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GATE 2023 CY – Question 43

Physical Chemistry · Kinetics and Reaction Dynamics: Elementary, parallel, opposing, and consecutive reactions. Steady state approximation. Mechanism of complex reactions. Unimolecular reactions. Potential energy surface and classical trajectories, Concept of saddle points, Transition state theory: Eyring equation, thermodynamic aspects. Kinetics of polymerization. Catalysis concepts and enzyme catalysis. Kinetic isotope effects. Fast reaction kinetics: relaxation and flow methods. Diffusion controlled reactions. Kinetics of unimolecular and bimolecular photophysical processes, Quantum yield calculation, static and dynamic quenching. · 2 marks · Multiple choice

For consecutive reactions $P\xrightarrow{k_1}Q\xrightarrow{k_2}R$, $[Q]=\dfrac{k_1[P]_0}{k_2-k_1}(e^{-k_1t}-e^{-k_2t})$, where $[P]_0$ is initial P concentration. With $k_2=25$ s$^{-1}$, which $k_1$ gives the longest waiting time for Q to reach its maximum?

  1. 20 s$^{-1}$
  2. 25 s$^{-1}$
  3. 30 s$^{-1}$
  4. 35 s$^{-1}$

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Correct answer: (A) 20 s$^{-1}$

Explanation

Differentiating the given concentration gives tmax=ln(k2/k1)/(k2−k1). For k1=20,25,30,35/s, the times are 0.04463,0.04000,0.03646,0.03365 s, respectively; the equal-rate limit is 1/k2. The longest time is therefore for **20/s, A**.