction of $C$ formed $=\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
$=\frac{[C]}{[B]+[C]}$
$\Rightarrow [C]$
$=\frac{5\times 10^{-6}}{20\times 10^{-6}}(0.0756)$
$=1.89\times 10^{-2}M$