Question

Concentration of Q in a consecutive reaction \(P \xrightarrow{k_1} Q \xrightarrow{k_2} R\) is given by \([Q]=\dfrac{k_1[P]_0}{k_2-k_1}\left(e^{-k_1 t}-e^{-k_2 t}\right)\). If \(k_2=25~\mathrm{s^{-1}}\), the value of \(k_1\) that leads to the \emph{longest waiting time for Q to reach its maximum is}

• A. \(k_1=20~\mathrm{s^{-1}}\)
• B. \(k_1=25~\mathrm{s^{-1}}\)
• C. \(k_1=30~\mathrm{s^{-1}}\)
• D. \(k_1=35~\mathrm{s^{-1}}\)

Hint:

For \(P \to Q \to R\), \(t_{\max}\) is largest when formation of \(Q\) is slow relative to its consumption (\(k_1\) as small as possible vs fixed \(k_2\)).
If \(k_1 \to k_2\), \(t_{\max} \to 1/k_2\) (l’Hôpital’s rule).

Answer 1

The Correct Option is A

Step 1: Time for \([Q]\) to be maximum. Differentiating and setting \(\dfrac{d[Q]}{dt}=0\) gives the time of maximum: \[ t_{\max}=\frac{\ln\!\left(\dfrac{k_2}{k_1}\right)}{k_2-k_1} \quad (k_1\neq k_2). \]

Step 2: Monotonicity with respect to \(k_1\) (for fixed \(k_2\)). For \(k_11\). Using \(\ln x \le x-1\) (equality only at \(x=1\)), \[ \frac{k_1-k_2}{k_1}+\ln\!\left(\frac{k_2}{k_1}\right)= -\left(x-1-\ln x\right)\le 0, \] so \(t_{\max}\) \emph{decreases} as \(k_1\) increases. Therefore, the longest waiting time occurs at the smallest \(k_1\). 

Step 3: Choose from options. Among the choices, the smallest \(k_1\) is \(20~\mathrm{s^{-1}}\). Hence the longest waiting time is obtained for \(k_1=20~\mathrm{s^{-1}}\).