Chemical KineticshardNUMERICAL

For a theoretical consecutive sequence defined by , the exact formula dictating the timestamp where Chemical Kinetics Chemistry Question

Question

For a theoretical consecutive sequence defined by $A \xrightarrow{k_1} B \xrightarrow{k_2} C$, the exact formula dictating the timestamp where intermediate B hits maximum saturation is $t_{max} = \frac{\ln(k_1/k_2)}{k_1 - k_2}$. Applying L'Hôpital's rule or limit theory, for a hypothetical reaction where $k_1$ approaches exactly $k_2$, the limit formula for $t_{max}$ brilliantly collapses precisely to $1/k$. If $k_1$ is identically equal to $k_2$ at exactly $0.05 \text{ min}^{-1}$, calculate the exact elapsed $t_{max}$ native value in minutes.

Answer: 20

💡 Solution & Explanation

When the formation rate constant exactly equals the decay rate constant ($k_1 = k_2 = k$), applying the limit as $k_1 \rightarrow k_2$ to the standard $t_{max}$ formula resolves to $t_{max} = 1/k$. Given that $k = 0.05 \text{ min}^{-1}$, substituting this value directly yields $t_{max} = 1 / 0.05 = 20 \text{ minutes}$.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full NEET Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry