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 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

To solve the problem of determining the value of \(k_1\) that leads to the longest waiting time for Q to reach its maximum concentration, we start by analyzing the given concentration function of Q: \([Q]=\dfrac{k_1[P]_0}{k_2-k_1}\left(e^{-k_1 t}-e^{-k_2 t}\right)\).

The maximum concentration of Q occurs when the derivative of \([Q]\) with respect to \(t\) is zero. Therefore, we need to find \(\frac{d[Q]}{dt}\) and solve \(\frac{d[Q]}{dt}=0\). Calculating the derivative:

\(\frac{d[Q]}{dt}=\dfrac{k_1[P]_0}{k_2-k_1}(-k_1 e^{-k_1 t} + k_2 e^{-k_2 t})\).

Setting \(\frac{d[Q]}{dt}=0\) gives:

\(-k_1 e^{-k_1 t} + k_2 e^{-k_2 t}=0\)

Rearranging terms:

\(k_2 e^{-k_2 t}=k_1 e^{-k_1 t}\)

Taking the natural logarithm of both sides:

\(k_2 ( -k_2 t) = \ln(k_1) + k_1 ( -t)\)

\((k_1-k_2)t = \ln\left(\frac{k_2}{k_1}\right)\)

\(t = \dfrac{\ln\left(\frac{k_2}{k_1}\right)}{k_1-k_2}\)

To maximize \(t\), we must maximize the absolute value of \(\dfrac{\ln\left(\frac{k_2}{k_1}\right)}{k_1-k_2}\). This occurs when \(k_1\) is close to \(k_2\), but because \(k_2=25~\mathrm{s^{-1}}\), and is greater than all provided options, selecting a value of \(k_1\) less than 25 will lead to a longer waiting time.

Hence, the longest waiting time occurs when \(k_1=20~\mathrm{s^{-1}}\).

\(k_1=20~\mathrm{s^{-1}}\)