For a competitive parallel reaction network, and . The total overall effective half-life of reactant — Chemical Kinetics Chemistry Question
Question
For a competitive parallel reaction network, $A \xrightarrow{k_1} B$ and $A \xrightarrow{k_2} C$. The total overall effective half-life of reactant A is experimentally logged as exactly $12 \text{ hours}$. If it is analytically known that the rate of formation of product C strictly constitutes exactly $60\%$ of the total overall rate of decomposition of A, what is the specific half-life of A natively expressed in hours if one were solely isolating and considering its conversion branch into B?
💡 Solution & Explanation
The overall half-life is $t_{1/2} = 0.693 / (k_1+k_2) = 12 \text{ hours}$. Since the rate of formation of C is $60\%$ of total consumption, $k_2 = 0.60 (k_1+k_2)$. Because fractions must sum to $1$, $k_1 = 0.40 (k_1+k_2)$. The specific half-life for solely the B pathway is defined as $0.693 / k_1$. Substituting $k_1$, we get $0.693 / [0.40 (k_1+k_2)]$. This equates exactly to $\frac{0.693 / (k_1+k_2)}{0.40} = 12 / 0.40 = 30 \text{ hours}$.