Question

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

If initial concentration of $A$ is $0.25\, M$, 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, $2.303 \log \frac{[ A ]_{0}}{[ A ]_{ t }}=\lambda \times t$
$\therefore 2.303 \log \frac{0.25}{[ A ]_{ t }}=20 \times 10^{-6} \times 5 \times 60 \times 60$
$\Rightarrow \log \frac{0.25}{[ A ]_{ t }}=0.1563$
$\therefore [ A ]_{ t }=0.1744 M =0.25-[ B ]-[ C ]$
$\therefore [ B ]+[ C ]=0.25-0.1744=0.0756 M$
Fraction of $C$ formed $=\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}$
$=\frac{[ C ]}{[B ]+[ C ]}$
$\Rightarrow [ C ]$
$=\frac{5 \times 10^{-6}}{20 \times 10^{-6}}(0.0756)$
$=1.89 \times 10^{-2} M$