Question

Consider the following parallel reactions being given by $A \left( t _{1 / 2}=\right.$ $1.386 \times 10^{2}$ hours), each path being $1^{\text {st }}$ order.

If distribution of $A$ in the product mixture is $50\%$, calculate partial half-life of $A$ for the conversion into $B$.

• A. 346.5 h
• B. 231 h
• C. 154 h
• D. 92.4 h

Answer 1

Answer: (B)

$\frac{2 k _{1}}{2 k _{1}+3 k _{2}}=0.5$ and $k _{1}+ k _{2}=\frac{0.693}{1.386 \times 10^{2}}=5 \times 10^{-3} h ^{-1}$
Solving
$\frac{ k _{1}}{ k _{2}}=\frac{3}{2}$ and $k _{1}=3 \times 10^{-3} h ^{-1}$ and $k _{2}=2 \times 10^{-3} h ^{-1}$
$t _{1 / 2 (A \rightarrow B) }=\frac{0.693}{ k _{1}}=\frac{0.693}{3 \times 10^{-3}}=231 h$