Concentration of Q in a consecutive reaction \(\rm P \xrightarrow{k_1}Q\xrightarrow{k_2}R\) is given by \(\rm [Q]=\frac{k_1]p]_0}{k_2-k_1}[e^{-k_1t}-e^{-k_2t}]\), where [𝑃]₀ is the initial concentration of P. If the value of 𝑘₂ = 25 s⁻¹ , the value of 𝑘₁ that leads to the longest waiting time for Q to reach its maximum is 

Concept:-

Consecutive Reactions: A consecutive (or sequential) reaction includes a series of reactions where the product of one reaction is the reactant of the next.

Rate Constants and Reaction Kinetics: The rate constant (k) of a reaction is a measure of how fast a reaction occurs. For a given set of reaction conditions, the rate constant is independent of the concentrations of the reactant substances.

First-order reactions: In a first-order reaction, the reaction rate is directly proportional to the concentration of only one reactant. This implies that the rate of the reaction depends on the concentration of a single species.

Explanation:-

For reaction \(\rm P \xrightarrow{k_1}Q\xrightarrow{k_2}R\) 

Waiting time, \(t_{max} = \frac{1}{K_1-K_2} ln\frac{K_1}{K_2}\)

 As per the given values of K1 the possible values of t_max are:|
 

Thus, the maximum value of t_max is K₁ = 20s^-1

Conclusion:-

So, the value of  tmax is K1 = 20s-1